为什么下面的代码不打印测试？

Question

Why the code below doesn't show an output. It shoudl do arithmetic converion by why not:

#include <stdio.h>

int main(void)
{
    int b=2147483647;
    if((b+1)==2147483648u)
    {
        printf("TEST\n");
    }
    return 0;
}

Here is

printf("%d\n",b+1); // shows -2147483648

printf("%u\n",(unsigned int)b+1); // shows 2147483648

Answer 1

The expression b+1 is an overflow, because 2147483647 is the maximum value of a signed int (assuming 32-bit ints). Overflowing a signed expression is undefined behavior, so you can't reason about anything that happens after that.

If you cast b to unsigned int before incrementing it, it should work correctly.

Reference: C11 6.5/5:

If an exceptional condition occurs during the evaluation of an expression (that is, if the result is not mathematically defined or not in the range of representable values for its type), the behavior is undefined.

Answer 2

2147483647 is the largest number that can be stored in an int data-type variable, as a signed integer variable has 32 bits lf memory allocated to it. The last bit, the 32nd bit, is allocated for the sign of the number. Therefore, that means you only have 31 bits of memory assigned for the number itself, making the largest number 2^31, which is 2147483647.

Doing 2147483648, which is greater than 2147483647, is referred to as integer overflow, which will result to undefined behaviour.

When you do 2147483648u, you store the number as an unsigned number, hence deallocating ghe 32nd bit originally meant for the sign, to store a number instead. This means that you can now store the largest number: 2^32, which equals 4294967296. That is the reason why you can now store 2147483648 without invoking integer overflow, and causing undefined behaviour.

What you should do is:

• cast your variable to unsigned int
• add a "u" to the end of the number, like 2147483648u