如果我需要c++字符串内容的可变原始副本，那么我会这样做:

std::string str = "string";
char* chr = strdup(str.c_str());

后来:

free(chr); 

So why don't I fiddle with std::vector or new[] like anyone else? Because when I need a mutable C-style raw char* string, then because I want to call C code which changes the string and C code deallocates stuff with free() and allocates with malloc() (strdup uses malloc). So if I pass my raw string to some function X written in C it might have a constraint on it's argument that it has to allocated on the heap (for example if the function might want to call realloc on the parameter). But it is highly unlikely that it would expect an argument allocated with (some user-redefined) new[]!

(这个答案只适用于c++ 98。)

请不要使用原始的char*。

std::string str = "string";
std::vector<char> chars(str.c_str(), str.c_str() + str.size() + 1u);
// use &chars[0] as a char*

这可能是对bobobobo的回答的一个更好的评论，但我没有那个代表。它完成了同样的事情，但是使用了更好的实践。

虽然其他答案很有用，但如果你需要显式地将std::string转换为char*而不使用const, const_cast是你的朋友。

std::string str = "string";
char* chr = const_cast<char*>(str.c_str());

注意，这不会为您提供数据的副本;它会给你一个指向字符串的指针。因此，如果修改chr的一个元素，就会修改str。

If you just want a C-style string representing the same content: char const* ca = str.c_str(); If you want a C-style string with new contents, one way (given that you don't know the string size at compile-time) is dynamic allocation: char* ca = new char[str.size()+1]; std::copy(str.begin(), str.end(), ca); ca[str.size()] = '\0'; Don't forget to delete[] it later. If you want a statically-allocated, limited-length array instead: size_t const MAX = 80; // maximum number of chars char ca[MAX] = {}; std::copy(str.begin(), (str.size() >= MAX ? str.begin() + MAX : str.end()), ca);

string不会隐式转换为这些类型，原因很简单，需要这样做通常是一种设计风格。确保你真的需要它。

如果你确实需要一个char*，最好的方法可能是:

vector<char> v(str.begin(), str.end());
char* ca = &v[0]; // pointer to start of vector

orlp的char*答案的安全版本使用unique_ptr:

std::string str = "string";
auto cstr = std::make_unique<char[]>(str.length() + 1);
strcpy(cstr.get(), str.c_str());

你可以使用迭代器。

std::string str = "string";
std::string::iterator p=str.begin();
char* chr = &(*p);

祝你好运。