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

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开发者_如何学运维 123.456, but the supposedly it is only 8 bits.

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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.

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To be horribly pedantic sizeof return the size of the operand in multiples of sizeof char.

On all common platforms that means that sizeof returns in bytes.