INF: sizeof(char Expression) Same as sizeof(int)

ID Number: Q68389

6.00 6.00a 6.00ax | 6.00

MS-DOS | OS/2

Summary:

In Microsoft C versions 6.0, 6.0a, and 6.0ax, expressions involving

variables of type 'char' are promoted to type 'int'.

More Information:

This is ANSI-specified behavior. Below is section 3.3.7 from the ANSI

specifications, which details the semantics of the shift operator:

Semantics

The integral promotions are performed on each of the operands.

The type of the result is that of the promoted left operand. If

the value of the right operand is negative or is greater than or

equal to the width in bits of the promoted left operand, the

behavior is undefined.

This means that chars are promoted to integers by default. If you

really want a char result, you must cast the final result.

The ANSI-specified semantics of all operators specify promotion from

char to int, so the size of any char expression will be the sizeof

int. This was also the case for Kernighan and Ritchie (K & R) C.

The sizes of the int and long expressions stay the same because no

promotion takes place.

Note that if int is the same size as long rather than short in this

implementation, the sizeof both a short expression and a char

expression will be 4, as will be the sizeof both an int and a long

expression.

Sample Code

-----------

#include <stdio.h>

void main(void)

{

short si;

long li;

char sc;

unsigned char uc;

printf("Signed char width: %d\n",

sizeof((char)(sc<<1)); // 1 byte

printf("Signed char width: %d\n",

sizeof(sc<<1)); // 2 bytes

printf("Unsigned char width: %d\n",

sizeof((unsigned char)uc<<1)); // 1 byte

printf("Unsigned char width: %d\n",

sizeof(uc<<1)); // 2 bytes

printf("Short width: %d\n",

sizeof(si<<1)); // 2 bytes

printf("Long width: %d\n",

sizeof(li<<1)); // 4 bytes

}

Additional reference words: 6.00 6.00a 6.00ax