c ++将函数作为参数传递给另一个带有void指针的函数

Question

我正在尝试将函数作为参数传递给另一个带有void指针的函数,但它不起作用

#include <iostream>
using namespace std;

void print()
{
    cout << "hello!" << endl;
}

void execute(void* f()) //receives the address of print
{
    void (*john)(); // declares pointer to function
    john = (void*) f; // assigns address of print to pointer, specifying print returns nothing
    john(); // execute pointer
}

int main()
{
    execute(&print); // function that sends the address of print
    return 0;
}

事情是void函数指针,我可以做一个更简单的代码

#include <iostream>
using namespace std;

void print();
void execute(void());

int main()
{
    execute(print); // sends address of print
    return 0;
}

void print()
{
    cout << "Hello!" << endl;
}

void execute(void f()) // receive address of print
{
    f();
}

但我想知道我是否可以使用void指针

这是为了实现这样的东西

void print()
{
    cout << "hello!" << endl;
}

void increase(int& a)
{
    a++;
}

void execute(void *f) //receives the address of print
{
    void (*john)(); // declares pointer to function
    john = f; // assigns address of print to pointer
    john(); // execute pointer
}

int main()
{
    int a = 15;
    execute(increase(a));
    execute(&print); // function that sends the address of print
    cout << a << endl;
    return 0;
}

Answer 1

用gcc test.cpp我得到:

test.cpp: In function ‘void execute(void* (*)())’:
test.cpp:12:22: error: invalid conversion from ‘void*’ to ‘void (*)()’ [-fpermissive]
test.cpp: In function ‘int main()’:
test.cpp:18:19: error: invalid conversion from ‘void (*)()’ to ‘void* (*)()’ [-fpermissive]
test.cpp:9:6: error:   initializing argument 1 of ‘void execute(void* (*)())’ [-fpermissive]

f参数的签名不正确.你需要使用

void execute(void (* f)())

代替.因此,在分配给时,您不需要演员john:

john = f

此外,您可以通过f直接调用来简化:

f(); // execute function pointer

编辑:因为你想使用void指针,你需要f作为一个void指针传递:

void execute(void *f)

在这里,您将需要分配john,但是f已经是void *您不需要演员.

注意:假设您正在传递一个void指针,该execute函数将接受任何内容,如果您传递错误的东西,您将遇到运行时错误.例如:

void print_name(const char *name)
{
    printf("%s", name);
}

void execute1(void *f);
void execute2(void (*f)());

int main()
{
    int value = 2;
    execute1(&value); // runtime crash
    execute1(&print_name); // runtime crash
    execute2(&value); // compile-time error
    execute2(&print_name); // compile-time error
}

使用特定定义的函数指针,编译器会在您传递错误参数类型的位置生成错误.这在运行时崩溃是首选,因为运行时崩溃可能被利用为安全漏洞,需要进行大量测试以确保不会发生此错误.