C ++对象和C风格的问题 [英] C++ object and C style cast question

问题描述

我有以下代码由gcc编译：

I have the following code compiled by gcc:

#include <iostream>
using namespace std;


class Buffer {
public:
   operator char *() { cout << "operator const * called" << endl; return buff; }
private:
    char buff[1024];
};


int main(int, char**) {
    Buffer b;
    (char *)b;  // Buffer::operator char * is called here

    return 0;
}

我看到的是Buffer :: operator char *

What I see is that Buffer::operator char * is called on line:

(char *)b;

为什么C风格的转型调用Buffer :: operator char *

Why C style cast calls Buffer::operator char * is called here?

我认为

static_cast<char *>(b);

来显式调用Buffer :: operator char *。

should be used in order to invoke explicitly Buffer::operator char *.

推荐答案

如果你做了（char *）（& b）样式转换和 operator char * 将不会被调用。这里你试图将一个对象转换为char *。由于没有自动转换编译器查找由您提供的operator char *。如果你没有提供它，你将得到一个编译器错误，说 Buffer 不能转换为 char *

If you had done (char *)(&b) it would have been C style cast and operator char* will not be called. Here you are trying to cast an object into char*. Since there is no automatic conversion compiler looks for operator char* provided by you. If you had not provided it, you'll get a compiler error saying that Buffer can not be converted into char*

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