访问 std::string 中的空终止字符(字符串下标超出范围) [英] Accessing null-termination character in std::string (string subscript out of range)

问题描述

考虑以下非常简单的文本示例:

Consider the following very simple text example:

#include <stdio.h>
#include <string>
int main() {
    std::string x("ugabuga");
    int i=0;
    while (x[i]) {
        ++i;
    }
    printf("%d\n",i); //should print 7
    return 0;
}

我希望程序遍历字符串的所有字符，然后到达中断循环的空终止字符并正确到达程序结束.但是，当我尝试在 Visual Studio 2010 下以调试模式编译它时，我遇到了异常字符串下标超出范围".在发布模式下编译时，该程序通过了，但我基于此行为的更大项目崩溃了 - 可能是因为这个问题.

I would expect the program to iterate over all characters of the string, then reach the null-terminating character breaking the loop and reach the program end correctly. However, when I tried compiling it in debug mode under Visual Studio 2010 I am reaching an exception "string subscript out of range". When compiling in release mode this program passes, but my bigger project depending on this behaviour crashes - perhaps because of this issue.

但是，当我在 www.cplusplus.com，明确处理结束字符串:

However, when I checked the specification of std::string::operator[] at www.cplusplus.com, the end-character string is handled explicitly:

如果 pos 等于字符串长度，则函数返回对空字符('\0')的引用.

If pos is equal to the string length, the function returns a reference to a null character ('\0').

我想问一下:

• 我对std::string 规范的解释是否正确?还是我遗漏了什么?
• 如果问题出在 VS 的实现方面，我该如何轻松解决这个问题——希望每次使用 operator[] 时都不会调用 length()?例如使用 c_str()[i] 会安全吗?
• 如果问题出在 VS 的实现方面 - 您知道它是否在 VS 2012 中得到修复，或者未来是否会得到修复?

• Is my interpretation of specification of std::string correct? Or am I missing something?
• If the problem lies on VS side of implementation, how can I easily fix this - hopefully without calling length() each time I use the operator[]? e.g. will using c_str()[i] will be safe?
• If the problem lies on VS side of implementation - do you know if it is fixed in VS 2012 or pehaps will be fixed in the future?

推荐答案

这是 C++03 和 C++11 之间发生变化的事情之一.

This is one of the things that changed between C++03 and C++11.

这似乎是 C++03 中未定义的行为:

It seems it is undefined behaviour in C++03:

21.3.4 basic_string 元素访问[lib.string.access]

const_reference operator[](size_type pos) const;

reference operator[](size_type pos);

1 返回:如果 pos ，返回data()[pos].否则，如果 pos == size()，const 版本返回 charT().否则，行为未定义.

1 Returns: If pos < size(), returns data()[pos]. Otherwise, if pos == size(), the const version returns charT(). Otherwise, the behavior is undefined.

虽然在 C++11 中是可以的.

while in C++11, it is OK.

21.4.5 basic_string 元素访问[string.access]

const_reference operator[](size_type pos) const;

reference operator[](size_type pos);

1 要求:pos <= size().

1 Requires: pos <= size().

2 返回:*(begin() + pos) 如果 pos ，否则引用类型为 T 且值为 charT(); 的对象，引用的值不得修改.

2 Returns: *(begin() + pos) if pos < size(), otherwise a reference to an object of type T with value charT(); the referenced value shall not be modified.

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