Why do I have an error when sizeof is used in a preprocessor directive?

Question

MathWorks Support Team 2011-2-24

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/91486-why-do-i-have-an-error-when-sizeof-is-used-in-a-preprocessor-directive

the sizeof() function cannot be used in a preprocessor directive, according to the Ansi C standard. But some compilers accept this syntax.

We see here how to bypass the problem.

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Answer 1

MathWorks Support Team 2011-2-24

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/91486-why-do-i-have-an-error-when-sizeof-is-used-in-a-preprocessor-directive#answer_100837

For the moment, the only solution is to add some code to bypass the #if sizeof(...) line.

So, if the is something like:

#if (sizeof(myttype) != 8)

// Part 1

// some code

#else

// Part 2

// some code again

#endif

then the user should first evaluate if the condition is true or not (so here if sizeof(mtype) is different from 8 ), and add only the part of code that is true.

So if we consider that sizeof(mtype) is different from 8 (Part 1 is true), the code can be changed to:

#ifndef POLYSPACE

#if (sizeof(mytype) != 8)

// Part 1

// some code

#else

// Part 2

// some code

#endif

#else // POLYSPACE

// Part 1

// some code

#endif

Of course, the flag POLYSPACE should be added to the list of compilation flags (-D) of the project. Hence, the #if (sizeof(mytype) != 8) / #else / #endif part will be ignored (since POLYSPACE is defined) and the part 1 will be the one considered by the preprocessor.