hoe getopt() deals with char type

Question

I have a question about the getopt function, as the following code shows, the type of "ch" is "int",but in the switch clause, it is regarded as "char" .. I am confused why?

Thansk!!

 int main(int argc, char **argv) 
{
extern int optind;
extern char * optarg;
int ch;
char * format = "f:hnBm:";

// Default makefile name will be Makefile
char szMakefile[64] = "Makefile";
char szTarget[64];
char szLog[64];

while((ch = getopt(argc, argv, format)) != -1) 
{
    switch(ch) 
    {
        case 'f':
            strcpy(szMakefile, strdup(optarg));
            break;
        case 'n':
            break;
        case 'B':
            break;
        case 'm':   

            strcpy(szLog, strdup(optarg));
            break;
        case 'h':
        default:
            show_error_message(argv[0]);
            exit(1);
    }
}

Answer 1

In C, a char is really just an integer of a certain size and an int can be implicitly converted into one, so it works transparently.