C++ revision help! - Page 2

Okay, so I've got that cleared. Now, I need to tackle all these 'floats', 'doubles', 'integers', etc. What are the names of these? Are they the 'return data types'?

On June 29 2009 00:16 Uligor wrote:


You're wrong, the compiler automatically makes the main function return 0 when it's specified as void. (xor eax, eax before ret)

On June 28 2009 23:49 konadora wrote:
Oh I see, so it's just that int main() allows you to identify whether your program has an error or not, but void main() can't?

On June 29 2009 00:13 konadora wrote:

I need help with explanation on the 'arrays' part, I can't find the notes online on my school's e-learning platform. Step-by-step explanation would help so much, I would love you for life

// nric[ i ] - '0' is a kind of kludge that will turn a character representation of a digit into a numerical represention of the same digit. That is '9' - '0' = 9 and '6' - '0' = 6, et cetera. Obviously '0' - '0' = 0.
// You then multiply each digit of the nric by some pre-determined "weight" and add them up.

for(int i = 0; i<=4; i++){
cout >> example[i] >> endl;
}

On June 29 2009 00:32 konadora wrote:
@ms291052:


Why is the -0 necessary? (Great first post btw, thanks lol)

@treason:
so if you put number[5], you'll have 5 different values, but if you want to access the nth number, you must use number[n-1]?

On June 29 2009 00:38 Uligor wrote:


The character '0' has a different value than int 0 (char '0' is in fact 48 int, '1' is 49 etc). But when you subtract the character with '0', you essentially do -48. So when the character is '1' (49), it becomes 49-48 = 1.

What data type do I use if I want to use a fraction? I'm currently doing this exercise where I'm supposed to convert inputted Farenheit temperature into centigrade and vice versa, and I need to use the fraction 5/9.

On June 29 2009 00:48 konadora wrote:
My next question:

On June 29 2009 00:16 Uligor wrote:


You're wrong, the compiler automatically makes the main function return 0 when it's specified as void. (xor eax, eax before ret)

voidmain.cpp:1: error: '::main' must return 'int'

On June 29 2009 00:53 konadora wrote:
Isn't double for numbers with decimal points? What's the difference between float and double?

(Sorry if these questions are very basic, but no revision + 1 month of missing school + bad teacher = sucky at language)

On June 29 2009 00:58 tec27 wrote:

Both double and float are for numbers with decimal points. A double has more precision than a float (IE: can store longer decimals), but is also therefore larger in memory.

On June 29 2009 00:53 konadora wrote:
Isn't double for numbers with decimal points? What's the difference between float and double?

(Sorry if these questions are very basic, but no revision + 1 month of missing school + bad teacher = sucky at language)

On June 29 2009 00:56 tec27 wrote:

I'm sorry, but you're wrong except for very specific compilers. Microsoft's compiler does this, but the standards do not support that. It was merely done to make bad code that they/other people had written in the past still work.

If you hop over to g++, however, you'll find that you can't even compile a program with 'void main':


Now, what is within the standards is to auto-return 0 for main functions declared as type int (IE: if no return statement is specified inside). While this is still bad practice, it should work on all standards-compliant compilers.

In short, don't encourage someone to not follow standards on a simple matter like this just because a certain compiler will fix the mistake for them. Its incredibly bad practice, and this is the kind of thing that, even though its small, will make lots of people have zero respect for you and your code.

Maybe I misunderstood your question, you wanted to know what datatype you should use in an algorithm to convert Farenheit temperature into centigrade and viceversa?

#include <iostream>
using namespace std;

int main()
{
double c,f;
cout << "Please enter the temperature in Centigrade." << endl;
cin >> c;
f = ((c*1.8)+32);
cout << "The temperature is " << f << " degrees Farenheit." << endl;

cout << "Please enter the temperature in Farenheit." << endl;
cin >> f;
c = ((f-32)*5/9);
cout << "The temperature is " << c << " degrees Centigrade." << endl;

return 0;
}

if(myChar == 'c' ) {
// do Celsius -> Farenheit calculation
}
else if(myChar == 'f' ) {
// do Farenheit -> Celsius calculation
}
else {
// Tell the user to follow instructions, better luck next time
}