CODEWITHSUNDEEP
cc-preprocessorconditional-compilation
michael
michael

Reputation: 110600

Need help in understand C code

Hi I need some help in understand some C code:

#if 0
   some C code 1
#elif 0
   static int8 arry[10];
#pragma Align_to(32, arry)
   ASSERT(((int8ptr_t)arry) & 15) == 0)
#else
   ASSERT(((int8ptr_t)arry) & 15) == 0)
#endif

My questions:

  1. Is only the #else part compiled?

  2. What is the meaning of #pragma Align_to(32, arry) in the #elif 0 case?

Upvotes: 2

Views: 193

Answers (3)

chux
chux

Reputation: 154305

Yes, the #else part is what is compiled.

The #pragma directive is a compiler specific directive. As you compiler was not specified, it could mean anything.

In your case #pragma Align_to(32, arry), likely tells the compiler to insure variable 'arry' is place in memory on a 32 byte boundary. This is typically for performance reasons or compatibility concerns. You may also want to look into the keyword __attribute__ use to control similar variable attributes.

Upvotes: 0

Jens
Jens

Reputation: 72697

Answer to 1: Yes, but note that even the parts within #if 0 etc. must consist of valid preprocessing tokens. This means that this will fail with a diagnostic:

#if 0
That's what C is all about
#endif

because there's an unterminated character constant introduced by the lone '. Same goes for unterminated string literals.

Answer to 2: The pragma is a hint to the compiler that the address of arry shall be aligned on a multiple of 32.

Upvotes: 0

Dewfy
Dewfy

Reputation: 23634

Actually better way to answer is ask compiler - use g++ -E or MSVC: cl /EP to print what is really compiled

Upvotes: 2

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