为什么在 returns 字符串不起作用的函数上调用 std::string.c_str()？

Question

我有以下代码：

std::string getString() {
    std::string str("hello");
    return str;
}

int main() {
    const char* cStr = getString().c_str();
    std::cout << cStr << std::endl; // this prints garbage
}

我认为 getString() 会 return copy of str (getString() returns 按值）；因此，str 的副本将在 main() 中保留 "alive"，直到 main() return 秒。这将使 cStr 指向一个有效的内存位置：char[] 或 char*（或其他）str return 的副本的底层14=] 保留在 main().

但是，显然不是这样的，因为程序输出的是垃圾。那么，问题是，str 何时销毁，为什么？

Answer 1

getString() would return a copy of str (getString() returns by value);

没错。

thus, the copy of str would stay "alive" in main() until main() returns.

不，返回的副本是临时的 std::string，它将在创建它的语句结束时销毁，即在 std::cout << cStr << std::endl; 之前。然后 cStr 变得悬空，对它的取消引用导致 UB，一切皆有可能。

您可以将返回的临时变量复制到命名变量，或将其绑定到 const 左值引用或右值引用（临时变量的生命周期将延长，直到引用超出范围） .如：

std::string s1 = getString();    // s1 will be copy initialized from the temporary
const char* cStr1 = s1.c_str();
std::cout << cStr1 << std::endl; // safe

const std::string& s2 = getString(); // lifetime of temporary will be extended when bound to a const lvalue-reference
const char* cStr2 = s2.c_str();
std::cout << cStr2 << std::endl; // safe

std::string&& s3 = getString();  // similar with above
const char* cStr3 = s3.c_str();
std::cout << cStr3 << std::endl; // safe

或者在临时对象被销毁之前使用指针。例如

std::cout << getString().c_str() << std::endl;  // temporary gets destroyed after the full expression

这是来自 [The.C++.Programming.Language.Special.Edition] 10.4.10 临时对象 [class.temp]] 的解释：

Unless bound to a reference or used to initialize a named object, a temporary object is destroyed at the end of the full expression in which it was created. A full expression is an expression that is not a subexpression of some other expression.

The standard string class has a member function c_str() that returns a C-style, zero-terminated array of characters (§3.5.1, §20.4.1). Also, the operator + is defined to mean string concatenation. These are very useful facilities for strings . However, in combination they can cause obscure problems. For example:
void f(string& s1, string& s2, string& s3)
{

    const char* cs = (s1 + s2).c_str();
    cout << cs ;
    if (strlen(cs=(s2+s3).c_str())<8 && cs[0]==´a´) {
        // cs used here
    }

}
Probably, your first reaction is "but don’t do that," and I agree. However, such code does get written, so it is worth knowing how it is interpreted.

A temporary object of class string is created to hold s1 + s2 . Next, a pointer to a C-style string is extracted from that object. Then – at the end of the expression – the temporary object is deleted. Now, where was the C-style string allocated? Probably as part of the temporary object holding s1 + s2 , and that storage is not guaranteed to exist after that temporary is destroyed. Consequently, cs points to deallocated storage. The output operation cout << cs might work as expected, but that would be sheer luck. A compiler can detect and warn against many variants of this problem.

Answer 2

这里的问题是您要返回一个临时变量并超过您正在执行的那个临时变量 c_str 函数。

"c_str() function Returns a pointer to an array that contains a null-terminated sequence of characters (i.e., a C-string) representing the current value of the string object( [http://www.cplusplus.com/reference/string/string/c_str/][1]).

在这种情况下，您的指针指向现在不存在的内存位置。

std::string getString() {
        std::string str("hello");
        return str; // Will create Temporary object as it's return by value}

    int main() {
         const char* cStr = getString().c_str(); // Temporary object is destroyed
        std::cout << cStr << std::endl; // this prints garbage }

解决方案是将您的临时对象正确复制到内存位置（通过创建本地副本），然后对该对象使用 c_str。

Answer 3

正如其他人所提到的，您在临时指针已被删除后使用它 - 这是 免费使用后堆的经典示例。

我可以添加到其他人的答案中的是，您可以使用 gcc's or clang's 地址清理器轻松检测到此类用法。

示例：

#include <string>
#include <iostream>

std::string get()
{
  return "hello";
}

int main()
{
  const char* c = get().c_str();
  std::cout << c << std::endl;
}

消毒剂输出：

=================================================================
==2951==ERROR: AddressSanitizer: heap-use-after-free on address 0x60300000eff8 at pc 0x7f78e27869bb bp 0x7fffc483e670 sp 0x7fffc483de20
READ of size 6 at 0x60300000eff8 thread T0
    #0 0x7f78e27869ba in strlen (/usr/lib64/libasan.so.2+0x6d9ba)
    #1 0x39b4892ba0 in std::basic_ostream<char, std::char_traits<char> >& std::operator<< <std::char_traits<char> >(std::basic_ostream<char, std::char_traits<char> >&, char const*) (/usr/lib64/libstdc++.so.6+0x39b4892ba0)
    #2 0x400dd8 in main /tmp/tmep_string/main.cpp:12
    #3 0x39aa41ed5c in __libc_start_main (/lib64/libc.so.6+0x39aa41ed5c)
    #4 0x400c48  (/tmp/tmep_string/a.out+0x400c48)

0x60300000eff8 is located 24 bytes inside of 30-byte region [0x60300000efe0,0x60300000effe)
freed by thread T0 here:
    #0 0x7f78e27ae6ea in operator delete(void*) (/usr/lib64/libasan.so.2+0x956ea)
    #1 0x39b489d4c8 in std::basic_string<char, std::char_traits<char>, std::allocator<char> >::~basic_string() (/usr/lib64/libstdc++.so.6+0x39b489d4c8)
    #2 0x39aa41ed5c in __libc_start_main (/lib64/libc.so.6+0x39aa41ed5c)

previously allocated by thread T0 here:
    #0 0x7f78e27ae1aa in operator new(unsigned long) (/usr/lib64/libasan.so.2+0x951aa)
    #1 0x39b489c3c8 in std::string::_Rep::_S_create(unsigned long, unsigned long, std::allocator<char> const&) (/usr/lib64/libstdc++.so.6+0x39b489c3c8)
    #2 0x400c1f  (/tmp/tmep_string/a.out+0x400c1f)

SUMMARY: AddressSanitizer: heap-use-after-free ??:0 strlen
Shadow bytes around the buggy address:
  0x0c067fff9da0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c067fff9db0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c067fff9dc0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c067fff9dd0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c067fff9de0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
=>0x0c067fff9df0: fa fa fa fa fa fa fa fa fa fa fa fa fd fd fd[fd]
  0x0c067fff9e00: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c067fff9e10: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c067fff9e20: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c067fff9e30: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c067fff9e40: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
Shadow byte legend (one shadow byte represents 8 application bytes):
  Addressable:           00
  Partially addressable: 01 02 03 04 05 06 07 
  Heap left redzone:       fa
  Heap right redzone:      fb
  Freed heap region:       fd
  Stack left redzone:      f1
  Stack mid redzone:       f2
  Stack right redzone:     f3
  Stack partial redzone:   f4
  Stack after return:      f5
  Stack use after scope:   f8
  Global redzone:          f9
  Global init order:       f6
  Poisoned by user:        f7
  Container overflow:      fc
  Array cookie:            ac
  Intra object redzone:    bb
  ASan internal:           fe
==2951==ABORTING

为什么在 returns 字符串不起作用的函数上调用 std::string.c_str()？

Why does calling std::string.c_str() on a function that returns a string not work?

c++

string

temporary

object-lifetime