C++ function pointer parameter passing mechanism

The function pointer is passed as a parameter in C: the function pointer is passed as a constant pointer, a copy is created during the transfer process, the received function formal parameter points to the copy, and the dereferenced copy can call the underlying function.

C function pointer parameter passing mechanism

In C, function pointers can be passed to functions as parameters. This allows us to dynamically encapsulate function calls in a callable object.

Passing mechanism

When passing a function pointer as a parameter, the following mechanism is followed:

1. A function pointer is essentially a pointer to a function address constant pointer.
2. When a function pointer is passed to a function as a parameter, a copy of the function pointer is created and passed to the function.
3. The formal parameters in the receiving function point to a copy of the passed function pointer. It can be dereferenced to call the underlying function.

Practical case

The following is an example of a C program that uses function pointers as parameters:

#include <iostream>

// 一个接受函数指针作为参数的函数
void callFunction(void (*function)()) {
    function();  // 调用通过函数指针传递的函数
}

// 一个示例函数
void printMessage() {
    std::cout << "Hello, world!" << std::endl;
}

int main() {
    // 定义一个函数指针指向 printMessage 函数
    void (*printMessageFunction)() = &printMessage;

    // 将函数指针传递给 callFunction 函数
    callFunction(printMessageFunction);

    return 0;
}

In the above example,callFunction The function accepts a function pointer (void (*function)() as a parameter, which points to a function that does not accept parameters and does not return a value. main Function A function pointer named printMessageFunction is defined that points to the printMessage function, which is then passed to the callFunction function. The callFunction function dereferences the function pointer (function()) and calls the underlying printMessage function to output "Hello, world!"

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