Char array subscript warning

Question

when I use char array subscript as in this example:

int main(){
    char pos=0;
    int array[100]={};

    for(pos=0;pos<100;pos++)
        printf(\"%i\\         


        

Answer 1

This is because int is always signed.

char doesn't have to.

char can be signed or unsigned, depending on implementation. (there are three distinct types - char, signed char, unsigned char)

But what's the problem? I can just use values from 0 to 127. Can this hurt me silently?

Oh, yes it can.

//depending on signedess of char, this will
//either be correct loop,
//or loop infinitely and write all over the memory
char an_array[50+1];
for(char i = 50; i >= 0; i--)
{
    an_array[i] = i;
    // if char is unsigned, the i variable can be never < 0
    // and this will loop infinitely
}