Does C support overloading?

by

[*]

No, C doesn’t support any form of overloading (unless you count the fact that the built-in operators are overloaded already, to be a form of overloading).

printf works using a feature called varargs. You make a call that looks like it might be overloaded:

printf("%d", 12); // int overload?
printf("%s", "hi"); // char* overload?

Actually it isn’t. There is only one printf function, but the compiler uses a special calling convention to call it, where whatever arguments you provide are put in sequence on the stack[*]. printf (or vprintf) examines the format string and uses that to work out how to read those arguments back. This is why printf isn’t type-safe:

char *format = "%d";
printf(format, "hi"); // undefined behaviour, no diagnostic required.

[*] the standard doesn’t actually say they’re passed on the stack, or mention a stack at all, but that’s the natural implementation.

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