用字符串文字初始化非const参数 [英] Initializing non-const parameter with string literal

问题描述

所以我有这个代码：

class ConstTest {
public:
    explicit ConstTest(char* name) {}
};

int main() {
    ConstTest t("blarghgh");
}

显然编译，即使我认为它不应该。由于C ++中的字符串文字具有类型 const char [] 和 ConstTest 构造函数需要一个const- < c> char * - not const char * 。并且通过C ++隐式生成 const 指向非const常量的指针。

It obviously compiles, even though I thought that it shouldn't. As string literals in C++ have type const char[], and ConstTest constructor requires a const-less char* — not const char*. And casting a const pointer to a non-const one isn't something usually done by C++ implicitly.

，我错了？为什么要编译？我可以在构造函数中合法修改取消引用的指针吗？

So, where I'm wrong? Why it's compiling? Can I legally modify the dereferenced pointer inside the constructor?!

推荐答案

错误？为什么要编译？

So, where I'm wrong? Why it's compiling?

这是编译，因为你的编译器太宽容，而且你的编译器太宽容了，因为在C ++ 03从字符串字面值到 char * 的隐式转换仅被 废弃，无效。

It is compiling because your compiler is too permissive, and your compiler is too permissive because in C++03 the implicit conversion from a string literal to char* was only deprecated, not invalid.

原理是向后兼容旧版C API。按照C ++ 03标准的第4.2 / 2节：

The rationale was backward compatibility with legacy C APIs. Per paragraph 4.2/2 of the C++03 Standard:

字符串文字（2.13.4）可以转换为类型的指针
char 的右值;一个宽字符串文字可以转换为指向 wchar_t 类型的右值。在任何一种情况下，
的结果都是指向数组的第一个元素的指针。只有当有一个
显式的适当的指针目标类型，而不是一般需要从一个左值转换到一个
值时，才会考虑这种转换。 [ 注意：此转化已弃用。请参阅附录D。]

A string literal (2.13.4) that is not a wide string literal can be converted to an rvalue of type "pointer to char"; a wide string literal can be converted to an rvalue of type "pointer to wchar_t". In either case, the result is a pointer to the first element of the array. This conversion is considered only when there is an explicit appropriate pointer target type, and not when there is a general need to convert from an lvalue to an rvalue. [Note: this conversion is deprecated. See Annex D.]

然而，在C ++ 11中，隐式转换是非法的

In C++11, however, the implicit conversion is illegal (the above paragraph has been removed altogether).

我可以在构造函数中合法解引用和修改指针吗？！

Can I legally dereference-and-modify the pointer inside the constructor?!

可以，但不能修改取消引用的对象。这样做是未定义的行为，因为对象的类型是 const -qualified。

You can, but you cannot modify the dereferenced object. Doing so would be undefined behavior, since the type of the object is const-qualified.

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