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[英]C++: how can I overload a function so that it can take any functor, including pointer to member functions?
[英]How do I declare a C++ prototype with a void * pointer so that it can take any pointer type?
我想在C ++中創建一個函數原型,以便有一個void *
參數可以接受任何類型的指針。 我知道這在C語言中是可能的。在C ++中是否可能?
[編輯]這是我嘗試使用的代碼的簡化版本:
#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);
}
這是編譯器的輸出:
$ 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*))’
$ %
您可以像使用C一樣使用void *,但是在調用它時需要轉換參數。 我建議您使用模板功能
template<typename T>
void doSomething(T* t) {...}
怎么樣:
void func(void *);
完全像在C中一樣? :P
是。
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 explicitly 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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