C中的“ pow”功能打印的值不正确

Question

Why does the below code gives 127 as output, when it has to be 128. i have even tried to figure out, but I don't understand why 127?

#include<stdio.h>
#include<math.h>
int main()
{
signed char ch;

int size,bits;

size = sizeof(ch);

bits = size * 8;
printf("totals bits is : %d\n",bits);

printf("Range is : %u\n", (char)(pow((double)2, (double)(7))));

}

Answer 1

If you want 128 as result then typecast pow() result as int instead of char . for eg

printf("Range is : %u\n", (int)(pow((double)2, (double)(7)))); /* this print 128 */

Why this

printf("Range is : %u\n", (char)(pow((double)2, (double)(7))));

prints 127 as pow((double)2,(double)7) is 128 but at same time that whole result vale explicitly type cast ed as char and default char is signed which ranges from -128 to +127 , hence it prints 127 .

Side note , pow() is floating point function as @lundin suggested & same you can find here . you can use

unsigned char ch = 1 << 7;

to get the same in particular case.