What is the difference between the following two declarations:
1. int foo(int);
2. int foo(int());
I am not sure if both the declarations are equivalent. What makes (2) different from (1)?
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A:
int foo(int); is the declaration of a function taking an integer as an argument and returning an integer as well
int foo(int()); declares a function taking as an argument "a pointer to a function returning int and taking {no arguments[in C++] and unspecified number of arguments[in C]} " and returning an integer.
(2) is equivalent to int foo(int (*pf)()) and int foo(int f())
Prasoon Saurav
2010-10-04 14:15:45
Taking no arguments? I believe you mean taking unspecified but not variadic arguments. `(void)` is the way to specify "no arguments".
R..
2010-10-04 14:19:36
@R I belive that is C only, not C++
Tom
2010-10-04 14:20:26
@R: Sorry I missed that. Added to my post.
Prasoon Saurav
2010-10-04 14:21:52
I think you mean "`(2)` is equivalent to `int foo(int (*pf)())`..."
John Bode
2010-10-04 14:22:01
@John Bode: Ohh! I need a cup of tea `:)`
Prasoon Saurav
2010-10-04 14:23:17
To clarify, what @R means is that a specific invokation of `foo` can only call `pf` with a certain but unspecified numbre of arguments. But `foo` can take functions that accept different number of arguments in *different* invocations of `foo`. I.e I think this is valid: `void f(void a(), int i) { if (i == 0) a(); else a(1); } void g() { } void h(int i) { } int main() { f(g, 0); f(h, 1); }` but this is undefined behavior: `void j(int a()) { a(); a(1); }` as is this: `void j(int x, ...) { } int main() { f(j, 1); }`
Johannes Schaub - litb
2010-10-04 14:26:34
@Johannes: That's roughly correct. What I mean is that the function pointed to cannot be a variadic function, but the pointer is able to point to a function with arbitrary argument types compatible with default promotions as long as the correct types corresponding to the actual function pointed to are used when making the call.
R..
2010-10-04 17:15:11