在C ++中初始化结构体 [英] Initializing structs in C++

问题描述

作为此问题的补充，这里发生了什么：

As an addendum to this question, what is going on here:

#include <string>
using namespace std;

struct A {
    string s;
};

int main() {
    A a = {0};
}

显然，您不能将std :: string设置为零。有人可以提供一个解释（支持参考C ++标准，请）关于实际应该发生在这里？然后解释例子）：

Obviously, you can't set a std::string to zero. Can someone provide an explanation (backed with references to the C++ Standard, please) about what is actually supposed to happen here? And then explain for example):

int main() {
    A a = {42};
}

这两个是否都很好定义？

Are either of these well-defined?

对我来说一个令人尴尬的问题 - 我总是给我的structs构造函数，所以这个问题从来没有出现过。

Once again an embarrassing question for me - I always give my structs constructors, so the issue has never arisen before.

推荐答案

您的结构是一个聚合，所以聚合初始化的普通规则对它起作用。该过程在8.5.1中描述。基本上整个8.5.1是专门为它，所以我没有看到理由在这里复制整个东西。一般的想法是几乎一样的C，只是适应C ++：你从右边的初始化，你从左边取一个成员，你用该初始化器初始化成员。根据8.5 / 12，这将是一个复制初始化。

Your struct is an aggregate, so the ordinary rules for aggregate initialization work for it. The process is described in 8.5.1. Basically the whole 8.5.1 is dedicated to it, so I don't see the reason to copy the whole thing here. The general idea is virtually the same it was in C, just adapted to C++: you take an initializer from the right, you take a member from the left and you initialize the member with that initializer. According to 8.5/12, this shall be a copy-initialization.

当你做

A a = { 0 };

你基本上是以 0 ，即 as 它在语义上等同于

you are basically copy-initializing a.s with 0, i.e. for a.s it is semantically equivalent to

string s = 0;

上述编译是因为 std :: string 可以从 const char * 指针转换。 （这是未定义的行为，因为空指针在这种情况下不是有效的参数。）

The above compiles because std::string is convertible from a const char * pointer. (And it is undefined behavior, since null pointer is not a valid argument in this case.)

您的 42 版本将不会编译为同样的原因

Your 42 version will not compile for the very same reason the

string s = 42;

无法编译。 42 不是空指针常数， std :: string 没有从 int 类型。

will not compile. 42 is not a null pointer constant, and std::string has no means for conversion from int type.

PS >在C ++中不是递归的（例如，与POD的定义相反）。 std :: string 不是一个聚合，但它不会改变你的 A 的任何东西。 A 仍为汇总。

P.S. Just in case: note that the definition of aggregate in C++ is not recursive (as opposed to the definition of POD, for example). std::string is not an aggregate, but it doesn't change anything for your A. A is still an aggregate.

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