在成员变量中无法从initializer-string中推导出数组大小的原因是什么？ [英] What is the reason for not being able to deduce array size from initializer-string in member variable?

问题描述

请考虑代码：

struct Foo
{
    const char str[] = "test";
};

int main()
{
    Foo foo;
}

无法使用g ++和clang ++编译， >

It fails to compile with both g++ and clang++, spitting out essentially

错误：无法从类初始化器中推导出数组约束

我知道这是标准可能说的，原因？因为我们有一个字符串字面量，似乎编译器应该能够推导出大小没有任何问题，类似于当你简单地声明一个out-of-class const C类空结束字符串。

I understand that this is what the standard probably says, but is there any particular good reason why? Since we have a string literal it seems that the compiler should be able to deduce the size without any problem, similarly to the case when you simply declare an out-of-class const C-like null terminated string.

推荐答案

原因是你总是有可能重写一个类初始化列表构造函数。所以我猜这最后可能很混乱。

The reason is that you always have the possibility to override an in class initializer list in the constructor. So I guess that in the end it could be very confusing.

struct Foo
{
   Foo() {} // str = "test\0";

   // Implementing this is easier if I can clearly see how big `str` is, 
   Foo() : str({'a','b', 'c', 'd'}) {} // str = "abcd0"
   const char str[] = "test";
};

请注意，将 const char 替换为 static constexpr 工作完美，可能这是你想要的

Notice that replacing const char with static constexpr works perfectly, and probably it is what you want anyway

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