What is the difference between #pragma
and _Pragma()
in C?
syntax:
#pragma arg
and
_Pragma(arg)
When should I use _Pragma(arg)
?
_Pragma
operator introduced in C99
. _Pragma(arg)
is an operator, much like sizeof
or defined
, and can be embedded in a macro.
According to cpp.gnu.org reference:
Its syntax is
_Pragma (string-literal)
, where string-literal can be either a normal or wide-character string literal. It is destringized, by replacing all\\
with a single\
and all\"
with a"
. The result is then processed as if it had appeared as the right hand side of a#pragma
directive. For example,_Pragma ("GCC dependency \"parse.y\"")
has the same effect as
#pragma GCC dependency "parse.y"
. The same effect could be achieved using macros, for example#define DO_PRAGMA(x) _Pragma (#x) DO_PRAGMA (GCC dependency "parse.y")
According to IBM tutorial:
The _Pragma operator is an alternative method of specifying #pragma directives. For example, the following two statements are equivalent:
#pragma comment(copyright, "IBM 2010") _Pragma("comment(copyright, \"IBM 2010\")")
The string IBM 2010 is inserted into the C++ object file when the following code is compiled:
_Pragma("comment(copyright, \"IBM 2010\")") int main() { return 0; }
For more information about _pragma with example.
From here:
Pragma directives specify machine- or operating-specific compiler features. The __pragma keyword, which is specific to the Microsoft compiler, enables you to code pragma directives within macro definitions.
Also (same link):
The __pragma() Keyword
Microsoft specific
The compiler also supports the
__pragma
keyword, which has the same functionality as the#pragma
directive, but can be used inline in a macro definition. The#pragma
directive cannot be used in a macro definition because the compiler interprets the number sign character ('#') in the directive to be the stringizing operator (#).
So basically you can always use #pragma
instead of __pragma()
. There is no need to use __pragma()
, but it can be used sometimes.
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