Not being able to use pow()

[ihsh]

It's me again.

I was trying to write the code of a binary converter. But I always got strange outputs so I tested my codes and found out that the problem was with the function pow ().
At first I thought it was a variable problem again, but it wasn't! I wrote a new program to test the function and the output was still zero.

code:

#include <math.h> #include <stdio.h> int main () {         printf ("%d", pow(2,3));         return (0); }

Thanks for your time.

[michaelp]

The format specifier "%d" is for an integer, but the function "pow" returns a double.
So, changing %d to %f will then print "8.000000" (or something similar). Changing that to "%.f" will then just print out 8.
Also, using %d, you can use:

code:

printf("%d", (int)pow(2, 3) );

And you will also get 8.
Here is a list of format specifiers in C:
http://www.mekong.net/tech/printf.htm

Also, when I compiled this using GCC, I got the

code:

warning: format ?%d? expects type ?int?, but argument 2 has type ?double

Not sure if you got that when you compiled your code.

[ihsh]

Thanks, program works now.

Quote:

Also, when I compiled this using GCC, I got the
code:
warning: format ?%d? expects type ?int?, but argument 2 has type ?double

Not sure if you got that when you compiled your code.

No I didn't get any warning when compiling the code.

[DtY]

michaelp @ Mon Sep 06, 2010 8:50 pm wrote:

Also, when I compiled this using GCC, I got the

code:

warning: format ?%d? expects type ?int?, but argument 2 has type ?double

Not sure if you got that when you compiled your code.

I think that warning is only in pretty new versions of gcc.

e; Also, you should avoid using pow() unless the exponent rather large or a variable. a*a*a will be faster than pow(a,3).