Can the name of function and function-like macro be same?

Question

Can the name of function and function-like macro be same?
Wouldn't this cause any problem?

This will cause lots of problems and confusion. Don't do this. — , Feb 04 '14 at 14:54
you could use #undef macro-name if you want to make sure the function is always called. — tesseract, Feb 04 '14 at 15:01
Read this also [What do the parentheses around a function name mean](http://stackoverflow.com/questions/13600790/what-do-the-parentheses-around-a-function-name-mean) — Grijesh Chauhan, Feb 04 '14 at 15:10

score 8 · Accepted Answer · edited Feb 04 '14 at 14:58

8

They could be the same. Depending on how you use the name, either it gets replaced by preprocessor or not. For example

//silly but just for demonstration.
int addfive(int n)
{
    return n + 5;
}
#define addfive(n) ((n) + 5)

int main(void)
{
    int a;
    a = addfive(2); //macro
    a = (addfive)(2); //function
}

for ex. MS says that: http://msdn.microsoft.com/en-us/library/aa272055(v=vs.60).aspx

edited Feb 04 '14 at 14:58

haccks

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answered Feb 04 '14 at 14:55

Hayri Uğur Koltuk

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I simply write this code and it fails to compile: `#include #define max(a,b) ((a > b) ? a : b) int max(int a, int b) { return ((a > b) ? a : b); } int main(){ int a = 5, b = 3; printf("%d\n", max(a,b)); return 0; } ` – haccks Feb 04 '14 at 14:57
2

because #define is before function definition – Hayri Uğur Koltuk Feb 04 '14 at 14:58
+1 because I learned something new. But please don't ever do this in real code. – Fred Larson Feb 04 '14 at 15:03
1

@FredLarson, AhmedHandy http://stackoverflow.com/questions/13600790/what-do-the-parentheses-around-a-function-name-mean – Grijesh Chauhan Feb 04 '14 at 15:09
@msam it should work regardless of the compiler (given it's c90 compliant) – Hayri Uğur Koltuk Feb 04 '14 at 15:09
OK. So the extra paren along with function call (`a = (addfive)(2);`) does the magic here! o.O. +1. Thanks @GrijeshChauhan for the link. – haccks Feb 04 '14 at 15:27

score 1 · Answer 2 · answered Feb 04 '14 at 14:55

http://gcc.gnu.org/onlinedocs/cpp/Function-like-Macros.html#Function-like-Macros

Here you can see that calling the function, of which a macro with the same name exists, calls the macro instead :) For the gcc at least!

This would cause no problem, but quite some confusions. I wouldn't recommend this.

score 1 · Answer 3 · edited May 23 '17 at 12:21

I will explain via cases:
If you declared the function first then the function like macro second, macro will over take the function. i.e. it will be called always instead of the function.

//Function
double squar(double x)
{
    return x*x;
}

//Macro
#define squar(x) (x*x)

On the other hand if you declare the macro first then the function later, an exception will be arise, you wont be able to build

//Macro
#define squar(x) (x*x)

//Function
double squar(double x)
{
    return x*x;
}

At the end, in the first case, you still call the function like @Hayri Uğur Koltuk said here in his answer by (squar)(5)

Can the name of function and function-like macro be same?

3 Answers3