cout vs printf - 执行顺序 [英] cout vs printf -- order of execution

问题描述

可能重复：



cout<

1. 这：

     int k = 3; 
    printf（％d％d％d，k ++，k，++ k）; 
     

   输出为 4 4 4 被推入堆栈中：

    ％d％d％d 
    
    4  - 用于k ++ 
    4 --for k 
    4 --for ++ k 
     

   右




2. 这：

     int k = 
    cout<< k ++<< k<< ++ k; 
     

   实际上是重复的函数调用，因此它等效于：

    （（（cout< k ++）<< k）< 
     

   所以，我想首先 k ++ 那么 k 然后 ++ k 必须始终按此顺序执行，对吗？我相信函数调用是一个序列点，但是输出在不同的实现上有所不同。为什么会这样呢？

解决方案

点之间，在printf语句中。没有序列点，编译器可以自由地对内存位置 k 按顺序写入。

可能想知道什么是地狱点是一个seqeunce点，为什么它是相关的？基本上，序列点是代码中的一个点，其中所讨论的存储器位置，在这种情况下 k 已被修改最多一次。此处有更完整的说明： http：// www.parashift.com/c++-faq-lite/misc-technical-issues.html#faq-39.16

从FAQ中可以看出， $ ， c>

cout 这是不同的，因为有3个函数调用 operator>> 。函数调用引入了序列点，因此对内存位置 k 的修改具有定义的顺序。然而（这是我错过了一点，但Cubbi指出），因为C / C ++没有定义函数参数的评价顺序，这些参数，即使它们是函数，可以按照编译器定义的任何顺序进行评估。所以在表达式中：

f（h（），g（））

是否首先评估h（）或g（）是否未定义： http://www2.research.att.com/~bs/bs_faq2.html#evaluation-order 。所以这就是为什么即使在cout的情况下，你从不同的编译器得到不同的结果，原因是因为 cout < k ++<< k<< ++ k 转换为 cout.operator<<（k ++）。operator<<（k）.operator（++ k）实际上是这样的表达式： f（h（g（cout，k ++），k），++ k），并且每个函数参数以未指定的顺序计算。 / p>

Possible Duplicate:
Compilers and argument order of evaluation in C++
cout << order of call to functions it prints?

1. This:

   int k=3;
   printf("%d %d %d",k++,k,++k);
   

   Gives output as 4 4 4 because they are pushed into the stack as:

   %d%d%d
   
   4  -- for k++    
   4  --for k    
   4  --for ++k
   

   Right?

2. This:

   int k = 3;
   cout << k++ << k << ++k;
   

   Is actually repeated function calls, so it's equivalent to:

   ( ( (cout << k++) << k) << ++k);
   

   So, I suppose first of all k++ then k and then ++k must always be executed in this order, right? I believe a function call is a sequence point, but the outputs vary on different implementations. Why is this so?

解决方案

This is undefined because there is no sequence point between the , in the printf statement. Without a sequence point the compiler is free to order writes to the memory location k as it wills.

Now you may be wondering 'what the hell is a seqeunce point' and why is it relevant? Basically a sequence point is a point in the code where the memory location in question, in this case k has been modified at most once. There is a fuller description here: http://www.parashift.com/c++-faq-lite/misc-technical-issues.html#faq-39.16

As you can see from the FAQ, the , in the printf does not introduce a sequence point.

In the case of cout this is different because there are 3 function calls to operator >>. A function call introduces a sequence point therefore the modifications to memory location k have a defined order. However (and this was a point I missed but Cubbi pointed out) because C/C++ doesn't define the order of evaluation of function arguments those arguments, even if they are functions, can be evaluated in any order the compiler defines. So in the expression:

f(h(), g())

Whether h() or g() is evaluated first is undefined: http://www2.research.att.com/~bs/bs_faq2.html#evaluation-order. So this is the reason why even in the case of cout you are getting different results from different compilers, basically because the cout << k++ << k << ++k translates to cout.operator<<(k++).operator<<(k).operator(++k) which is effectively an expression like this: f(h(g(cout, k++), k), ++k) and each function argument is evaluated in an unspecified order.

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