2.67 Int Size is 32

★★

Problem:

You are given the task of writing a procedure int_size_is_32() that yields 1 when run on a machine for which an int is 32 bits, and yields 0 otherwise. You are not allowed to use the sizeof operator. Here is a first attempt:

/* The following code does not run properly on some machines */
int bad_int_size_is_32() {
    /* Set most significant bit (msb) of 32-bit machine */
    int set_msb = 1 << 31;
    /* Shift past msb of 32-bit word */
    int beyond_msb = 1 << 32;
    /* set_msb is nonzero when word size >= 32
       beyond_msb is zero when word size <= 32 */
    return set_msb && !beyond_msb;
}

When compiled and run on a 32-bit SUN SPARC, however, this procedure returns 0. The following compiler message gives us an indication of the problem:

warning: left shift count >= width of type

A. In what way does our code fail to comply with the C standard?

If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand, the behavior it undefined.

B. Modify the code to run properly on any machine for which data type int is at least 32 bits.

C. Modify the code to run properly on any machine for which data type int is at least 16 bits.

Code:

#include <stdio.h>
#include <assert.h>

int int_size_is_32() {
    int set_msb = 1 << 31;
    int beyond_msb = set_msb << 1;
    return set_msb && !beyond_msb;
}

int int_size_is_32_16bits() {
    int set_msb = 1 << 15 << 15 << 1;
    int beyond_msb = set_msb << 1;
    return set_msb && !beyond_msb;
}

int main() {
    assert(int_size_is_32());
    assert(int_size_is_32_16bits());
}