does printf dont take integer constant as arguement?

Question

What's the problem with this code ?

   printf("%d", pow(2, 10));

as format specifier used is integral , it should give an integer value . But it don't . Why so ?

output - 0

Expected output - 1024

Answer 1

pow return double . You are using wrong specifier to print a double value. This will invoke undefined behavior and you may get either expected or unexpected result. Use %f instead.

 printf("%f", pow(2, 10));

Answer 2

The pow function from <math.h> returns type double .

The format specifier for double is %f . You used the wrong specifier, so your program causes undefined behaviour . Literally anything could happen.

To fix this, use %f instead.

BTW if you want to compute 2 to the power 10 in integers you can write 1 << 10 .

Answer 3

转换为int

printf("%d", (int)pow(2, 10));