How to achieve function overloading in C?

Question

Is there any way to achieve function overloading in C? I am looking at simple functions to be overloaded like

foo (int a)  
foo (char b)  
foo (float c , int d)

I think there is no straight forward way; I'm looking for workarounds if any exist.

Answer 1

In the sense you mean — no, you cannot.

You can declare a va_arg function like

void my_func(char* format, ...);

, but you'll need to pass some kind of information about number of variables and their types in the first argument — like printf() does.

Answer 2

Here is the clearest and most concise example I've found demonstrating function overloading in C:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int addi(int a, int b) {
    return a + b;
}

char *adds(char *a, char *b) {
    char *res = malloc(strlen(a) + strlen(b) + 1);
    strcpy(res, a);
    strcat(res, b);
    return res;
}

#define add(a, b) _Generic(a, int: addi, char*: adds)(a, b)

int main(void) {
    int a = 1, b = 2;
    printf("%d\n", add(a, b)); // 3

    char *c = "hello ", *d = "world";
    printf("%s\n", add(c, d)); // hello world

    return 0;
}

https://gist.github.com/barosl/e0af4a92b2b8cabd05a7

Answer 3

Try to declare these functions as extern "C++" if your compiler supports this, http://msdn.microsoft.com/en-us/library/s6y4zxec(VS.80).aspx

Answer 4

There are few possibilities:

1. printf style functions (type as an argument)
2. opengl style functions (type in function name)
3. c subset of c++ (if You can use a c++ compiler)

Answer 5

Yes, sort of.

Here you go by example :

void printA(int a){
printf("Hello world from printA : %d\n",a);
}

void printB(const char *buff){
printf("Hello world from printB : %s\n",buff);
}

#define Max_ITEMS() 6, 5, 4, 3, 2, 1, 0 
#define __VA_ARG_N(_1, _2, _3, _4, _5, _6, N, ...) N
#define _Num_ARGS_(...) __VA_ARG_N(__VA_ARGS__) 
#define NUM_ARGS(...) (_Num_ARGS_(_0, ## __VA_ARGS__, Max_ITEMS()) - 1) 
#define CHECK_ARGS_MAX_LIMIT(t) if(NUM_ARGS(args)>t)
#define CHECK_ARGS_MIN_LIMIT(t) if(NUM_ARGS(args) 
#define print(x , args ...) \
CHECK_ARGS_MIN_LIMIT(1) printf("error");fflush(stdout); \
CHECK_ARGS_MAX_LIMIT(4) printf("error");fflush(stdout); \
({ \
if (__builtin_types_compatible_p (typeof (x), int)) \
printA(x, ##args); \
else \
printB (x,##args); \
})

int main(int argc, char** argv) {
    int a=0;
    print(a);
    print("hello");
    return (EXIT_SUCCESS);
}

It will output 0 and hello .. from printA and printB.

Answer 6

Can't you just use C++ and not use all other C++ features except this one?

If still no just strict C then I would recommend variadic functions instead.