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that's sexy americanumlaut!
i was thinking bitwise shift operator and modulo, but it's probably not a portable solution, yours seems solid though.
it works for any number in the digit, decimal or not as well.
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On February 14 2013 02:56 supereddie wrote:Show nested quote +On February 13 2013 13:10 AmericanUmlaut wrote:On February 13 2013 07:43 darkness wrote:On February 13 2013 05:32 supereddie wrote:On February 13 2013 04:18 darkness wrote:Hey guys, thanks for helping me last time. I have another thing that I struggle with. Say you have a double variable with this value: double value = a / b; let's say 'value = 2.283214570' after division 'a' and 'b' are both integers. Now, my problem is that I need to print the n-th digit of this double variable. E.g. n = 2 Then output would be '8' because it's the 2nd digit after decimal point. I've thought about double -> string, but my implementation doesn't work for integers that are up to 60000 which is a requirement. Just use math? Multiply 'value' with 10^decimalnumber (so you have 228.3214570). Take the int part of it (=228). Do it again with decimalnumber - 1 and take int part: 22. Multiply by 10 = 220. Then just substract the two: 228 - 220 = 8  Something like
double value = a / b; // 2.283214570
double x = Math.Truncate(value * Math.Pow(10, decimalnumber)); // 228
double y = Math.Truncate(value * Math.Pow(10, decimalnumber - 1)) * 10; // 220
double digit = x - y; // 228 - 220 = 8;
Be careful about overflows etc. Awesome. This code only needs slight modification to make it work though. Math.Truncate = trunc in C Math.Pow = pow in C But yeah, this is irrelevant. Thanks. Edit: Actually, it seems I've forgotten to say it's for C, but yeah... it's obvious how to fix this. Edit 2: You may also want to change 'digit' to: int digit = (int)x - (int)y; because it's more natural for this scenario. That's a lot more complicated than it needs to be:
int(value * pow(10, decimalnumber)) % 10
2.283214570 * 100 => 228.3214570 cast to int => 228 228 mod 10 => 8 Edit: Forgot to include "value" in the above :-/ Doh, ofcourse! How could I forget about the mod operator 
Unfortunately, I'm not sure if I understand his code.
Is all of this code needed to be executed?
int(value * pow(10, decimalnumber)) % 10 <-----
2.283214570 * 100 => 228.3214570
cast to int => 228
228 mod 10 => 8
If yes, what is int here? Maybe (int)?
Also, I am wondering now why my double 'value' sas only 6 decimal digits. This is what I do:
int a, b;
double value;
scanf("%d %d", &a, &b);
value = (double)a / (double) b; <---- only 6 decimal digits (e.g. 8/13)
Even if I change 'int' to 'float, it is still the same.
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Just a general advice
Multiple inheritance in programming. Avoid it as far as it is possible. While it may sound cool, it may also be very very confusing for someone who should work with your code and hasn't followed the development of it.
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Canada2068 Posts
On February 14 2013 04:29 darkness wrote:Show nested quote +On February 14 2013 02:56 supereddie wrote:On February 13 2013 13:10 AmericanUmlaut wrote:On February 13 2013 07:43 darkness wrote:On February 13 2013 05:32 supereddie wrote:On February 13 2013 04:18 darkness wrote:Hey guys, thanks for helping me last time. I have another thing that I struggle with. Say you have a double variable with this value: double value = a / b; let's say 'value = 2.283214570' after division 'a' and 'b' are both integers. Now, my problem is that I need to print the n-th digit of this double variable. E.g. n = 2 Then output would be '8' because it's the 2nd digit after decimal point. I've thought about double -> string, but my implementation doesn't work for integers that are up to 60000 which is a requirement. Just use math? Multiply 'value' with 10^decimalnumber (so you have 228.3214570). Take the int part of it (=228). Do it again with decimalnumber - 1 and take int part: 22. Multiply by 10 = 220. Then just substract the two: 228 - 220 = 8 Something like
double value = a / b; // 2.283214570
double x = Math.Truncate(value * Math.Pow(10, decimalnumber)); // 228
double y = Math.Truncate(value * Math.Pow(10, decimalnumber - 1)) * 10; // 220
double digit = x - y; // 228 - 220 = 8;
Be careful about overflows etc. Awesome. This code only needs slight modification to make it work though. Math.Truncate = trunc in C Math.Pow = pow in C But yeah, this is irrelevant. Thanks. Edit: Actually, it seems I've forgotten to say it's for C, but yeah... it's obvious how to fix this. Edit 2: You may also want to change 'digit' to: int digit = (int)x - (int)y; because it's more natural for this scenario. That's a lot more complicated than it needs to be:
int(value * pow(10, decimalnumber)) % 10
2.283214570 * 100 => 228.3214570 cast to int => 228 228 mod 10 => 8 Edit: Forgot to include "value" in the above :-/ Doh, ofcourse! How could I forget about the mod operator Also, I am wondering now why my double 'value' sas only 6 decimal digits. This is what I do: int a, b; double value; scanf("%d %d", &a, &b); value = (double)a / (double) b; <---- only 6 decimal digits (e.g. 8/13) Even if I change 'int' to 'float, it is still the same.
the double has a lot higher precision than gets displayed. printf will display 6 digits after the decimal point by default. if you want to display more decimal digits, you have to specify i.e. printf("%.9f", value) will give you 9 digits after the decimal point.
int(value * pow(10, n)) % 10
you just need the one line that AmericanUmlaut provided. it does 3 things in this order:
1. multiply value by 10^n where n refers to the n'th digit you want (look up what pow function does if you don't know)
2. cast to int - he uses a different notation than the one you use. (int) x and int(x) mean the same thing
3. do % 10 which in math means "mod 10"
1. 2.283214570 * 100 => 228.3214570
2. cast to int => 228
3. 228 mod 10 => 8
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On February 13 2013 22:56 waxypants wrote:Show nested quote +On February 13 2013 02:40 Snuggles wrote:No it wants me to write out the Class, not use it. At some point I remember getting some feedback showing the input, how it runs through my methods. Here's the problem before that one that I managed to get right. Problem + Show Spoiler +Write the definition of a class Counter containing:
An instance variable named counter of type int .
A constructor that takes one int argument and assigns its value to counter
A method named increment that adds one to counter . It does not take parameters or return a value.
A method named decrement that subtracts one from counter . It also does not take parameters or return a value.
A method named get_value that returns the value of the instance variable counter .
Answer + Show Spoiler +class Counter:
def __init__(self):
self.counter = 0
def increment(self):
self.counter += 1
def decrement(self):
self.counter -= 1
def get_value (self):
return self.counter They took that one in real easily. I really don't know where the hell I messed up in the Player problem... Actually the __init__ should be something like:
def __init__(self, counter):
self.counter = counter
If I did that would I have to give the counter a default value? I feel like self.counter = counter wouldn't work unless the user inputted a value for counter.
def__init__(self,counter = 0):
self.counter = counter
Like say what would happen if someone did xxx = Counter()?
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On February 14 2013 04:58 Snuggles wrote:Show nested quote +On February 13 2013 22:56 waxypants wrote:On February 13 2013 02:40 Snuggles wrote:No it wants me to write out the Class, not use it. At some point I remember getting some feedback showing the input, how it runs through my methods. Here's the problem before that one that I managed to get right. Problem + Show Spoiler +Write the definition of a class Counter containing:
An instance variable named counter of type int .
A constructor that takes one int argument and assigns its value to counter
A method named increment that adds one to counter . It does not take parameters or return a value.
A method named decrement that subtracts one from counter . It also does not take parameters or return a value.
A method named get_value that returns the value of the instance variable counter .
Answer + Show Spoiler +class Counter:
def __init__(self):
self.counter = 0
def increment(self):
self.counter += 1
def decrement(self):
self.counter -= 1
def get_value (self):
return self.counter They took that one in real easily. I really don't know where the hell I messed up in the Player problem... Actually the __init__ should be something like:
def __init__(self, counter):
self.counter = counter
If I did that would I have to give the counter a default value? I feel like self.counter = counter wouldn't work unless the user inputted a value for counter.
def__init__(self,counter = 0):
self.counter = counter
Like say what would happen if someone did xxx = Counter()?
You are forced to give a value along when you create the object tho since the constructor takes the value as a argument?
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On February 14 2013 04:29 darkness wrote:Show nested quote +On February 14 2013 02:56 supereddie wrote:On February 13 2013 13:10 AmericanUmlaut wrote:On February 13 2013 07:43 darkness wrote:On February 13 2013 05:32 supereddie wrote:On February 13 2013 04:18 darkness wrote:Hey guys, thanks for helping me last time. I have another thing that I struggle with. Say you have a double variable with this value: double value = a / b; let's say 'value = 2.283214570' after division 'a' and 'b' are both integers. Now, my problem is that I need to print the n-th digit of this double variable. E.g. n = 2 Then output would be '8' because it's the 2nd digit after decimal point. I've thought about double -> string, but my implementation doesn't work for integers that are up to 60000 which is a requirement. Just use math? Multiply 'value' with 10^decimalnumber (so you have 228.3214570). Take the int part of it (=228). Do it again with decimalnumber - 1 and take int part: 22. Multiply by 10 = 220. Then just substract the two: 228 - 220 = 8 Something like
double value = a / b; // 2.283214570
double x = Math.Truncate(value * Math.Pow(10, decimalnumber)); // 228
double y = Math.Truncate(value * Math.Pow(10, decimalnumber - 1)) * 10; // 220
double digit = x - y; // 228 - 220 = 8;
Be careful about overflows etc. Awesome. This code only needs slight modification to make it work though. Math.Truncate = trunc in C Math.Pow = pow in C But yeah, this is irrelevant. Thanks. Edit: Actually, it seems I've forgotten to say it's for C, but yeah... it's obvious how to fix this. Edit 2: You may also want to change 'digit' to: int digit = (int)x - (int)y; because it's more natural for this scenario. That's a lot more complicated than it needs to be:
int(value * pow(10, decimalnumber)) % 10
2.283214570 * 100 => 228.3214570 cast to int => 228 228 mod 10 => 8 Edit: Forgot to include "value" in the above :-/ Doh, ofcourse! How could I forget about the mod operator Unfortunately, I'm not sure if I understand his code. Is all of this code needed to be executed? int(value * pow(10, decimalnumber)) % 10 <----- 2.283214570 * 100 => 228.3214570 cast to int => 228 228 mod 10 => 8 If yes, what is int here? Maybe (int)?
It mostly depends on the language, but it generally means that you will be casting the result of 'value * pow(10, decimalnumber)' to an integer. Casting to a 'lower' datatype means you're losing data, but also that you might get overflows when the result of (value * pow(10, decimalnumber)) > Int32.MaxValue. Therefore, it may be more prudent to convert the result of the entire operation (including the mod 10) to an integer. There are still changes of an overflow for the double for very large values, but I think those are negligible. Also, casting to an integer means losing precision; it will not round the number in any way; 1.9 becomes 1 (and not 2).
I suggest converting to an integer (or byte, as the number can only be between 0 - 9) as late as possible to avoid overflows.
For very robust programming you can remove unneeded digits as they are not relevant for your particular problem. With that I mean the for the number 12345.678, you can actually remove everything before the decimal point so the number you work with becomes 0.678 (something like value = value - Math.Floor(value)). That should solve the overflow problems.
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On February 14 2013 04:36 CountChocula wrote:Show nested quote +On February 14 2013 04:29 darkness wrote:On February 14 2013 02:56 supereddie wrote:On February 13 2013 13:10 AmericanUmlaut wrote:On February 13 2013 07:43 darkness wrote:On February 13 2013 05:32 supereddie wrote:On February 13 2013 04:18 darkness wrote:Hey guys, thanks for helping me last time. I have another thing that I struggle with. Say you have a double variable with this value: double value = a / b; let's say 'value = 2.283214570' after division 'a' and 'b' are both integers. Now, my problem is that I need to print the n-th digit of this double variable. E.g. n = 2 Then output would be '8' because it's the 2nd digit after decimal point. I've thought about double -> string, but my implementation doesn't work for integers that are up to 60000 which is a requirement. Just use math? Multiply 'value' with 10^decimalnumber (so you have 228.3214570). Take the int part of it (=228). Do it again with decimalnumber - 1 and take int part: 22. Multiply by 10 = 220. Then just substract the two: 228 - 220 = 8 Something like
double value = a / b; // 2.283214570
double x = Math.Truncate(value * Math.Pow(10, decimalnumber)); // 228
double y = Math.Truncate(value * Math.Pow(10, decimalnumber - 1)) * 10; // 220
double digit = x - y; // 228 - 220 = 8;
Be careful about overflows etc. Awesome. This code only needs slight modification to make it work though. Math.Truncate = trunc in C Math.Pow = pow in C But yeah, this is irrelevant. Thanks. Edit: Actually, it seems I've forgotten to say it's for C, but yeah... it's obvious how to fix this. Edit 2: You may also want to change 'digit' to: int digit = (int)x - (int)y; because it's more natural for this scenario. That's a lot more complicated than it needs to be:
int(value * pow(10, decimalnumber)) % 10
2.283214570 * 100 => 228.3214570 cast to int => 228 228 mod 10 => 8 Edit: Forgot to include "value" in the above :-/ Doh, ofcourse! How could I forget about the mod operator Also, I am wondering now why my double 'value' sas only 6 decimal digits. This is what I do: int a, b; double value; scanf("%d %d", &a, &b); value = (double)a / (double) b; <---- only 6 decimal digits (e.g. 8/13) Even if I change 'int' to 'float, it is still the same. the double has a lot higher precision than gets displayed. printf will display 6 digits after the decimal point by default. if you want to display more decimal digits, you have to specify i.e. printf("%.9f", value) will give you 9 digits after the decimal point. int(value * pow(10, n)) % 10 you just need the one line that AmericanUmlaut provided. it does 3 things in this order: 1. multiply value by 10^n where n refers to the n'th digit you want (look up what pow function does if you don't know) 2. cast to int - he uses a different notation than the one you use. (int) x and int(x) mean the same thing 3. do % 10 which in math means "mod 10" 1. 2.283214570 * 100 => 228.3214570 2. cast to int => 228 3. 228 mod 10 => 8
Thanks. However, there may be a bug. It says -8 for decimalnumber >= 10. I guess there aren't enough decimal places?
(int)(value * pow(10, decimalnumber)) % 10
Edit: C language
Edit 2: My lecturer replied with:
The precision requirement is very high, so built-in variables are not
good enough.
Could you give me a hint what to look for? GMP?
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On February 13 2013 23:19 alwinuz wrote:If you want to use a proxy server, you can assign the default proxy to your web request or something simililar. Maybe it is also possible to set the proxy for your code that makes the FTP connection. We had a similar problem where we could not connect to a old asmx webservice because we didn't use the system proxy. Now we have this code in the client to connect to the webservice (ws variable): // Make sure that webservice calls can pass through intranet proxy
ws.Credentials = CredentialCache.DefaultCredentials;
ws.Proxy = WebRequest.DefaultWebProxy;
if (ws.Proxy != null) ws.Proxy.Credentials = CredentialCache.DefaultCredentials;
I hope this helps, not sure if I understood your question correctly. Else you can maybe try to start up ping.exe with Process.Start() and parse the output?
This got me in the right direction. I found that I can setup a proxy on an ftpwebrequest. I'll have to play around with it, but I'm sure I can get it to work.
Parsing the output in ping won't work because on our network you can't ping any internet hostname. Can't ping google.com, cnn.com, or the like. Asinine, isn't it!
Thanks
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Canada2068 Posts
On February 14 2013 06:34 darkness wrote:Show nested quote +On February 14 2013 04:36 CountChocula wrote:On February 14 2013 04:29 darkness wrote:On February 14 2013 02:56 supereddie wrote:On February 13 2013 13:10 AmericanUmlaut wrote:On February 13 2013 07:43 darkness wrote:On February 13 2013 05:32 supereddie wrote:On February 13 2013 04:18 darkness wrote:Hey guys, thanks for helping me last time. I have another thing that I struggle with. Say you have a double variable with this value: double value = a / b; let's say 'value = 2.283214570' after division 'a' and 'b' are both integers. Now, my problem is that I need to print the n-th digit of this double variable. E.g. n = 2 Then output would be '8' because it's the 2nd digit after decimal point. I've thought about double -> string, but my implementation doesn't work for integers that are up to 60000 which is a requirement. Just use math? Multiply 'value' with 10^decimalnumber (so you have 228.3214570). Take the int part of it (=228). Do it again with decimalnumber - 1 and take int part: 22. Multiply by 10 = 220. Then just substract the two: 228 - 220 = 8 Something like
double value = a / b; // 2.283214570
double x = Math.Truncate(value * Math.Pow(10, decimalnumber)); // 228
double y = Math.Truncate(value * Math.Pow(10, decimalnumber - 1)) * 10; // 220
double digit = x - y; // 228 - 220 = 8;
Be careful about overflows etc. Awesome. This code only needs slight modification to make it work though. Math.Truncate = trunc in C Math.Pow = pow in C But yeah, this is irrelevant. Thanks. Edit: Actually, it seems I've forgotten to say it's for C, but yeah... it's obvious how to fix this. Edit 2: You may also want to change 'digit' to: int digit = (int)x - (int)y; because it's more natural for this scenario. That's a lot more complicated than it needs to be:
int(value * pow(10, decimalnumber)) % 10
2.283214570 * 100 => 228.3214570 cast to int => 228 228 mod 10 => 8 Edit: Forgot to include "value" in the above :-/ Doh, ofcourse! How could I forget about the mod operator Also, I am wondering now why my double 'value' sas only 6 decimal digits. This is what I do: int a, b; double value; scanf("%d %d", &a, &b); value = (double)a / (double) b; <---- only 6 decimal digits (e.g. 8/13) Even if I change 'int' to 'float, it is still the same. the double has a lot higher precision than gets displayed. printf will display 6 digits after the decimal point by default. if you want to display more decimal digits, you have to specify i.e. printf("%.9f", value) will give you 9 digits after the decimal point. int(value * pow(10, n)) % 10 you just need the one line that AmericanUmlaut provided. it does 3 things in this order: 1. multiply value by 10^n where n refers to the n'th digit you want (look up what pow function does if you don't know) 2. cast to int - he uses a different notation than the one you use. (int) x and int(x) mean the same thing 3. do % 10 which in math means "mod 10" 1. 2.283214570 * 100 => 228.3214570 2. cast to int => 228 3. 228 mod 10 => 8 Thanks. However, there may be a bug. It says -8 for decimalnumber >= 10. I guess there aren't enough decimal places? (int)(value * pow(10, decimalnumber)) % 10 Edit: C language
it's a sign that int is overflowing, which makes sense because int's range is from -2147483648 to 2147483647. if you try to assign 2147483647 + 1 to an int, it will loop back to -2147483648.
since you're dealing with overflows you should use supereddie's implementation, because it protects you from overflow problems.
edit: btw what's the biggest decimalnumber you need to use?
the precision of a double is about 15 digits, so if you need decimalnumber = 15 or >15 you might need to do something else (GMP or your own implementation of a division algorithm). i'm not sure but this might be what your teacher is hinting at. in any case, you should try supereddie's solution first to see if that solves your problems.
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I have a question(not homework, just curious)
Is there a way that, if say someone put in David Jacob Young
that I could use strings in C to find the initials and use a printf to print them out to the user?
David Jacob Young
DJY
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On February 14 2013 07:03 CountChocula wrote:Show nested quote +On February 14 2013 06:34 darkness wrote:On February 14 2013 04:36 CountChocula wrote:On February 14 2013 04:29 darkness wrote:On February 14 2013 02:56 supereddie wrote:On February 13 2013 13:10 AmericanUmlaut wrote:On February 13 2013 07:43 darkness wrote:On February 13 2013 05:32 supereddie wrote:On February 13 2013 04:18 darkness wrote:Hey guys, thanks for helping me last time. I have another thing that I struggle with. Say you have a double variable with this value: double value = a / b; let's say 'value = 2.283214570' after division 'a' and 'b' are both integers. Now, my problem is that I need to print the n-th digit of this double variable. E.g. n = 2 Then output would be '8' because it's the 2nd digit after decimal point. I've thought about double -> string, but my implementation doesn't work for integers that are up to 60000 which is a requirement. Just use math? Multiply 'value' with 10^decimalnumber (so you have 228.3214570). Take the int part of it (=228). Do it again with decimalnumber - 1 and take int part: 22. Multiply by 10 = 220. Then just substract the two: 228 - 220 = 8 Something like
double value = a / b; // 2.283214570
double x = Math.Truncate(value * Math.Pow(10, decimalnumber)); // 228
double y = Math.Truncate(value * Math.Pow(10, decimalnumber - 1)) * 10; // 220
double digit = x - y; // 228 - 220 = 8;
Be careful about overflows etc. Awesome. This code only needs slight modification to make it work though. Math.Truncate = trunc in C Math.Pow = pow in C But yeah, this is irrelevant. Thanks. Edit: Actually, it seems I've forgotten to say it's for C, but yeah... it's obvious how to fix this. Edit 2: You may also want to change 'digit' to: int digit = (int)x - (int)y; because it's more natural for this scenario. That's a lot more complicated than it needs to be:
int(value * pow(10, decimalnumber)) % 10
2.283214570 * 100 => 228.3214570 cast to int => 228 228 mod 10 => 8 Edit: Forgot to include "value" in the above :-/ Doh, ofcourse! How could I forget about the mod operator Also, I am wondering now why my double 'value' sas only 6 decimal digits. This is what I do: int a, b; double value; scanf("%d %d", &a, &b); value = (double)a / (double) b; <---- only 6 decimal digits (e.g. 8/13) Even if I change 'int' to 'float, it is still the same. the double has a lot higher precision than gets displayed. printf will display 6 digits after the decimal point by default. if you want to display more decimal digits, you have to specify i.e. printf("%.9f", value) will give you 9 digits after the decimal point. int(value * pow(10, n)) % 10 you just need the one line that AmericanUmlaut provided. it does 3 things in this order: 1. multiply value by 10^n where n refers to the n'th digit you want (look up what pow function does if you don't know) 2. cast to int - he uses a different notation than the one you use. (int) x and int(x) mean the same thing 3. do % 10 which in math means "mod 10" 1. 2.283214570 * 100 => 228.3214570 2. cast to int => 228 3. 228 mod 10 => 8 Thanks. However, there may be a bug. It says -8 for decimalnumber >= 10. I guess there aren't enough decimal places? (int)(value * pow(10, decimalnumber)) % 10 Edit: C language it's a sign that int is overflowing, which makes sense because int's range is from -2147483648 to 2147483647. if you try to assign 2147483647 + 1 to an int, it will loop back to -2147483648. since you're dealing with overflows you should use supereddie's implementation, because it protects you from overflow problems. edit: btw what's the biggest decimalnumber you need to use? the precision of a double is about 15 digits, so if you need decimalnumber = 15 or >15 you might need to do something else (GMP or your own implementation of a division algorithm). i'm not sure but this might be what your teacher is hinting at. in any case, you should try supereddie's solution first to see if that solves your problems.
I wasn't told how big, but the sample double is: 0.615384615384615384615384... so yeah, it looks huge. No idea how to implement this (int)(value * pow(10, decimalnumber)) % 10 via GMP for higher precision.
On February 14 2013 08:43 3FFA wrote:
I have a question(not homework, just curious)
Is there a way that, if say someone put in David Jacob Young
that I could use strings in C to find the initials and use a printf to print them out to the user?
David Jacob Young
DJY
You may want to read this
http://www.dreamincode.net/forums/topic/242738-full-names-to-initials-keeping-surname/
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On February 14 2013 06:34 darkness wrote:Show nested quote +On February 14 2013 04:36 CountChocula wrote:On February 14 2013 04:29 darkness wrote:On February 14 2013 02:56 supereddie wrote:On February 13 2013 13:10 AmericanUmlaut wrote:On February 13 2013 07:43 darkness wrote:On February 13 2013 05:32 supereddie wrote:On February 13 2013 04:18 darkness wrote:Hey guys, thanks for helping me last time. I have another thing that I struggle with. Say you have a double variable with this value: double value = a / b; let's say 'value = 2.283214570' after division 'a' and 'b' are both integers. Now, my problem is that I need to print the n-th digit of this double variable. E.g. n = 2 Then output would be '8' because it's the 2nd digit after decimal point. I've thought about double -> string, but my implementation doesn't work for integers that are up to 60000 which is a requirement. Just use math? Multiply 'value' with 10^decimalnumber (so you have 228.3214570). Take the int part of it (=228). Do it again with decimalnumber - 1 and take int part: 22. Multiply by 10 = 220. Then just substract the two: 228 - 220 = 8 Something like
double value = a / b; // 2.283214570
double x = Math.Truncate(value * Math.Pow(10, decimalnumber)); // 228
double y = Math.Truncate(value * Math.Pow(10, decimalnumber - 1)) * 10; // 220
double digit = x - y; // 228 - 220 = 8;
Be careful about overflows etc. Awesome. This code only needs slight modification to make it work though. Math.Truncate = trunc in C Math.Pow = pow in C But yeah, this is irrelevant. Thanks. Edit: Actually, it seems I've forgotten to say it's for C, but yeah... it's obvious how to fix this. Edit 2: You may also want to change 'digit' to: int digit = (int)x - (int)y; because it's more natural for this scenario. That's a lot more complicated than it needs to be:
int(value * pow(10, decimalnumber)) % 10
2.283214570 * 100 => 228.3214570 cast to int => 228 228 mod 10 => 8 Edit: Forgot to include "value" in the above :-/ Doh, ofcourse! How could I forget about the mod operator Also, I am wondering now why my double 'value' sas only 6 decimal digits. This is what I do: int a, b; double value; scanf("%d %d", &a, &b); value = (double)a / (double) b; <---- only 6 decimal digits (e.g. 8/13) Even if I change 'int' to 'float, it is still the same. the double has a lot higher precision than gets displayed. printf will display 6 digits after the decimal point by default. if you want to display more decimal digits, you have to specify i.e. printf("%.9f", value) will give you 9 digits after the decimal point. int(value * pow(10, n)) % 10 you just need the one line that AmericanUmlaut provided. it does 3 things in this order: 1. multiply value by 10^n where n refers to the n'th digit you want (look up what pow function does if you don't know) 2. cast to int - he uses a different notation than the one you use. (int) x and int(x) mean the same thing 3. do % 10 which in math means "mod 10" 1. 2.283214570 * 100 => 228.3214570 2. cast to int => 228 3. 228 mod 10 => 8 Thanks. However, there may be a bug. It says -8 for decimalnumber >= 10. I guess there aren't enough decimal places? (int)(value * pow(10, decimalnumber)) % 10 Edit: C language Edit 2: My lecturer replied with: Show nested quote +The precision requirement is very high, so built-in variables are not
good enough. Could you give me a hint what to look for? GMP?
I'd just write my own division function that works like long division and stops after finding the decimalnumber-th digit.
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@darkness
Having only programmed in C, and this being my first time learning, I didn't understand most of that. Was that perhaps C++? (Has never seen strings declared that way before, with '<<')
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On February 13 2013 13:10 AmericanUmlaut wrote:Show nested quote +On February 13 2013 07:43 darkness wrote:On February 13 2013 05:32 supereddie wrote:On February 13 2013 04:18 darkness wrote:Hey guys, thanks for helping me last time. I have another thing that I struggle with. Say you have a double variable with this value: double value = a / b; let's say 'value = 2.283214570' after division 'a' and 'b' are both integers. Now, my problem is that I need to print the n-th digit of this double variable. E.g. n = 2 Then output would be '8' because it's the 2nd digit after decimal point. I've thought about double -> string, but my implementation doesn't work for integers that are up to 60000 which is a requirement. Just use math? Multiply 'value' with 10^decimalnumber (so you have 228.3214570). Take the int part of it (=228). Do it again with decimalnumber - 1 and take int part: 22. Multiply by 10 = 220. Then just substract the two: 228 - 220 = 8 Something like
double value = a / b; // 2.283214570
double x = Math.Truncate(value * Math.Pow(10, decimalnumber)); // 228
double y = Math.Truncate(value * Math.Pow(10, decimalnumber - 1)) * 10; // 220
double digit = x - y; // 228 - 220 = 8;
Be careful about overflows etc. Awesome. This code only needs slight modification to make it work though. Math.Truncate = trunc in C Math.Pow = pow in C But yeah, this is irrelevant. Thanks. Edit: Actually, it seems I've forgotten to say it's for C, but yeah... it's obvious how to fix this. Edit 2: You may also want to change 'digit' to: int digit = (int)x - (int)y; because it's more natural for this scenario. That's a lot more complicated than it needs to be:
int(value * pow(10, decimalnumber)) % 10
2.283214570 * 100 => 228.3214570 cast to int => 228 228 mod 10 => 8 Edit: Forgot to include "value" in the above :-/
I like this solution, assuming that we don't have to solve cases where iee754 fails. For example on the edge case of the original question of 1/60000 = 1.66666666666666674653622670643E-5, you get a failure for most values of n > 20, because you lose the precision. So if the requirement includes handling a or b > 60000 and n > 20, americanumlaut's solution fails. (or if the requirement is a, b ~3 and n > 60,000, it also fails - weren't exactly clear on which is which).
If the requirement is only b ~ 60000 and n < 20, you're probably ok, but you might want to check the power multiplication edge cases, because you'll lose some precision there too.
I like del's solution better, and it's probably what the prof is going for in this case. If the integer requirements are only up to 60k, you won't have any precision problems if you roll your own long division, whereas a 64 bit double isn't sufficient.
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For you software engineers im curious, what is the common IQ of programmers? Are there a lot of idiots in the profession?
May seem like a blunt question but I'm going into IT and I have around 120 IQ; gotta wonder where that puts me.
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IQ is pretty meaningless for software engineering. Solving logic puzzles only gets you so far.
Also IT is as very general term that sometimes includes software engineering and sometimes does not.
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Yeah fair enough
For all you programmers~
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On February 14 2013 13:07 Oaky wrote:
For you software engineers im curious, what is the common IQ of programmers? Are there a lot of idiots in the profession?
May seem like a blunt question but I'm going into IT and I have around 120 IQ; gotta wonder where that puts me.
There are a lot of idiots in every walk of life. There are idiotic aspects to almost every personality. The question of paramount importance is: does it blend?
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On February 14 2013 13:26 JeanLuc wrote:Show nested quote +On February 14 2013 13:07 Oaky wrote:
For you software engineers im curious, what is the common IQ of programmers? Are there a lot of idiots in the profession?
May seem like a blunt question but I'm going into IT and I have around 120 IQ; gotta wonder where that puts me. There are a lot of idiots in every walk of life. There are idiotic aspects to almost every personality. The question of paramount importance is: does it blend?
You just killed me.
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EPT
