为什么我不能在 C 中将“char**"转换为“const char* const*"? [英] Why can't I convert 'char**' to a 'const char* const*' in C?
问题描述
以下代码片段(正确地)在 C 中给出警告，在 C++ 中给出错误(分别使用 gcc 和 g++，用 3.4.5 和 4.2.1 版测试；MSVC 似乎并不关心):
The following code snippet (correctly) gives a warning in C and an error in C++ (using gcc & g++ respectively, tested with versions 3.4.5 and 4.2.1; MSVC does not seem to care):
char **a;
const char** b = a;
我能理解并接受这一点.
这个问题的 C++ 解决方案是将 b 更改为 const char * const *，这不允许重新分配指针并防止您规避 const 正确性(C++ 常见问题解答).
I can understand and accept this.
The C++ solution to this problem is to change b to be a const char * const *, which disallows reassignment of the pointers and prevents you from circumventing const-correctness (C++ FAQ).
char **a;
const char* const* b = a;
但是，在纯 C 中，更正后的版本(使用 const char * const *)仍然给出警告，我不明白为什么.有没有办法在不使用演员表的情况下解决这个问题?
However, in pure C, the corrected version (using const char * const *) still gives a warning, and I don't understand why. Is there a way to get around this without using a cast?
澄清:
1) 为什么这会在 C 中产生警告?它应该是完全常量安全的，并且 C++ 编译器似乎可以识别它.
2)在说(并让编译器强制执行)我不会修改它指向的字符时，接受这个 char** 作为参数的正确方法是什么?例如，如果我想写一个函数:
To clarify:
1) Why does this generate a warning in C? It should be entirely const-safe, and the C++ compiler seems to recognize it as such.
2) What is the correct way to go about accepting this char** as a parameter while saying (and having the compiler enforce) that I will not be modifying the characters it points to?
For example, if I wanted to write a function:
void f(const char* const* in) {
// Only reads the data from in, does not write to it
}
我想在字符**上调用它，参数的正确类型是什么?
And I wanted to invoke it on a char**, what would be the correct type for the parameter?
推荐答案
几年前我也遇到过同样的问题，这让我很恼火.
I had this same problem a few years ago and it irked me to no end.
C 中的规则表述得更简单(即它们没有列出像将 char** 转换为 const char*const* 这样的例外情况).结果，这是不允许的.在 C++ 标准中，他们包含了更多规则来允许这样的情况.
The rules in C are more simply stated (i.e. they don't list exceptions like converting char** to const char*const*). Consequenlty, it's just not allowed. With the C++ standard, they included more rules to allow cases like this.
归根结底只是C标准的问题.我希望下一个标准(或技术报告)能够解决这个问题.
In the end, it's just a problem in the C standard. I hope the next standard (or technical report) will address this.
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