[완료]Pointer to function 부분에서 갑자기 혼란이 옵니다!

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http://kldp.org/node/4103

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답변이 맞는건 같은데;;;;. 알면. 참 좋을텐데... 영어해석이 어렵네요.
http://www.velocityreviews.com/forums/t715007-re-how-to-understand-this-cpp-syntax.html

#define SIG_ERR ((void (*) ())-1)
#define SIG_DFL ((void (*) ())0)
#define SIG_IGN ((void (*) ())1)

Each definition is an integer constant expression converted,
via a cast, to a pointer-to-function type. The cast is to type
``void(*)()'', i.e., pointer to function returning void.

A modern implementation would probably define the
parameter type:

#define SIG_ERR ((void (*) (int))-1)
#define SIG_DFL ((void (*) (int))0)
#define SIG_IGN ((void (*) (int))1)

SIG_ERR, SIG_DFL, and SIG_IGN have to be function pointers, but
they merely have to be unique values; it needn't be possible to
call anything using their values. Converting a constant 0 to a
pointer-to-function type yields a null pointer, so SIG_DFL happens
to expand to an expression that evaluates to a null pointer value;
(SIG_DFL == NULL) is true. Converting any other integer value to
a pointer-to-function type yields some value that's not specified
by the standard (I'm not sure whether it's implementation-defined
or unspecified). The point is that, for the implementation for
which the above definitions were written, the three expressions
yield pointer values that are distinguishable from any pointer to
an actual function.