c++ - 为什么不允许从字符数组进行 std::string 初始化？

Question

在 C++ 中，您可以从 char * 和 const char * 初始化一个 std::string 对象，这隐含地假设字符串将在指针后找到的第一个 NUL 字符处结束。

然而，在 C++ 中，字符串文字是数组，即使字符串文字包含嵌入的 NUL，也可以使用模板构造函数来获取正确的大小。例如，参见以下玩具实现:

#include <stdio.h>
#include <string.h>
#include <vector>
#include <string>

struct String {
    std::vector<char> data;
    int size() const { return data.size(); }

    template<typename T> String(const T s);

    // Hack: the array will also possibly contain an ending NUL
    // we don't want...
    template<int N> String(const char (&s)[N])
        : data(s, s+N-(N>0 && s[N-1]=='\0')) {}

    // The non-const array removed as probably a lot of code
    // builds strings into char arrays and the convert them
    // implicitly to string objects.
    //template<int N> String(char (&s)[N]) : data(s, s+N) {}
};

// (one tricky part is that you cannot just declare a constructor
// accepting a `const char *` because that would win over the template
// constructor... here I made that constructor a template too but I'm
// no template programming guru and may be there are better ways).
template<> String::String(const char *s) : data(s, s+strlen(s)) {}

int main(int argc, const char *argv[]) {
    String s1 = "Hello\0world\n";
    printf("Length s1 -> %i\n", s1.size());
    const char *s2 = "Hello\0world\n";
    printf("Length s2 -> %i\n", String(s2).size());
    std::string s3 = "Hello\0world\n";
    printf("std::string size = %i\n", int(s3.size()));
    return 0;
}

是否有任何特定的技术原因导致标准未考虑此方法，而是嵌入 NUL 的字符串文字在用于初始化 std 时最终被截断: :string 对象?

Answer 1

C++14 为字符串文字引入了一个后缀，使它们成为 std::string 对象，因此主要用例不再相关。

#include <iostream>
#include <string>
using namespace std;
using namespace std::literals;

int main() {
    string foo = "Hello\0world\n";
    string bar = "Hello\0world\n"s;
    cout << foo.size() << " " << bar.size() << endl; // 5 12
    cout << foo << endl; // Hello
    cout << bar << endl; // Helloworld
    return 0;
}