ivy
ivy

Reputation: 1485

Function pointers - pass arguments to a function pointer

I have a problem with the code that uses function pointers, take a look:

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

typedef void (*VFUNCV)(void);

void fun1(int a, double b) { printf("%d %f fun1\n", a, b); }
void fun2(int a, double b) { printf("%d %f fun2\n", a, b); }

void call(int which, VFUNCV* fun, int a, double b)
{
    fun[which](a, b);
}

int main()
{
    VFUNCV fun[2] = {fun1, fun2};
    call(0, fun, 3, 4.5);
    return 0;
}

And it produces errors:

/home/ivy/Desktop/CTests//funargs.c||In function ‘call’:|
/home/ivy/Desktop/CTests//funargs.c|11|error: too many arguments to function ‘*(fun + (unsigned int)((unsigned int)which * 4u))’|
/home/ivy/Desktop/CTests//funargs.c||In function ‘main’:|
/home/ivy/Desktop/CTests//funargs.c|16|warning: initialization from incompatible pointer type [enabled by default]|
/home/ivy/Desktop/CTests//funargs.c|16|warning: (near initialization for ‘fun[0]’) [enabled by default]|
/home/ivy/Desktop/CTests//funargs.c|16|warning: initialization from incompatible pointer type [enabled by default]|
/home/ivy/Desktop/CTests//funargs.c|16|warning: (near initialization for ‘fun[1]’) [enabled by default]|
||=== Build finished: 1 errors, 4 warnings ===|

I used a Code::Blocks to compile it.

Its simple, when I dont have any arguments but with some, I got confused:

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

typedef void (*VFUNCV)(void);

void fun1() { printf("fun1\n"); }
void fun2() { printf("fun2\n"); }

void call(int which, VFUNCV* fun)
{
    fun[which]();
}

int main()
{
    VFUNCV fun[2] = {fun1, fun2};
    call(1, fun);
    return 0;
}

Upvotes: 2

Views: 5405

Answers (2)

P0W
P0W

Reputation: 47844

Fix typedef to

typedef void (*VFUNCV)(int , double );

as fun1 and fun2 accepts two argument of type int and double

Upvotes: 6

Bort
Bort

Reputation: 2491

Your function pointer does not fit to your function declarations. Try defining it as

typedef void (*VFUNCV)(int, double);

Upvotes: 8

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