CODEWITHSUNDEEP
c
Matt
Matt

Reputation: 1704

Issue passing a function as a parameter in C

I'm new to C and stuck on this. The answer here seems widely accepted, but I believe I'm following the correct format and still getting issues.

Here's the offending line in main() (in main.c)

PrintWrapper(PrintFunction, *decoded); // *decoded is a char

The PrintFunction (also in main.c):

void PrintFunction(char c)
{
    printf("%c", c);
}

And the prototype declaration (in p1a2.h - in the same directory as main.c)

extern void PrintWrapper(void (*)(char), char c);

The actual source is hidden away in printwrapper.o (same directory as main.c). This is for an assignment and is a very contrived usage of passing a function as a parameter. I compile the program with 

g++ main.c printwrapper.o -Wall -g -o tnine

and get the compiler error (the backtick is not a typo if that is relevant)

<path>/main.c:42: undefined reference to `PrintWrapper(void (*)(char), char'

Why does this happen? I'm hoping it's me overlooking something simple.

Upvotes: 2

Views: 160

Answers (3)

Grzegorz Szpetkowski
Grzegorz Szpetkowski

Reputation: 37954

You should use gcc instead of g++ to compile C code. The latter front-end is for C++ code, which symbols are subject to name mangling. Assuming that you are really asking about C language, the proper way to compile your project is:

gcc main.c printwrapper.o -Wall -g -o tnine

Here is some working sample:

main.c

#include <stdio.h>
#include "pla2.h"

void PrintFunction(char c)
{
    printf("%c\n", c);
}

int main(void)
{
    char c = 'a';
    char *decoded = &c;

    PrintWrapper(PrintFunction, *decoded);

    return 0;
}

pla2.h

extern void PrintWrapper(void (*)(char), char c);

printwrapper.c

extern void PrintWrapper(void (*f)(char), char c)
{
    f(c);
}

then:

$ gcc -c printwrapper.c 

$ gcc main.c printwrapper.o -Wall -g -o tnine

$ ./tnine 
a

Upvotes: 1

ababababanababab
ababababanababab

Reputation: 186

As you are using g++, the compiler expects the function name to be mangled (for more information see this wikipedia article.
To fix the issue, use gcc instead of g++ or try the following:

extern "C" void PrintWrapper(void (*)(char), char*);

Upvotes: 4

HolyBlackCat
HolyBlackCat

Reputation: 96941

The code you showed us is valid.

Looks like printwrapper.o just don't contain such function.

Upvotes: 1

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