如何将“指针转换为指针类型"常量? [英] How to convert "pointer to pointer type" to const?

问题描述

用下面的代码

void TestF(const double ** testv){;}
void callTest(){
    double** test;
    TestF(test);
}

我明白了:

'TestF' : cannot convert parameter 1 from 'double **' to 'const double **'

我不明白为什么.为什么 test 不能被无声地转换为 const double**?我为什么要明确地这样做?我知道

I cannot understand why. Why test cannot be silently casted to const double**? Why should I do it explicitly? I know that

TestF(const_cast<const double**>(test)) 

使我的代码正确，但我觉得这应该是不必要的.

makes my code correct, but I feel this should be unnecessary.

我缺少一些关于 const 的关键概念吗?

Are there some key concepts about const that I'm missing?

推荐答案

语言允许从 double ** 到 const double *const * 的隐式转换，但不能到const double **.您尝试的转换将隐含违反 const 正确性规则，即使它不是立即显而易见的.

The language allows implicit conversion from double ** to const double *const *, but not to const double **. The conversion you attempt would implicitly violate the rules of const correctness, even though it is not immediately obvious.

[de-facto standard] C++ FAQ 中的示例说明了这个问题

The example in the [de-facto standard] C++ FAQ illustrates the issue

https://isocpp.org/wiki/faq/const-correctness#constptrptr-conversion

基本上，规则是:在某个间接级别添加 const 后，您必须将 const 一直添加到所有间接级别.例如 int ***** 不能隐式转换为 int **const ***，但可以隐式转换为 int **const*const *const *

Basically, the rule is: once you add const at some level of indirection, you have to add const to all levels of indirection all the way to the right. For example, int ***** cannot be implicitly converted to int **const ***, but it can be implicitly converted to int **const *const *const *

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