C ++：char **到const char **转换 [英] C++: char** to const char** conversion

问题描述

在C ++中，为什么不能将 char ** 作为参数传递给接受 const char ** ，从 char * 到 const char * 的转换是可能的， p>



In C++, why is it not possible to pass a char** as an argument to a function that accepts const char** , when a conversion from char* to const char* is possible, as shown below

void f1(const char** a)
{

}

void f2(const char* b)
{

}

int main(int argc, char const *argv[])
{
   char* c;

   f1(&c); // doesn't work
   f2(c); //works

   return 0;
}

编译器输出为

test.cpp: In function 'int main(int, const char**)':
test.cpp:15:10: error: invalid conversion from 'char**' to 'const char**' [-fpermissive]
test.cpp:1:6: error:   initializing argument 1 of 'void f1(const char**)' [-fpermissive]

推荐答案

您需要在指针的两个取消引用层次上保护内容。使用 const char ** ，您可以实际修改第一个引用的内容。

You need to protect contents on both levels of dereference of the pointer. With const char** you could actually modify contents on the 1st dereference.

char *tmp = "foo"; //Deprecated but it's ok for the example
void f1(const char** a)
{
  a[0] = tmp;     //this is legal
  a[0][1] = 'x';  //this is not
}

这很可能不是想要的。它应该看起来像这样：

And this is most probably not intended. It should look like this:

char *tmp = "foo"; //Deprecated but it's ok for the example
void f1(char const* const* a)
{
  a[0] = tmp;    // this is not legal any more
  a[0][1] = 'x'; // this still upsets compiler
}

编译器不允许隐式转换为这样的部分保护的指针类型。允许此类转换可能会产生讨厌的后果，如 c ++ faq 在@zmb的评论中指出。此回答还引用了如何允许违反对象的 constness （如果允许），使用 char 示例。

The compiler does not allow implicit conversion to such "partially" protected pointer types. Allowing such conversion could have nasty consequences as discussed in c++faq pointed out in comment by @zmb. This answer also cites how could you violate constness of an object if this was allowed, using char examples.

然而，可以隐式转换为第二个代码示例中所示的完全以下代码编译。

One can however implicitly convert to a "fully" protected pointer as shown in the 2nd code example so below code compiles.

void f1(char const* const* a){}
void f2(const char* b){}
int main(int argc, char const *argv[])
{
   char* c;
   f1(&c); // works now too!
   f2(c);  // works
   return 0;
}

实际上有很多关于这件事的问题和答案。例如：

Actually there is a bunch of questions and answers on this matter lying around. E.g:

1. 从char **无效转换为const char **

1. invalid conversion from ‘char**’ to ‘const char**’
2. Why am I getting an error converting a ‘float**’ to ‘const float**’?

编辑：我第一个例子错了一点。感谢指出！

I got the 1st example wrong by a bit. Thank for pointing it out!

这篇关于C ++：char **到const char **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IT屋！