当使用g ++编译C ++时，“隐藏构造函数”警告是什么意思？ [英] What does the 'hides constructor for' warning mean when compiling C++ with g++?

问题描述

使用以下代码：

#include <stdio.h>


struct my_struct {
        int a;
        int b;
        my_struct();
};

my_struct::my_struct(void)
{
        printf("constructor\n");
}

void my_struct(void)
{
        printf("standard function\n");
}

int main (int argc, char *argv[])
{
        struct my_struct s;
        s.a = 1;
        s.b = 2;

        printf("%d-%d\n", s.a, s.b);

        return 0;
}

我使用g ++ -Wshadow main.cpp编译警告：

I get a warning compiling with g++ -Wshadow main.cpp:

main.cpp:15:20: warning: ‘void my_struct()’ hides constructor for ‘struct my_struct’

如果void my_struct函数实际上替换了my_struct :: my_struct，那么我会对该警告确定。但它似乎并不是这样的。如果我运行程序，我得到：

I would be ok with that warning if the void my_struct function actually replaced the my_struct::my_struct one. But it does not appears to be the case. If I run the program, I get:

constructor
1-2

任何想法这个警告是什么意思？特别是当我把C头文件包含到C ++代码中时，这很烦人。

Any idea what this warning mean ? It is quite annoying especially when I include C headers into C++ code

推荐答案

警告指出， my_struct（）函数与 my_struct 结构具有相同的名称。这意味着你将无法写下：

The warning points out that the my_struct() function has the same name as the my_struct structure. It means you will be unable to write:

my_struct s;         // Error.

因为编译器会认为你使用函数作为类型。但是，您可能已经意识到，您仍然可以使用 struct 关键字实例化您的结构：

Because the compiler will think that you're using a function as a type. However, as you probably realized, you can still instantiate your structure with the struct keyword:

struct my_struct s;  // Valid.

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