What does the "different width due to prototype" warning mean?

Question

The code generating the warning, passing argument 1 of 'isHex' with different width due to prototype:

/* Checks if a character is either 0-9 or A-F */
int isHex(char ch) {
    return isdigit(ch) || (ch >= 65 && ch <= 70);
}

/* Checks if a string only contains numeric characters or A-F */
int strIsHex(char * str) {
    char *ch;
    size_t len = strlen(str);
    for(ch=str;ch<(str+len);ch++) {
        if (!isHex(*ch)) return 0;
    }
    return 1;
}

What does this mean, shouldn't the char values be the be same width? How can I cast them to the same width to prevent this warning?

By the way, the gcc command was: gcc.exe -std=c99 -fgnu89-inline -pedantic-errors -Wno-long-long -Wall -Wextra -Wconversion -Wshadow -Wpointer-arith -Wcast-qual -Wcast-align -Wwrite-strings -fshort-enums -gstabs -l"C:\program files\quincy\mingw\include" -o main.o -c main.c

I can't remove any of the warning options from gcc as one of the marking criteria for an assignment is that there are no errors or warnings with this command.

Thanks.

Answer 1

This is due to the -Wconversion flag on the command line. It pops up "if a prototype causes a type conversion that is different from what would happen to the same argument in the absence of a prototype." Since the default integral argument type is int, you can't declare isHex(char ch) without triggering this.

You have two options, I think: declare isHex(int ch) and let it be widened in the call or else declare it isHex(char *ch) and change the call.

P.S. If this is homework, it should be tagged as such.