Integer conversion to double in C yielding zero values - c

Question

I originally had a function that uses pow(), and I noticed it was returning zero values every time. While debugging it, I found that this seems to be the source of the problem:
#include <math.h>
#include <stdio.h>
#include <string.h>
int foo(int n)
{
printf("%d is number passed in \n", n);
double base = (double) n;
printf("%d is base \n", base);
printf("%d is power\n", pow(base ,2));
return (1/2 *( pow( (double) n, 2)));
}
If I call foo on any integer value (say, 8), the second printf statement prints zero every time. (And of course, the third one does as well). Shouldn't conversion of double to int be straightforward?

Answer 1

You got several things wrong.
First of all, you're using the %d specifier in printf, but you're passing to it a double value; for doubles, you should use %f1, otherwise the printf will try to interpret the bytes of the double as an int, leading to unexpected results (in your case 0).
Notice that this kind of error is usually noticed by recent compilers with warning level high enough.
Moreover, in the return statement you wrote
1/2 *( pow( (double) n, 2))
This expression will always yield zero, because of the 1/2 factor. Both 1 and 2 are ints, thus an integer division is performed, and obviously it results in 0, that kills the whole expression.
Solution: change the 1/2 to 1/2. (where 2. denotes 2 as a double value, which promotes the operation to a floating-point division), or simply replace it with 0.5, or (even simpler) perform a division by two at the end of the expression.
Trivia: I always had the doubt that %f was for float, and something else was used for doubles; instead, it turns out that there's no specifier for floats, since they are automatically converted to doubles when passed to a function as "variable arguments" (so this applies to printf).

Answer 2

printf("%g is base \n", base);
printf("%g is power\n", pow(base ,2));
Your second problem is that you're doing this:
return (1/2 *( pow( (double) n, 2)));
But have defined your function to return an int. You need to define it to return double and not use integer math there :)

Answer 3

Your version
with correct conversion specifications
with no casts
with 1.0 in the final return

Answer 4

to output a double in c via printf, you should use %g or %lf (lf stands for "long float" which is a double, in fact float stands for single precision :D )
%g truncate to the least significant number, %lf do not.
You can even use something like
%.20g, so you are asking to write the number with 20 decimal number after dot (or comma).
If you want to get a float result from your function, you shoud change the function declaration to
double foo(int n)
I suggest to change the return line to this
return (0.5 * pow( (double) n, 2)));
0.5 is the right way to write 1/2 in floating point.
But if you only want an integer, then you should do this:
return ((int)pow( (double) n, 2)/2);
In fact pow return a double, and you should cast it to integer
Hope that's usefull :)