CODEWITHSUNDEEP
cvoid-pointers
Honza B
Honza B

Reputation:

error: invalid conversion from 'void (*)(...)' to 'void (*)()'

having this problem on Mac with gcc 4.0.1 build 5370, XCode 2.5. The code snippet is:

there is a declared function, the second parameter cause the problem:

void ffi_call(ffi_cif *cif, void (*fn)(void), void *rvalue, void **avalue)

calling it like this:

typedef void (*FuncPtr)();
FuncPtr func = some_function_pointer;
ffi_call(null, func, ...);

cause the error on the third line. Looks like 'void func(void)' is different from 'void func()' for gcc 4.0.1

Any ideas about any gcc switch or just gcc update would help? Thanks for help Honza B.

Upvotes: 3

Views: 8611

Answers (3)

David R Tribble
David R Tribble

Reputation: 12204

Yes, void func(void) is different from void func().

But any function pointer can be typecast into void (*)func() and back to its original type. So you could change the function pointer parameter's type in your C function.

Or you could just force the function pointer to be the right type with an explicit cast. On most systems, all function pointers smell the same, so even though it's not portable (or kosher), it should still work on most systems.

Upvotes: 0

DigitalRoss
DigitalRoss

Reputation: 146053

There are two kinds of C, and it will get really confusing to mix them.

C compilers need to be backwards compatible because a lot of code was written in early C. It's possible to mix the two styles, but that can cause confusion and even runtime errors, because the rules are only consistent for programs that are entirely one style or entirely the other.

In legacy C ("K&R C") have things like: f(){/* stuff */} and

double f(d)
double d;
{ stuff }

A forward declaration in the legacy style might well use () even though it's really (d) because there was no way to specify the type, so little point in having a formal parameter.

And in C89- and C99-conforming style you have the familiar double f(double); for the declaration and double f(double d) { /* stuff */} for the definition. double f(void) is the ANSI way of doing it, but double f() is still "legal", but an unwise blend of the two language levels.

Upvotes: 0

please delete me
please delete me

Reputation:

If you are programming in C, void func() and void func(void) are not the same. void func() is equivalent to void func(...). (This is different from C++.)

You could try making sure the code is being compiled as C++ (if that's what you're actually intending), or if not just change your FuncPtr typedef to typedef void (*FuncPtr)(void).

Upvotes: 3

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