如何将std::string转换为char*或const char*?
当前回答
C++17
c++ 17(即将推出的标准)修改了basic_string模板的概要，添加了非const重载data():
图表*数据()noexcept; 返回:一个指针p，对于[0,size()]中的每个i, p + i == &运算符。
CharT const * from std::basic_string<CharT>
std::string const cstr = { "..." };
char const * p = cstr.data(); // or .c_str()
从std::basic_string<图表>
std::string str = { "..." };
char * p = str.data();
C++11
CharT const * from std::basic_string<CharT>
std::string str = { "..." };
str.c_str();
从std::basic_string<图表>
从c++ 11开始，标准说:
The char-like objects in a basic_string object shall be stored contiguously. That is, for any basic_string object s, the identity &*(s.begin() + n) == &*s.begin() + n shall hold for all values of n such that 0 <= n < s.size(). const_reference operator[](size_type pos) const; reference operator[](size_type pos); Returns: *(begin() + pos) if pos < size(), otherwise a reference to an object of type CharT with value CharT(); the referenced value shall not be modified. const charT* c_str() const noexcept;const charT* data() const noexcept; Returns: A pointer p such that p + i == &operator[](i) for each i in [0,size()].
有多种方法可以获得非const字符指针。
1. 使用c++ 11的连续存储
std::string foo{"text"};
auto p = &*foo.begin();
Pro
简单而简短 快速(唯一不涉及复制的方法)
Cons
Final '\0'不会被改变/不一定是非const内存的一部分。
2. 使用std::向量<图>
std::string foo{"text"};
std::vector<char> fcv(foo.data(), foo.data()+foo.size()+1u);
auto p = fcv.data();
Pro
简单的 自动内存处理 动态
Cons
需要字符串复制
3.使用std::array<CharT, N>如果N是编译时间常数(并且足够小)
std::string foo{"text"};
std::array<char, 5u> fca;
std::copy(foo.data(), foo.data()+foo.size()+1u, fca.begin());
Pro
简单的 堆栈内存处理
Cons
静态 需要字符串复制
4. 原始内存分配与自动存储删除
std::string foo{ "text" };
auto p = std::make_unique<char[]>(foo.size()+1u);
std::copy(foo.data(), foo.data() + foo.size() + 1u, &p[0]);
Pro
内存占用小 自动删除 简单的
Cons
需要字符串复制 静态(动态使用需要大量代码) 特征比向量或数组少
5. 使用手动处理的原始内存分配
std::string foo{ "text" };
char * p = nullptr;
try
{
p = new char[foo.size() + 1u];
std::copy(foo.data(), foo.data() + foo.size() + 1u, p);
// handle stuff with p
delete[] p;
}
catch (...)
{
if (p) { delete[] p; }
throw;
}
Pro
最大的“控制”
Con
需要字符串复制 错误的最大责任/易感性 复杂的
其他回答
我正在使用一个API，其中有很多函数获得char*作为输入。
我创建了一个小类来处理这类问题，并且实现了RAII习惯用法。
class DeepString
{
DeepString(const DeepString& other);
DeepString& operator=(const DeepString& other);
char* internal_;
public:
explicit DeepString( const string& toCopy):
internal_(new char[toCopy.size()+1])
{
strcpy(internal_,toCopy.c_str());
}
~DeepString() { delete[] internal_; }
char* str() const { return internal_; }
const char* c_str() const { return internal_; }
};
你可以这样使用它:
void aFunctionAPI(char* input);
// other stuff
aFunctionAPI("Foo"); //this call is not safe. if the function modified the
//literal string the program will crash
std::string myFoo("Foo");
aFunctionAPI(myFoo.c_str()); //this is not compiling
aFunctionAPI(const_cast<char*>(myFoo.c_str())); //this is not safe std::string
//implement reference counting and
//it may change the value of other
//strings as well.
DeepString myDeepFoo(myFoo);
aFunctionAPI(myFoo.str()); //this is fine
我将这个类称为DeepString，因为它正在创建一个现有字符串的深度且唯一的副本(DeepString不可复制)。
从c++ std字符串转换到c风格字符串现在真的很容易。
为此，我们有string::copy函数，它可以轻松地将std字符串转换为C风格字符串。参考
字符串::连续复制函数形参
Char字符串指针 字符串大小，b复制多少个字符 位置，从字符复制开始的位置
另一件重要的事，
此函数不会在操作结束时附加空字符。所以，我们需要手动放置它。
代码考试在下面
// char string
char chText[20];
// c++ string
string text = "I am a Programmer";
// conversion from c++ string to char string
// this function does not append a null character at the end of operation
text.copy(chText, text.size(), 0);
// we need to put it manually
chText[text.size()] = '\0';
// below statement prints "I am a Programmer"
cout << chText << endl;
反之亦然，从C样式字符串转换到c++ std字符串要容易得多
有三种方法可以将C样式字符串转换为c++ std字符串
第一个是使用构造函数，
char chText[20] = "I am a Programmer";
// using constructor
string text(chText);
第二个是使用string::assign方法
// char string
char chText[20] = "I am a Programmer";
// c++ string
string text;
// convertion from char string to c++ string
// using assign function
text.assign(chText);
第三个是赋值操作符(=)，其中字符串类使用操作符重载
// char string
char chText[20] = "I am a Programmer";
// c++ string
// convertion from char string to c++ string using assignment operator overloading
string text = chText;
第三个也可以写成如下-
// char string
char chText[20] = "I am a Programmer";
// c++ string
string text;
// convertion from char string to c++ string
text = chText;
char* result = strcpy((char*)malloc(str.length()+1), str.c_str());
鉴于说…
std::string x = "hello";
从string对象中获取char*或const char*
如何获得一个字符指针，是有效的，而x仍然在范围内，并没有进一步修改
c++ 11简化了事情;下面这些都可以访问相同的内部字符串缓冲区:
const char* p_c_str = x.c_str();
const char* p_data = x.data();
char* p_writable_data = x.data(); // for non-const x from C++17
const char* p_x0 = &x[0];
char* p_x0_rw = &x[0]; // compiles iff x is not const...
以上所有指针都将保存相同的值——缓冲区中第一个字符的地址。即使是空字符串也有一个“缓冲区中的第一个字符”，因为c++ 11保证总是在显式分配的字符串内容之后保留一个额外的NUL/0结束符(例如std::string("this\0that"， 9)将有一个缓冲区保存"this\0that\0")。
给定以上任意一个指针:
char c = p[n]; // valid for n <= x.size()
// i.e. you can safely read the NUL at p[x.size()]
仅适用于非const指针p_writable_data和来自&x[0]:
p_writable_data[n] = c;
p_x0_rw[n] = c; // valid for n <= x.size() - 1
// i.e. don't overwrite the implementation maintained NUL
在字符串的其他地方写入NUL不会改变字符串的大小();string对象允许包含任意数量的NULs——std::string对它们没有特殊处理(c++ 03中相同)。
在c++ 03中，事情要复杂得多(重点区别突出显示):
x.data() returns const char* to the string's internal buffer which wasn't required by the Standard to conclude with a NUL (i.e. might be ['h', 'e', 'l', 'l', 'o'] followed by uninitialised or garbage values, with accidental accesses thereto having undefined behaviour). x.size() characters are safe to read, i.e. x[0] through x[x.size() - 1] for empty strings, you're guaranteed some non-NULL pointer to which 0 can be safely added (hurray!), but you shouldn't dereference that pointer. &x[0] for empty strings this has undefined behaviour (21.3.4) e.g. given f(const char* p, size_t n) { if (n == 0) return; ...whatever... } you mustn't call f(&x[0], x.size()); when x.empty() - just use f(x.data(), ...). otherwise, as per x.data() but: for non-const x this yields a non-const char* pointer; you can overwrite string content x.c_str() returns const char* to an ASCIIZ (NUL-terminated) representation of the value (i.e. ['h', 'e', 'l', 'l', 'o', '\0']). although few if any implementations chose to do so, the C++03 Standard was worded to allow the string implementation the freedom to create a distinct NUL-terminated buffer on the fly, from the potentially non-NUL terminated buffer "exposed" by x.data() and &x[0] x.size() + 1 characters are safe to read. guaranteed safe even for empty strings (['\0']).
访问外部法律索引的后果
无论以何种方式获取指针，都不能访问指针以外的内存，不能访问上述描述中保证存在的字符。尝试这样做会有未定义的行为，即使对于读，也有非常真实的应用程序崩溃和垃圾结果的机会，而且对于写，还会有大量数据、堆栈损坏和/或安全漏洞。
这些指针什么时候失效?
如果调用某个字符串成员函数来修改字符串或保留进一步的容量，则上述任何方法之前返回的任何指针值都将无效。您可以再次使用这些方法来获取另一个指针。(规则与迭代器到字符串的规则相同)。
请参见如何在x离开作用域或在....下面进一步修改后仍使字符指针有效
那么，用哪个更好呢?
从c++ 11开始，使用.c_str()表示ASCIIZ数据，使用.data()表示“二进制”数据(下文将进一步解释)。
在c++ 03中，使用.c_str()，除非确定.data()是足够的，并且优先使用.data()而不是&x[0]，因为它对空字符串....是安全的
.．.尝试充分理解程序，以便在适当的时候使用data()，否则您可能会犯其他错误……
由.c_str()保证的ASCII NUL '\0'字符被许多函数用作标记值，表示相关且可安全访问的数据的结束。这既适用于c++函数，如fstream::fstream(const char* filename，…)，也适用于与C语言共享的函数，如strchr()和printf()。
鉴于c++ 03的.c_str()对返回缓冲区的保证是.data()的超集，您总是可以安全地使用.c_str()，但人们有时不会这样做，因为:
using .data() communicates to other programmers reading the source code that the data is not ASCIIZ (rather, you're using the string to store a block of data (which sometimes isn't even really textual)), or that you're passing it to another function that treats it as a block of "binary" data. This can be a crucial insight in ensuring that other programmers' code changes continue to handle the data properly. C++03 only: there's a slight chance that your string implementation will need to do some extra memory allocation and/or data copying in order to prepare the NUL terminated buffer
作为进一步的提示，如果函数的形参需要(const) char*，但不坚持获取x.s size()，则该函数可能需要一个ascii输入，因此.c_str()是一个很好的选择(函数需要知道文本以某种方式在何处结束，因此如果它不是一个单独的形参，则只能是一个约定，如长度前缀或哨兵或一些固定的预期长度)。
如何得到一个字符指针有效，即使x离开范围或进一步修改
你需要将字符串x的内容复制到x外部的一个新的内存区域。这个外部缓冲区可能在很多地方，比如另一个字符串或字符数组变量，由于在不同的作用域(例如命名空间，全局，静态，堆，共享内存，内存映射文件)，它可能有或没有与x不同的生命周期。
将std::string x中的文本复制到一个独立的字符数组中:
// USING ANOTHER STRING - AUTO MEMORY MANAGEMENT, EXCEPTION SAFE
std::string old_x = x;
// - old_x will not be affected by subsequent modifications to x...
// - you can use `&old_x[0]` to get a writable char* to old_x's textual content
// - you can use resize() to reduce/expand the string
// - resizing isn't possible from within a function passed only the char* address
std::string old_x = x.c_str(); // old_x will terminate early if x embeds NUL
// Copies ASCIIZ data but could be less efficient as it needs to scan memory to
// find the NUL terminator indicating string length before allocating that amount
// of memory to copy into, or more efficient if it ends up allocating/copying a
// lot less content.
// Example, x == "ab\0cd" -> old_x == "ab".
// USING A VECTOR OF CHAR - AUTO, EXCEPTION SAFE, HINTS AT BINARY CONTENT, GUARANTEED CONTIGUOUS EVEN IN C++03
std::vector<char> old_x(x.data(), x.data() + x.size()); // without the NUL
std::vector<char> old_x(x.c_str(), x.c_str() + x.size() + 1); // with the NUL
// USING STACK WHERE MAXIMUM SIZE OF x IS KNOWN TO BE COMPILE-TIME CONSTANT "N"
// (a bit dangerous, as "known" things are sometimes wrong and often become wrong)
char y[N + 1];
strcpy(y, x.c_str());
// USING STACK WHERE UNEXPECTEDLY LONG x IS TRUNCATED (e.g. Hello\0->Hel\0)
char y[N + 1];
strncpy(y, x.c_str(), N); // copy at most N, zero-padding if shorter
y[N] = '\0'; // ensure NUL terminated
// USING THE STACK TO HANDLE x OF UNKNOWN (BUT SANE) LENGTH
char* y = alloca(x.size() + 1);
strcpy(y, x.c_str());
// USING THE STACK TO HANDLE x OF UNKNOWN LENGTH (NON-STANDARD GCC EXTENSION)
char y[x.size() + 1];
strcpy(y, x.c_str());
// USING new/delete HEAP MEMORY, MANUAL DEALLOC, NO INHERENT EXCEPTION SAFETY
char* y = new char[x.size() + 1];
strcpy(y, x.c_str());
// or as a one-liner: char* y = strcpy(new char[x.size() + 1], x.c_str());
// use y...
delete[] y; // make sure no break, return, throw or branching bypasses this
// USING new/delete HEAP MEMORY, SMART POINTER DEALLOCATION, EXCEPTION SAFE
// see boost shared_array usage in Johannes Schaub's answer
// USING malloc/free HEAP MEMORY, MANUAL DEALLOC, NO INHERENT EXCEPTION SAFETY
char* y = strdup(x.c_str());
// use y...
free(y);
想要从字符串生成char*或const char*的其他原因
那么，上面你已经看到了如何获取(const) char*，以及如何独立于原始字符串复制文本，但是你可以用它做什么呢?一些随机的例子……
give "C" code access to the C++ string's text, as in printf("x is '%s'", x.c_str()); copy x's text to a buffer specified by your function's caller (e.g. strncpy(callers_buffer, callers_buffer_size, x.c_str())), or volatile memory used for device I/O (e.g. for (const char* p = x.c_str(); *p; ++p) *p_device = *p;) append x's text to an character array already containing some ASCIIZ text (e.g. strcat(other_buffer, x.c_str())) - be careful not to overrun the buffer (in many situations you may need to use strncat) return a const char* or char* from a function (perhaps for historical reasons - client's using your existing API - or for C compatibility you don't want to return a std::string, but do want to copy your string's data somewhere for the caller) be careful not to return a pointer that may be dereferenced by the caller after a local string variable to which that pointer pointed has left scope some projects with shared objects compiled/linked for different std::string implementations (e.g. STLport and compiler-native) may pass data as ASCIIZ to avoid conflicts
C++17
c++ 17(即将推出的标准)修改了basic_string模板的概要，添加了非const重载data():
图表*数据()noexcept; 返回:一个指针p，对于[0,size()]中的每个i, p + i == &运算符。
CharT const * from std::basic_string<CharT>
std::string const cstr = { "..." };
char const * p = cstr.data(); // or .c_str()
从std::basic_string<图表>
std::string str = { "..." };
char * p = str.data();
C++11
CharT const * from std::basic_string<CharT>
std::string str = { "..." };
str.c_str();
从std::basic_string<图表>
从c++ 11开始，标准说:
The char-like objects in a basic_string object shall be stored contiguously. That is, for any basic_string object s, the identity &*(s.begin() + n) == &*s.begin() + n shall hold for all values of n such that 0 <= n < s.size(). const_reference operator[](size_type pos) const; reference operator[](size_type pos); Returns: *(begin() + pos) if pos < size(), otherwise a reference to an object of type CharT with value CharT(); the referenced value shall not be modified. const charT* c_str() const noexcept;const charT* data() const noexcept; Returns: A pointer p such that p + i == &operator[](i) for each i in [0,size()].
有多种方法可以获得非const字符指针。
1. 使用c++ 11的连续存储
std::string foo{"text"};
auto p = &*foo.begin();
Pro
简单而简短 快速(唯一不涉及复制的方法)
Cons
Final '\0'不会被改变/不一定是非const内存的一部分。
2. 使用std::向量<图>
std::string foo{"text"};
std::vector<char> fcv(foo.data(), foo.data()+foo.size()+1u);
auto p = fcv.data();
Pro
简单的 自动内存处理 动态
Cons
需要字符串复制
3.使用std::array<CharT, N>如果N是编译时间常数(并且足够小)
std::string foo{"text"};
std::array<char, 5u> fca;
std::copy(foo.data(), foo.data()+foo.size()+1u, fca.begin());
Pro
简单的 堆栈内存处理
Cons
静态 需要字符串复制
4. 原始内存分配与自动存储删除
std::string foo{ "text" };
auto p = std::make_unique<char[]>(foo.size()+1u);
std::copy(foo.data(), foo.data() + foo.size() + 1u, &p[0]);
Pro
内存占用小 自动删除 简单的
Cons
需要字符串复制 静态(动态使用需要大量代码) 特征比向量或数组少
5. 使用手动处理的原始内存分配
std::string foo{ "text" };
char * p = nullptr;
try
{
p = new char[foo.size() + 1u];
std::copy(foo.data(), foo.data() + foo.size() + 1u, p);
// handle stuff with p
delete[] p;
}
catch (...)
{
if (p) { delete[] p; }
throw;
}
Pro
最大的“控制”
Con
需要字符串复制 错误的最大责任/易感性 复杂的
