Basic and Advance C Question:
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I cant get va_arg to pull in an argument of type pointer-to-function.
Answer:
Try using a typedef for the function pointer type.
The type-rewriting games which the va_arg macro typically plays are stymied by overly-complicated types such as pointer-to-function. To illustrate, a simplified implementation of va_arg is
#define va_arg(argp, type)
(*(type *)(((argp) += sizeof(type)) - sizeof(type)))
where argp's type (va_list) is char *. When you attempt to invoke
va_arg(argp, int (*)())
the expansion is
(*(int (*)() *)(((argp) += sizeof(int (*)())) - sizeof(int (*)())))
which is a syntax error (the first cast (int (*)() *) is meaningless).
If you use a typedef for the function pointer type, however, all will be well. Given
typedef int (*funcptr)();
the expansion of
va_arg(argp, funcptr)
is
(*(funcptr *)(((argp) += sizeof(funcptr)) - sizeof(funcptr)))
which works correctly.
The type-rewriting games which the va_arg macro typically plays are stymied by overly-complicated types such as pointer-to-function. To illustrate, a simplified implementation of va_arg is
#define va_arg(argp, type)
(*(type *)(((argp) += sizeof(type)) - sizeof(type)))
where argp's type (va_list) is char *. When you attempt to invoke
va_arg(argp, int (*)())
the expansion is
(*(int (*)() *)(((argp) += sizeof(int (*)())) - sizeof(int (*)())))
which is a syntax error (the first cast (int (*)() *) is meaningless).
If you use a typedef for the function pointer type, however, all will be well. Given
typedef int (*funcptr)();
the expansion of
va_arg(argp, funcptr)
is
(*(funcptr *)(((argp) += sizeof(funcptr)) - sizeof(funcptr)))
which works correctly.
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