Why can't I use sizeof() in a #if?

Question

I have this:

#if sizeof(int)
    #error Can't use sizeof in a #if
#endif

I get this compiler error:

missing binary operator before token "("

Why can't I use the sizeof operator here?

This has been asked many times. Just do a search. Here's one answer, for example, http://stackoverflow.com/questions/1717665/c-throwing-compilation-error-on-sizeof-comparison-in-preprocessor-if — AnT stands with Russia, Feb 23 '10 at 15:54

score 12 · Accepted Answer · answered Feb 23 '10 at 15:53

Because sizeof() is calculated after the preprocessor is run, so the information is not available for #if.

C compilers are logically split into two phases, even if most modern compilers don't separate them. First, the source is preprocessed. This involves working out and substituting all the preprocessor conditionals (#if, #define, replacing defined words with their replacements). The source is then passed, processed, to the compiler itself. The preprocessor is only minimally aware of the structure of C, it has no type knowledge, so it can't handle compiler-level constructs like sizeof().

score 1 · Answer 2 · answered Feb 23 '10 at 15:53

1

Because you can only use literal constants in a preprocessor directive. Besides, sizeof(int) is always larger than 0, so I believe this #if would be true all the time anyway.

answered Feb 23 '10 at 15:53

Nick

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That was a bad example on my part. I was trying to strip all of my project-specific details out and came up with this example. – Robert Deml Feb 23 '10 at 15:56
2

Not strictly true: you can do boolean operations and there are some function-like calls (`defined()`). Some preprocessors allow a shed-load of extra stuff (I've seen people asking for log operators in the preprocessor, because they'd used them with an embedded compiler). – Andrew Aylett Feb 23 '10 at 16:01

score -3 · Answer 3 · answered Feb 28 '13 at 14:02

-3

just use ordinary if-else

if      (sizeof(x)==2)  {...}
else if (sizeof(x)==4)  {...}
else                    {...}

and compiler will optimize it in compile time...

answered Feb 28 '13 at 14:02

demiurg_spb

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1

the whole point is compile time declarations. – vitiral May 01 '15 at 15:55

score -3 · Answer 4 · answered Feb 23 '10 at 15:57

-3

Consider:

#if sizeof(MyClass) > 3
   #define MY_CONSTRAINT 2
#endif

class MyClass
{
   #if MY_CONSTRAINT == 3
      int myMember = 3;
   #endif
};

Now, this is prolly not written in the correct syntax as it's been a while since the last time I did C++, but the point still stands :)

answered Feb 23 '10 at 15:57

cwap

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Looking back on it, I really don't know :) – cwap Aug 29 '13 at 06:42
haha, I appreciate the honesty :P Glad you didn't interpret my comment as being hostile. – Ponkadoodle Aug 29 '13 at 07:21

Why can't I use sizeof() in a #if?

4 Answers4