Question

How does c store a double decimal in an 8 bit slot?

#include "stdio.h"

main(){
  double x = 123.456;
  printf("\n %d - %e \n",sizeof(x),x);
}

outputs:

8 - 23.456

The value of x is correct being 123.456, but the supposedly it is only 8 bits.

Answer 1

That’s not 8 bits. It’s 8 bytes. And each byte is at least 8 bits (and usually exactly 8 bits).

So it’s probably 8 * 8 = 64-bits for a double.

EDIT:

The sizeof() operator yields the size of an object in bytes.

A “byte” is by definition the size of a char. (That’s how the C standard defines the word “byte”; it may have different meanings in other contexts.)

The number of bits in a byte is specified by the macro CHAR_BIT, defined in <limits.h>. Almost any system you’re likely to encounter will have CHAR_BIT == 8, but I understand that some implementations for DSPs (Digital Signal Processors) have CHAR_BIT set to 16 or 32.