Replacing function call (and not declaration) with define C

Question

I try to use define to replace function call, but I can't find how to replace only the call and not the declaration.

IE:

#define test(); printf("worked\n");

void test()
{
printf("how sad ?\n");
}

int main()
{
test();
}

I can't create my define after the function (project rules)

The problem is : I expected the semicolon after "test()" in the define to replaces only the call, but it replaces also the declaration.

I tried to google and nothing on that, is it really possible ? It's weird that it is not taking the literal expression.

Attie · Accepted Answer · 2018-08-16 09:57:59Z

7

A few things to note:

#define doesn't require braces () - only use them if you need to handle parameters
The syntax doesn't include a semicolon: #define test printf
Calling a function like printf() like this (somewhat obscured) can be risky, especially if the caller doesn't expect that their string is used as a format string. Prefer #define test(msg) printf("%s", msg)
After a #define test ..., the pre-processor will dumbly replace all instances of test - thus the function declaration will actually read void printf("worked\n"); { ... }

The result should be either:

#include <stdio.h>

#define test(msg) printf("%s\n", msg)

void main(void) {
    test("hello");
}

or:

#include <stdio.h>

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

void main(void) {
    test("hello");
}

If you are trying to use a #define to redirect a function call, then you must use a different symbol... for example:

#include <stdio.h>

/* Comment / uncomment this to change behaviour */
#define REDIRECT_TEST

#ifdef REDIRECT_TEST
#  define my_test(msg) printf("REDIRECTED:%s\n", msg)
#else
#  define my_test      test    
#endif

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

void main(void) {
    my_test("hello");
}

edited Aug 16, 2018 at 9:57

answered Aug 16, 2018 at 9:52

Attie

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2 Comments

Malorke Over a year ago

Thanks for your answer. Sadly, my project really restrict me and I can't use your solution. I will have to do it manually.

Paul Stelian Over a year ago

#define test(); printf("worked\n"); actually means that when you write test() it will output ; printf("worked\n"); (notice that extra semicolon)... (or am I wrong?)

Dennis R · Accepted Answer · 2019-05-04 13:33:54Z

1

printf have a "variadic arguments". Sometimes уou will have a problem if you wont use any solution from that q/a: "How to wrap printf() into a function or macro?"

Some tip like this, for example:

#define MY_PRINTF(...) printf(__VA_ARGS__)

or this:

#define MY_PRINTF(f_, ...) printf((f_), ##__VA_ARGS__)

answered May 4, 2019 at 13:33

Dennis R

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Comments

Eliyahu Machluf · Accepted Answer · 2018-08-16 09:55:43Z

0

you should have the define at a separated header file. and the define shall not include the semicolon. so your code shall look like this:

replace_test.h:

#define test() dummy_test()

test.h:

void test();

test.c:

void test()
{ your test code}

dummy.c:

void dummy_test()
{
 your dummy code here (printf("worked!"); etc.
}

program.c:

//decide which behavior you want, either include replace_test.h or test.h header
#include "replace_test.h"
//#include "test.h"

int main()
{
   test();
}

answered Aug 16, 2018 at 9:55

Eliyahu Machluf

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6 Comments

Eliyahu Machluf Over a year ago

Can you explain why do you think the answer is completely wrong??? as far as I know, this the common way to use mocks at c with macros. having the define at a header, and using the dummy header as needed.

unalignedmemoryaccess Over a year ago

You did macro with name test and then function with the same name test. what?

Eliyahu Machluf Over a year ago

it is OK to have a macro and a function with the same name. It would cause problem only if the file test.c or test.h have include to replace_test.h. at the using code (program.c), the user can decide whether to include the real test.h header, and really call the function test(), or to include the replace_test.h header, and have the code which call 'test()' replaced with a call to 'dummy_test()' method.

Eliyahu Machluf Over a year ago

BTW, regarding mock methods I mentioned before, there is one way doing this with macros, as I mentioned, and another way doing this using function pointers, directing pointers to dummy implementations as needed.

unalignedmemoryaccess Over a year ago

Then explain properly.

|

chuckt · Accepted Answer · 2022-10-01 18:12:50Z

0

Here is an excerpt from my unit test framework. I believe this does what you want...

void *(*old_calloc)(size_t, size_t) = calloc;
#define calloc unit_calloc
void *unit_calloc(size_t num, size_t size) {
    ...
    void* retv = old_calloc(num, size);
    ...
    return retv;
}

This allows me to redirect a call to calloc() without modifying the code that I am trying to test.

answered Oct 1, 2022 at 18:12

chuckt

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