C++ ## macro doesn't work

Question

A macro with a ## will concatenate the two elements together, for example if you use #define sptember oct ## ober you will obtain october.

So my problem is: I have a macro like this #define getRegByPin(pin) set ## pin than I have from 1 to 19 some defines like this: #define set0 xxx and #define set1 xxx, etc.

But when I call my macro in code int p = getPinNo(pin); st(getRegByPin(p), p, to); it replaces getRegByPin(p) with setp instead of set0 or set13 or etc.

What can i do?

Thx for help! You are awesome! :)

You can use macros, but your example code doesn't make sense. What is `set0` supposed to be? — trojanfoe, Dec 24 '15 at 11:41
Possible duplicate of [Creating C macro with ## and \_\_LINE\_\_ (token concatenation with positioning macro)](http://stackoverflow.com/questions/1597007/creating-c-macro-with-and-line-token-concatenation-with-positioning-macr) — Dialecticus, Dec 24 '15 at 11:42
Why don't you use a function instead of a macro? With inlining and perhaps templates, it should be more type-safe, clean and similarly performant. — Ulrich Eckhardt, Dec 24 '15 at 12:01
Well, it is working. It just isn't supposed to do what you think it's supposed to do. — StoryTeller - Unslander Monica, Dec 24 '15 at 12:10

score 1 · Answer 1 · answered Dec 24 '15 at 12:00

The C preprocessor (and C++ has just inherited it), just does textual substitution. It knows nothing of variables. So given

#define getRegByPin(pin) set ## pin
const int p = 5;

getRegByPin(p);  // Will expand to setp, not set5

From the syntax, I guess that set0 to set13 are constants. Do they have values you can calculate? For example:

auto getRegByPin(int pin) { return set0+pin; }  // or (set0 << pin)

If not, you are going to need a constant array which you can index:

auto getRegByPin(int pin) {
    static const setType_t pins[16] = { set0, set1, set2 ... set15};
    return pins[pin];
}

If they are not constants, but functions, your array will need to be an array of function pointers.

Prefer to use functions than the preprocessor.

C++ ## macro doesn't work

1 Answers1