C ++：std :: cout的顺序 [英] C++: The order of std::cout

问题描述

这里，我写了一个简单的队列类模板：

Here, I wrote a simple Queue Class Template:

#include "stdafx.h"
#include "iostream"
#include "conio.h"

template <class T> class Queue {
private:
    struct Node {
        T value;
        Node *next;
    };
    Node *first, *last;
public:

    Queue () { //Initialization
        first = 0;
        last = 0;
    };

    void push(T value) { //Put new value into Queue
        Node *p = new Node;
        p->value = value;

        if (!first) //If Queue is not empty
            first = p;
        else
            last->next = p;

        last = p; //Put the value to the last of the Queue
    };

    T get() { //Get the first value of the Queue
        return (first) ? first->value : NULL;
    };

    T pop() { //Take the first value out of Queue
        T result = first->value;
        Node *temp = first;
        first = first->next;
        delete temp;
        return result;
    };
};

int _tmain(int argc, _TCHAR* argv[])
{
    Queue<int> bag;
    bag.push(1); //Bag has value 1
    bag.push(2); //Bag has values: 1, 2

    std::cout << bag.get() << '\n' << bag.pop() << '\n' << bag.pop();
    _getch();
    return 0;
}

有一个问题 - 输出为：

There is a problem - the output is:

0
2
1

/*
Correct output should be:
1
1
2
*/

std :: cout 行，我发现程序调用 bag.pop（）在最右边，那么另一个 bag.pop（），最后是 bag.get（）。

When I debug the std::cout line, I found out that program call bag.pop() at the right most first, then the other bag.pop(), at last the bag.get(). Is that the right order?

 解决方案 

推荐答案

/获取队列的第一个值
return（！first）？ first-> value：NULL;
};

T get() { //Get the first value of the Queue return (!first) ? first->value : NULL; };

这是向后的。删除！。你是说如果第一是不非空，（即如果第一

This is backwards. Drop the !. You are saying "if first is not non-null, (ie if first is null), use it.

也就是说，函数参数的求值顺序是未指定的（编译器可以评估函数的参数这是 get（）， pop（）和 pop（）可以按任何顺序调用它们作为单独的语句调用：

That said, the order of evaluation of arguments to a function is unspecified (the compiler can evaluate the arguments in any order it feels like, as long as they are all done before the function itself starts). That is get(), pop() and pop() can be called in any order. Call them as separate statements:

int a = bag.get();
int b = bag.pop();
int c = bag.pop();
std::cout << a << b << c;

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