如何在C ++中使用弱函数 [英] How to use weak function in C++

问题描述

我正在尝试在C ++中的类中使用弱函数.下面是我写的:

I am trying to use weak function in a class in C++. Below is what I wrote:

#include <stdio.h>
#include <iostream>

class A {
    public:
    void func(int argc, char *argv[]) __attribute__((weak));
};

// optional definition:
#if 0
void A::func(int argc, char *argv[]) { 
    printf("In func()\n");
    for(int aa = 0; aa < argc; aa++){
        printf("arg %d = %s \n", aa, argv[aa]);
    }
}
#endif

int main(int argc, char *argv[]) {
    A a1;
    if (a1.func){ 
        a1.func(argc, argv); 
    } else {
        printf("func() not available\n");
    }
    return 0;
}

但这会导致以下编译错误:

But this gives below compilation error:

main.cpp: In function ‘int main(int, char**)’:
main.cpp:21:16: error: cannot convert ‘A::func’ from type ‘void (A::)(int, char**)’ to type ‘bool’
     if (a1.func){
                ^

如果我将func()移到类之外，并使用gcc而不是g ++，它可以正常编译并按预期工作.有人可以告诉我是什么问题.基本上，我想实现仅在某些类函数可用时(可选功能)而不在cpp文件中使用编译器标志的情况下调用它们.

If I move the func() outside a class and use gcc instead of g++, it compiles fine and works as expected. Can someone please tell what's the problem. Basically I want to achieve calling some class functions only if they are available (an optional feature) without using Compiler flags in cpp file.

推荐答案

C ++为此提供了一种标准机制.无需链接器技巧.

C++ has a standard mechanism for this. No need for linker tricks.

class Base {
 public:
    virtual void func(int argc, char *argv[]) 
    {
         printf("func() not available\n");
    }
};

class A : public Base {
    public:
#if 0
    void func(int argc, char *argv[]) override;
#endif
};

#if 0
void A::func(int argc, char *argv[]) { 
    printf("In func()\n");
    for(int aa = 0; aa < argc; aa++){
        printf("arg %d = %s \n", aa, argv[aa]);
    }
}
#endif

int main(int argc, char *argv[]) {
    A a1;
    a1.func(argc, argv); 
}

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