从 char** 到 const char** 的隐式转换

Implicit conversion from char** to const char**(从 char** 到 const char** 的隐式转换)

问题描述

为什么我的编译器 (GCC) 不会从 char** 隐式转换为 const char**?

Why my compiler(GCC) doesnt implicitly cast from char** to const char**?

以下代码:

#include <iostream>

void print(const char** thing) {
    std::cout << thing[0] << std::endl;
}

int main(int argc, char** argv) {
    print(argv);
}

出现以下错误:

oi.cpp: In function ‘int main(int, char**)’:
oi.cpp:8:12: error: invalid conversion from ‘char**’ to ‘const char**’ [-fpermissive]
oi.cpp:3:6: error:   initializing argument 1 of ‘void print(const char**)’ [-fpermissive]

推荐答案

这样的转换将允许您将 const char* 放入您的 char* 数组中，这将是不安全的.在 print 你可以这样做:

Such a conversion would allow you to put a const char* into your array of char*, which would be unsafe. In print you could do:

thing[0] = "abc";

现在 argv[0] 将指向一个无法修改的字符串文字，而 main 期望它是非常量的 (char*代码>).所以为了类型安全，这种转换是不允许的.

Now argv[0] would point to a string literal that cannot be modified, while main expects it to be non-const (char*). So for type safety this conversion is not allowed.

这篇关于从 char** 到 const char** 的隐式转换的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！