Print name of caller function

Question

I found __FUNCTION__ macro that will give the name of currently executing function, But is there any macro available to print the name of parent function (a function that invoked currently running function) ?

void print_error_message(const char *msg)
{
    printf("%s : %s\n",__FUNCTION__,msg);
}

int main()
{
    print_error_message("failed to connect");
    return 0;
}

The above program will give output :

print_error_message : failed to connect

But what I actually required is

main : failed to connect

because main() invoked print_error_message()

I don't want to pass one more argument to function print_error_message() like print_error_message(__FUNCTION__, msg), what I am expecting is a similar macro or function to retrieve the name of the caller function.

Answer 1

There's no standard way to get the name from the call-stack, and if debug information is stripped or otherwise unavailable there's possible no way at all to get the name.

As a simple workround, you could use macros:

#define print_error_message(msg) actual_print_error_message(__FUNCTION__, (msg))

void actual_print_error_message(const char *func, const char *msg)
{
    printf("%s : %s\n",func,msg);
}

Note that __FUNCTION__ is not standard C, it's an extension. For standards-compliance use the C99 special predefined variable __func__.