How does Variable > -1 exactly evaluate in C?

Question

#include <stdio.h>
#include <stdint.h>
int main()
{
    uint16_t peter = 8;
    uint32_t peter2 = 8;
    if(peter > -1)
    {
        printf("Peter true\n"); //expected
    } 
    if (peter2 > -1)
    {
        printf("Peter 2 true\n"); //wtf, why not
    }

    return 0;
}

Why does the first statement enter the clause and the second does not (on a 32bit architecture)?

Answer 1

The explanation for the difference between peter > -1 and peter2 > -1 are the notions of “integer promotion”, and also “usual arithmetic conversions”.

“Integer promotion” means that any integer type t narrower than int is “promoted” to either int (if int can contain all the values of t) or unsigned int otherwise. In your example and with your compiler, promotion is applied to peter, which is converted to int before being compared to -1 (which also has type int).

Then, the “usual arithmetic conversions” decide what happen when an arithmetic binary operation, here >, is applied to operands of different types. Between an unsigned and a signed integer type of the same width, these rules say that the signed operand is converted to unsigned before the operation takes place.

-1 has type int. When it is compared to peter2, the usual arithmetic conversions mean -1 is converted to uint32_t, becoming a large value (greater than 8).

Answer 2

It's all to do with implicit conversions.

For peter > -1 the value of peter is promoted to an int. But for peter2 > -1 it's the value of -1 which is converted to an uint32_t.

And in two's complement systems -1 is all ones, so is converted to the value 2147483647, which is quite a bit larger than the value 8 of peter2, making the condition false.