基于范围的在c ++ 11中 [英] range-based for in c++11

问题描述

如果我们有一个 set< int> S ;我们可以说：

  for（auto i：S）
 cout< i<< endl; 
  

但我们可以强制 i 迭代器，我的意思是写一个等价的代码：

  for（auto i = S.begin（）; i！ S.end（）; i ++）
 cout< （i！= s.begin（））？ ：< *一世; 
  

或者我们可以做一些我们可以理解 i 在集合（或向量）？

另一个问题是我们怎么可以不这样做的所有元素<$ c $

或者当我们有一个 S ，但前半部分或全部。 > vector< int> V ，并且要打印其第一个 n 值，我们应该怎么办？我知道我们可以创建一个新的向量，但需要时间将一个向量复制到一个新的向量。

解决方案

查看标准的内容：

语句的范围为
for（for-range-declaration：expression）statement
相当于

  {
 auto&& __range =（expression）; 
 for（auto __begin = begin-expr，__end = end-expr; __begin！= __end; ++ __ begin）{
 for-range-declaration = * __ begin; 
语句
} 
} 
  

其中__range，__begin和__end是只为exposition定义的变量

换句话说，它已经从 begin 到 end ，并且已经解除引用迭代器了，你永远不会看到。

in c++ 11 if we have a set<int> S; we could say:

for (auto i: S)
    cout << i << endl;

but can we force i to be a iterator, I mean write a code that is equivalent to:

for (auto i = S.begin(); i != S.end(); i++)
    cout << (i != s.begin()) ? " " : "" << *i;

or could we do something that we can understand the index of i in the set(or vector)?

and another question is how could we say that don't do this for all elements in S but for first half of them or all of them except the first one.

or when we have a vector<int> V, and want to print its first n values what should we do? I know we can create a new vector but it takes time to copy a vector to a new vector.

解决方案

No, unluckily. See what the standard says:

The range-based for statement for ( for-range-declaration : expression ) statement is equivalent to

{
    auto && __range = ( expression );
    for ( auto __begin = begin-expr, __end = end-expr; __begin != __end; ++__begin ) {
        for-range-declaration = *__begin;
        statement
    }
}

where __range, __begin, and __end are variables defined for exposition only

In other words, it already iterates from begin to end and already dereferences the iterator, which you never get to see.

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