Define a macro with defined(X)

Question

Here is a macro I want to use, if X_DEFINED is defined it will be evaluated to DEFAULT_X otherwise it will be evaluated to x

#define GET_X(x) (defined(X_DEFINED) ? DEFAULT_X : x)

It doesn't compile, with the error

error: 'X_DEFINED' was not declared in this scope

Any suggestions? I want to be able to select between a parameter and a global variable based on if X_DEFINED was defined or not

suggestion: Let us know what you actually want to use this for so we can tell you how to do it without macros — 463035818_is_not_an_ai
– 463035818_is_not_an_ai, Commented May 8, 2019 at 15:16
btw this line alone does not produce any error. Please provide a minimal reproducible example — 463035818_is_not_an_ai
– 463035818_is_not_an_ai, Commented May 8, 2019 at 15:17

HolyBlackCat · Accepted Answer · 2019-05-08 15:19:27Z

11

defined() only works in #if and similar preprocessor directives.

You want something like this:

#ifdef X_DEFINED
#define GET_X(x) DEFAULT_X
#else
#define GET_X(x) x
#endif

answered May 8, 2019 at 15:19

HolyBlackCat

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1 Comment

Pete Becker Over a year ago

Re: "defined() only works in #if" -- indeed. And the code could be written with #if defined(X_DEFINED) as well as with #ifdef X_DEFINED.

king_nak · Accepted Answer · 2019-05-08 15:20:47Z

2

You need to define 2 different macros, depending on whether X_DEFINED is defined:

#ifdef X_DEFINED
#   define GET_X(x) x
#else
#   define GET_X(x) DEFAULT_X
#endif

answered May 8, 2019 at 15:20

king_nak

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Comments

463035818_is_not_an_ai · Accepted Answer · 2019-05-08 15:21:22Z

1

Sloppy speaking you are mixing runtime stuff (evaluation of a ternary operator) with stuff that happens even before compilation (preprocessor). You can use #ifdef instead:

#ifdef X_DEFINED
    #define GET_X(x) DEFAULT_X
#else
    #define GET_X(x) x
#endif

answered May 8, 2019 at 15:21

463035818_is_not_an_ai

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