在C ++ 11和更高版本中，std :: string :: operator []做边界检​​查吗？ [英] In C++11 and beyond does std::string::operator[] do bounds checking?

问题描述

我已经看到很多次， std :: string :: operator [] 不做任何边界检查。即使是在C ++中，string :: at和string :: operator []？，在2013年问的答案说， operator [] 不做任何边界检查。

我的问题是，如果我看看标准（在这种情况下

 
 code> const_reference operator []（size_type pos）const; 
 reference operator []（size_type pos）; 
  

1. 需要： pos <= size（）。


2. 返回： *（begin if pos< size（）。否则，返回对 charT 的值为 charT（）的对象的引用，其中修改对象导致未定义行为。


3. 投掷没有任何东西。


4. >
   

这让我相信 operator [] 必须做一些边界检查以确定是否需要返回字符串的元素或默认 charT 。这个假设是否正确， operator [] 现在需要做边界检查吗？

解决方案

这说明：

• 前提条件是 [] 的参数是= n 或<
  。


• 假设前提条件满足：
  
  
  
  
  • 那么你会得到你要求的字符。
  
  
  • 否则（即如果是< $ c> charT（）（即空字符）。
  
  

但是没有定义违反前提条件的规则，并且可以通过实际存储来隐式满足对<= em> n 的检查（但没有明确强制要求）

$ 不需要执行任何边界检查…而常见的则不会。

I have seen many times that std::string::operator[] does not do any bounds checking. Even In C++, what is the difference between string::at and string::operator[]?, asked in 2013, the answers say that operator[] does not do any bounds checking.

My issue with this is if I look at the standard (in this case draft N3797) in [string.access] we have

const_reference operator[](size_type pos) const;
reference operator[](size_type pos);

1. Requires: pos <= size().
2. Returns: *(begin() + pos) if pos < size(). Otherwise, returns a reference to an object of type charT with value charT(), where modifying the object leads to undefined behavior.
3. Throws: Nothing.
4. Complexity: constant time.

This leads me to believe that operator[] has to do some sort of bounds checking to determine if it needs to return a element of the string or a default charT. Is this assumption correct and operator[] is now required to do bounds checking?

解决方案

The wording is slightly confusing, but if you study it in detail you'll find that it's actually very precise.

It says this:

• The precondition is that the argument to [] is either = n or it's < n.
• Assuming that precondition is satisfied:
  • If it's < n then you get the character you asked for.
  • "Otherwise" (i.e. if it's n) then you get charT() (i.e. the null character).

But no rule is defined for when you break the precondition, and the check for = n can be satisfied implicitly (but isn't explicitly mandated to be) by actually storing a charT() at position n.

So implementations don't need to perform any bounds checking… and the common ones won't.

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