Is there a way to both check a macro is defined and it equals a certain value at the same time

懵懂的女人 提交于 2019-12-29 06:39:29

问题


I regularly use object-like preprocessor macros as boolean flags in C code to turn on and off sections of code.

For example

#define DEBUG_PRINT 1

And then use it like

#if(DEBUG_PRINT == 1)
    printf("%s", "Testing");
#endif

However, it comes a problem if the header file that contains the #define is forgotten to be included in the source code. Since the macro is not declared, the preprocessor treats it as if it equals 0, and the #if statement never runs.

When the header file is forgotten to be included, non-expected, unruly behaviour can occur.

Ideally, I would like to be able to both check that a macro is defined, and check that it equals a certain value, in one line. If it is not defined, the preprocessor throws an error (or warning).

I'm looking for something along the lines of:

#if-def-and-true-else-throw-error(DEBUG_PRINT)
    ...
#endif

It's like a combination of #ifdef and #if, and if it doesn't exist, uses #error.

I have explored a few avenues, however, preprocessor directives can't be used inside a #define block, and as far as I can tell, there is no preprocessor option to throw errors/warnings if a macro is not defined when used inside a #if statement.


回答1:


This may not work for the general case (I don't think there's a general solution to what you're asking for), but for your specific example you might consider changing this sequence of code:

#if(DEBUG_PRINT == 1)
    printf("%s", "Testing");
#endif

to:

if (DEBUG_PRINT == 1) {
    printf("%s", "Testing");
}

It's no more verbose and will fail to compile if DEBUG_PRINT is not defined or if it's defined to be something that cannot be compared with 1.




回答2:


as far as I can tell, there is no preprocessor option to throw errors/warnings if a macro is not defined when used inside a #if statement.

It can't be an error because the C standard specifies that behavior is legal. From section 6.10.1/3 of ISO C99 standard:

After all replacements due to macro expansion and the defined unary operator have been performed, all remaining identifiers are replaced with the pp-number 0....

As Jim Balter notes in the comment below, though, some compilers (such as gcc) can issue warnings about it. However, since the behavior of substituting 0 for unrecognized preprocessor tokens is legal (and in many cases desirable), I'd expect that enabling such warnings in practice would generate a significant amount of noise.

There's no way to do exactly what you want. If you want to generate a compilation failure if the macro is not defined, you'll have to do it explicitly

#if !defined DEBUG_PRINT
#error DEBUG_PRINT is not defined.
#endif

for each source file that cares. Alternatively, you could convert your macro to a function-like macro and avoid using #if. For example, you could define a DEBUG_PRINT macro that expands to a printf call for debug builds but expands to nothing for non-debug builds. Any file that neglects to include the header defining the macro then would fail to compile.


Edit:

Regarding desirability, I have seen numerous times where code uses:

#if ENABLE_SOME_CODE
...
#endif

instead of:

#ifdef ENABLE_SOME_CODE
...
#endif

so that #define ENABLE_SOME_CODE 0 disables the code rather than enables it.




回答3:


Rather than using DEBUG_PRINT directly in your source files, put this in the header file:

#if !defined(DEBUG_PRINT)
    #error DEBUG_PRINT is not defined
#endif

#if DEBUG_PRINT
    #define PrintDebug([args]) [definition]
#else
    #define PrintDebug
#endif

Any source file that uses PrintDebug but doesn't include the header file will fail to compile.

If you need other code than calls to PrintDebug to be compiled based on DEBUG_PRINT, consider using Michael Burr's suggestion of using plain if rather than #if (yes, the optimizer will not generate code within a false constant test).

Edit: And you can generalize PrintDebug above to include or exclude arbitrary code as long as you don't have commas that look like macro arguments:

#if !defined(IF_DEBUG)
    #error IF_DEBUG is not defined
#endif

#if IF_DEBUG
    #define IfDebug(code) code
#else
    #define IfDebug(code)
#endif

Then you can write stuff like

IfDebug(int count1;)  // IfDebug(int count1, count2;) won't work
IfDebug(int count2;)
...
IfDebug(count1++; count2++;)



回答4:


Yes you can check both:

#if defined DEBUG  &&  DEBUG == 1
#  define D(...) printf(__VA_ARGS__)
#else
#  define D(...)
#endif

In this example even when #define DEBUG 0 but it is not equal to 1 thus nothing will be printed.

You can do even this:

#if defined DEBUG  &&  DEBUG
#  define D(...) printf(__VA_ARGS__)
#else
#  define D(...)
#endif

Here if you #define DEBUG 0 and then D(1,2,3) also nothing will be printed

DOC




回答5:


Simply create a macro DEBUG_PRINT that does the actual printing:

#define DEBUG_PRINT(n, str)    \
                               \
  if(n == 1)                   \
  {                            \
    printf("%s", str);         \
  }                            \
  else if(n == 2)              \
  {                            \
    do_something_else();       \
  }                            \
                               \
#endif


#include <stdio.h>

int main()
{
  DEBUG_PRINT(1, "testing");
}

If the macro isn't defined, then you will get a compiler error because the symbol is not recognized.




回答6:


#if 0 // 0/1
#define DEBUG_PRINT printf("%s", "Testing")
#else
#define DEBUG_PRINT printf("%s")
#endif

So when "if 0" it'll do nothing and when "if 1" it'll execute the defined macro.



来源:https://stackoverflow.com/questions/17160755/is-there-a-way-to-both-check-a-macro-is-defined-and-it-equals-a-certain-value-at

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!