What this trick is?

Question

#include <stdio.h>
#define PRINT(name) print ## name()

void printHE()
{
    printf("Hello");
}
void printWO()
{
    printf("World\n");
}


enum {
    HE,
    WO,
};

int main()
{
    PRINT(HE);
    PRINT(WO);
}

It works perfectly, but why?

What does ## in #define mean?

And why HE didn't convert to 0 ?

You don't show us the `#define` so we can only guess what you are doing. In the preprocessor `##` stands for concatenation. — Jens Gustedt, Sep 05 '13 at 14:49
Perhaps it's doing a [System.Console.WriteLine()](http://stackoverflow.com/q/18639239/656243) — Lynn Crumbling, Sep 05 '13 at 14:51
Note that the `enum { HE, WO };` is a red herring — it has no effect or use in the code shown. — Jonathan Leffler, Sep 05 '13 at 14:53
@trojanfoe `#define PRINT(name) print ## name` is not the defined? — Lidong Guo, Sep 05 '13 at 14:54
@dasblinkenlight: I inserted the macro (and header); I missed the parentheses invoking the built function name in the first pass. — Jonathan Leffler, Sep 05 '13 at 14:55
@LidongGuo: you definitely did not post all the code. What you posted would not compile — there is no function `PRINT`. You don't actually say what it means when "it works perfectly", but the presumption is that the program does compile and produces "HelloWorld" as output. For it to do that, you need a macro similar to the one I added which creates the function name and invokes it. — Jonathan Leffler, Sep 05 '13 at 14:59
@JonathanLeffler Oh, I have a marco just the same as you add, so I think I post all my codes. sorry about this . — Lidong Guo, Sep 05 '13 at 15:01

Sergey Kalinichenko · Accepted Answer · 2013-09-05T15:01:38.510

4

Given that you're asking about ## I am assuming that PRINT is defined as

#define PRINT(X) print##X()

The ## is a token-pasting operator, it connects two tokens to its left and to its right together, producing a single token.

When you write PRINT(HE), preprocessor converts that to printHE(), which is a regular function call.

since HE is a enum, should HE translate to 0

That's a very good question! The translation does not happen, because preprocessor runs before enums are interpreted, so the fact that HE and WO are enum members does not change anything.

edited Sep 05 '13 at 15:01

answered Sep 05 '13 at 14:52

Sergey Kalinichenko

714,442
84
1,110
1,523

I am confused about this : since `HE` is a enum, should `HE` translate to `0` ? – Lidong Guo Sep 05 '13 at 14:59
@LidongGuo No, as mentioned the `enum` plays no part. The pre-processor takes the term `print` and pastes it onto `HE` to produce `printHE`, which is the name of a function. – trojanfoe Sep 05 '13 at 15:01
@LidongGuo That's a very good question, please see the edit. – Sergey Kalinichenko Sep 05 '13 at 15:02

score 0 · Answer 2 · answered Sep 05 '13 at 14:53

0

It is the escape sequence in your expression. It concatenates the leftmost and the rightmost to produce the token.

## is token pasting operator

answered Sep 05 '13 at 14:53

Rahul Tripathi

168,305
31
280
331

score 0 · Answer 3 · answered Sep 05 '13 at 14:53

0

In #define PRINT(name) print ## name

## is token pasting operator, used to "glue" tokens together

answered Sep 05 '13 at 14:53

P0W

46,614
9
72
119

score 0 · Answer 4 · answered Sep 05 '13 at 14:55

0

On my computer, it doesn't work right. Print error like below:

two_sharp.c:(.text+0x3a): undefined reference to PRINT' two_sharp.c:(.text+0x46): undefined reference toPRINT' collect2: ld returned 1 exit status

answered Sep 05 '13 at 14:55

styshoo

111
4

What this trick is?

4 Answers4