Correction on float and int

valembe · 10-30-2008, 02:19 PM

Code:

#include "stdafx.h"
#include <iostream>
#include <fstream>
using namespace std;

int main()
{
	float d = 0.00;
	int x = 3, y = 9;
	
	d = x/y;
	
	cout<<d<<endl;
	return 0;
}

If i want a float result is the right way to do it??

CRC123 · 10-30-2008, 02:43 PM

Quote:

Originally Posted by valembe

Code:

#include "stdafx.h"
#include <iostream>
#include <fstream>
using namespace std;

int main()
{
	float d = 0.00;
	int x = 3, y = 9;
	
	d = x/y;
	
	cout<<d<<endl;
	return 0;
}

If i want a float result is the right way to do it??

No, x or y would need to be a float or cast to a float in order for d to have the actual floating point answer.

Search on google for 'implicit type conversion'

abolishtheun · 10-30-2008, 03:03 PM

The compiler first looks at "x/y", and performs integer division because neither x nor y are floats. Then after it gets that value, it sees the lvalue is a float, so it converts the result to a float.

dmail · 10-30-2008, 08:09 PM

Adding to what others have said there is a floating point cast happening here.

Code:

float d = 0.00;

This is a double cast to a float, although initialisation is not required here it would be

Code:

float d = 0.f;

Just for completeness

Code:

#include <iostream>
int main()
{
	int x = 3, y = 9;
	float d( static_cast<float>(x)/y );
	std::cout<<d<<std::endl;
}

abolishtheun · 10-30-2008, 09:04 PM

why not just (float)x/y?

dmail · 10-31-2008, 05:59 AM

"why not just (float)x/y?"
Because that is a C cast and the code is C++. It was late when I posted that and it really should be static_cast I will fix the post.

valembe · 11-01-2008, 12:02 PM

Quote:

Originally Posted by dmail

"why not just (float)x/y?"
Because that is a C cast and the code is C++. It was late when I posted that and it really should be static_cast I will fix the post.

Thanx guys!! (float)x/(float)y worked..

CRC123 · 11-01-2008, 12:38 PM

Quote:

Originally Posted by valembe

Thanx guys!! (float)x/(float)y worked..

You really only need to do it on one of them. As long as one of them is a float, it will do floating point division.

Sergei Steshenko · 11-01-2008, 12:47 PM

Quote:

Originally Posted by CRC123

You really only need to do it on one of them. As long as one of them is a float, it will do floating point division.

A better practice is to explicitly convert all the types involved.

dmail · 11-01-2008, 03:22 PM

Quote:

Originally Posted by Sergei Steshenko

A better practice is to explicitly convert all the types involved.

Can you please explain why you think this is "better practice" please?

Sergei Steshenko · 11-01-2008, 03:29 PM

Quote:

Originally Posted by dmail

Can you please explain why you think this is "better practice" please?

Because you do not have to remember what a compiler does for you and what it doesn't.

It's especially important if you deal with more than one language, and type conversion rules are different.

nishamathew1980 · 11-02-2008, 05:11 AM

Instead of doing a typecasting operation [ (float) x/ (float) y ] - why don't you define the variables itself as float [ float x = 3.00, y = 9.00; ]

As per me - that would be a better programming practice per se!

Linux Archive

Sergei Steshenko · 11-02-2008, 05:19 AM

Quote:

Originally Posted by nishamathew1980

Instead of doing a typecasting operation [ (float) x/ (float) y ] - why don't you define the variables itself as float [ float x = 3.00, y = 9.00; ]

As per me - that would be a better programming practice per se!

Strictly saying, you should write

float x = 3.00f

not

float x = 3.00

- the latter means "3.00" is considered to be double.

Better yet, this is more or less what I do:

Code:

#define MY_FLOAT_TYPE float
// #define MY_FLOAT_TYPE double
...
MY_FLOAT_TYPE x = (MY_FLOAT_TYPE)3;
MY_FLOAT_TYPE y = (MY_FLOAT_TYPE)4.1;

, etc.

The points are:

define a type macro (or a typdef);
always explicitly declare with the above;
always explicitly cast to the above.