#define var in another #define in C

Question

I used #define macro with variable from another #define, but the ## not work! Please help!

#define _var 0b1010
#define JOIN(var1, var2) var1 ## var2

int main()
{   
    int num1 = JOIN(0b1010, 1010); // <== This works!
    int num2 = JOIN(_var, 1010); // <== This does not work!
    return 0;
}

Define "not work". What does it do? What do you want it to do? There's no actual question in your question. — chris, Apr 02 '19 at 11:28
The C Standard does not define the `0b` prefix for binary values. — pmg, Apr 02 '19 at 11:30
FYI `_var` is an identifier reserved by the C **and** C++ standards for use in the scope you use it in. You can trigger undefined behavior. Avoid leading underscores at file/global scope and as macro identifiers. — StoryTeller - Unslander Monica, Apr 02 '19 at 11:34

HolyBlackCat · Accepted Answer · 2019-04-02T11:33:39.220

3

## doesn't expand macros in its operands. You need to wrap JOIN into yet another macro to fix this:

#define JOIN(var1, var2) JOIN_(var1, var2)
#define JOIN_(var1, var2) var1##var2

edited Apr 02 '19 at 11:33

answered Apr 02 '19 at 11:30

HolyBlackCat

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2

You're a macro. – Hatted Rooster Apr 02 '19 at 11:30
1

@SombreroChicken No you. – HolyBlackCat Apr 02 '19 at 11:30
1

@SombreroChicken - And your'e a `##`! – StoryTeller - Unslander Monica Apr 02 '19 at 11:32
3

You substituting token-paster! – Quentin Apr 02 '19 at 11:33
Great, this works! Thank you! If this is not useful for everyone, could i delete this? Sorry, i'm just beginner in C. But problem i could not find anywhere! – Nguyen Duc Linh Apr 02 '19 at 11:40

#define var in another #define in C

1 Answers1