arith-overflow

arithmetic.c

#include <stdio.h>
    
int main() {
    // Given an unsigned char, if we add 1 to the max value an unsigned char can have,
    // we strangely get the correct value, which should have overflown the char...
    unsigned char a = 255;
    printf("%lu + 1 == %d \n", a, a + 1);
    // Output: 255 + 1 == 256
    
    // When we take this value and assign it to an unsigned char, and look at the
    // value again we get ...
    unsigned char b = a + 1;
    printf("%lu + 1 == %d \n", a, b);
    // Output: 255 + 1 == 0
    
    // This is because in C all arithmetic UP CASTS if there is an overflow!
    // Looking at the type of the result of a + 1
    printf("sizeof a == %d; sizeof a+1 == %d\n", sizeof(a), sizeof(a+1));
    // Output: sizeof a == 1; sizeof a+1 == 4
    // As you can see the sizeof a is 1 byte, and the sizeof a+1 is 4 bytes
    //
    unsigned char i = 0;
    unsigned char j = 255;
    printf(" 0 - 255 == %d\n", i - j);
    // Output: 0 - 255 == -255
    printf(" 0 - 255 == %d\n", (unsigned char)(i - j));
    // Output: 0 - 255 == 1
    // You can see that this can get confusing if you do not explicitly cast arithmetic

    printf("-128/-1 == %d\n", (signed char)(-128/-1));
    // Output: -128/-1 == -128
    // WAT? :)

    return 0;
}