Implicit type promotion rules

Question

This post is meant to be used as a FAQ regarding implicit integer promotion in C, particularly implicit promotion caused by the usual arithmetic conversions and/or the i

Answer 1

According to the previous post, I want to give more information about each example.

Example 1)

int main(){
    unsigned char x = 0;
    unsigned char y = 1;
    printf("%u\n", x - y); 
    printf("%d\n", x - y);
}

Since unsigned char is smaller than int, we apply the integer promotion on them, then we have (int)x-(int)y = (int)(-1) and unsigned int (-1) = 4294967295.

The output from the above code:(same as what we expected)

4294967295
-1

How to fix it?

I tried what the previous post recommanded, but it doesn't really work. Here is the code based on the previous post:

change one of them to unsigned int

int main(){
    unsigned int x = 0;
    unsigned char y = 1;
    printf("%u\n", x - y); 
    printf("%d\n", x - y);
}

Since x is already an unsigned integer, we only apply the integer promotion to y. Then we get (unsigned int)x-(int)y. Since they still don't have the same type, we apply the usual arithmetic converions, we get (unsigned int)x-(unsigned int)y = 4294967295.

The output from the above code:(same as what we expected):

4294967295
-1

Similarly, the following code gets the same result:

int main(){
    unsigned char x = 0;
    unsigned int y = 1;
    printf("%u\n", x - y); 
    printf("%d\n", x - y);
}

change both of them to unsigned int

int main(){
    unsigned int x = 0;
    unsigned int y = 1;
    printf("%u\n", x - y); 
    printf("%d\n", x - y);
}

Since both of them are unsigned int, no integer promotion is needed. By the usual arithmetic converison(have the same type), (unsigned int)x-(unsigned int)y = 4294967295.

The output from the above code:(same as what we expected):

4294967295
-1

One of possible ways to fix the code:(add a type cast in the end)

int main(){
    unsigned char x = 0;
    unsigned char y = 1;
    printf("%u\n", x - y); 
    printf("%d\n", x - y);
    unsigned char z = x-y;
    printf("%u\n", z);
}

The output from the above code:

4294967295
-1
255

Example 2)

int main(){
    unsigned int a = 1;
    signed int b = -2;
    if(a + b > 0)
        puts("-1 is larger than 0");
        printf("%u\n", a+b);
}

Since both of them are integers, no integer promotion is needed. By the usual arithmetic conversion, we get (unsigned int)a+(unsigned int)b = 1+4294967294 = 4294967295.

The output from the above code:(same as what we expected)

-1 is larger than 0
4294967295

How to fix it?

int main(){
    unsigned int a = 1;
    signed int b = -2;
    signed int c = a+b;
    if(c < 0)
        puts("-1 is smaller than 0");
        printf("%d\n", c);
}

The output from the above code:

-1 is smaller than 0
-1

Example 3)

int main(){
    unsigned short a = 1;
    signed short b = -2;
    if(a + b < 0)
        puts("-1 is smaller than 0");
        printf("%d\n", a+b);
}

The last example fixed the problem since a and b both converted to int due to the integer promotion.

The output from the above code:

-1 is smaller than 0
-1

If I got some concepts mixed up, please let me know. Thanks~