I want to create a function prototype in C++ so that there is a void *
argument that can take pointers of any type. I know that this is possible in C. Is it possible in C++?
[EDIT] Here is a simplified version of the code that I am trying to get to work:
#include <stdio.h>
void func(void (f)(const void *))
{
int i = 3;
(*f)(&i);
}
void func_i(const int *i)
{
printf("i=%p\n",i);
}
void func_f(const float *f)
{
printf("f=%p\n",f);
}
void bar()
{
func(func_i);
}
And here is the compiler output:
$ g++ -c -Wall x.cpp
x.cpp: In function ‘void bar()’:
x.cpp:21: error: invalid conversion from ‘void (*)(const int*)’ to ‘void (*)(const void*)’
x.cpp:21: error: initializing argument 1 of ‘void func(void (*)(const void*))’
$ %
You may use void*, just as with C, but you'll need to cast your argument when calling it. I suggest you use a template function
template<typename T>
void doSomething(T* t) {...}
How about:
void func(void *);
exactly like in C? : P
Yes.
int i = 345;
void * ptr = &i;
int k = *static_cast< int* >(ptr);
UPDATE :: What you have shown in the code certainly cannot be done in C++. Casting between void and any other must always be done. Check these SO link for more details on what the C -standard has to say: 1) http://stackoverflow.com/questions/188839/function-pointer-cast-to-different-signature 2) http://stackoverflow.com/questions/559581/casting-a-function-pointer-to-another-type
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