std::cout statements evaluation order

Question

What is wrong with the pop() function why doesn't it work correctly?

class stack{
    int *p, *Cursor;
    int size ;
public:
    stack(int sz) {Cursor = p = new int[size=sz+1];} //consider the stack empty when its size is 1
    ~stack() {delete[] p;} //Cursor and P will be destroyed when the program finishes
    void push(int x) {Cursor+=1; *Cursor=x; size++;}
    int pop() {if(Cursor == p) return -1; int temp = *Cursor; Cursor--; size--; return (temp);}
    bool isEmpty(){return(Cursor == p);}
    bool isFull(){return(Cursor == p+size);}
};

here is my testing:

stack A(3);
    std::cout<<"Empty: "<<A.isEmpty()<<std::endl;
    std::cout<<"Full: "<<A.isFull()<<std::endl;
    A.push(10);
    A.push(20);
    A.push(30);
    std::cout<<std::endl;
    std::cout<<" 1st pop: "<<A.pop()<<std::endl<<" 2nd pop: " <<A.pop()<<std::endl<<" 3rd pop: " <<A.pop()<<std::endl<<" 4th pop: " <<A.pop()<<std::endl;

the output I get is:

1st pop: -1
2nd pop: 10
3rd pop: 20
4th pop: 30

while I should get sth like:

1st pop: 30
2nd pop: 20
3rd pop: 10
4th pop: -1

The question is where did I go wrong?

Separate your pop printouts (`cout << pop() << endl; cout << pop() << ...`) — Mat, Feb 07 '15 at 22:41

score 5 · Accepted Answer · edited May 23 '17 at 11:50

5

Nothing did go wrong, but the pops get (in your case) evaluated from right to left if you put them all in one std::cout line. In general. the evaluation order is unspecified. For more detail on this see here.

So you correctly get the elements in reverse order of insertion and then -1 and then print them reversed.

edited May 23 '17 at 11:50

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answered Feb 07 '15 at 22:45

Baum mit Augen

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Order of evaluation is unspecified, not right-to-left. – Mat Feb 07 '15 at 22:46
It's obviously right-to-left in OP's case. – zmbq Feb 07 '15 at 22:47
1

@Mat Should be right to left because each `< – Baum mit Augen Feb 07 '15 at 22:47
`f(a,f(b))` - you don't know whether `a` or `f(b)` will be evaluated first. It's not left-to-right. – Mat Feb 07 '15 at 22:49
http://stackoverflow.com/questions/7718508/order-of-evaluation-of-arguments-using-stdcout http://stackoverflow.com/questions/2129230/cout-order-of-call-to-functions-it-prints http://stackoverflow.com/questions/9377592/please-explain-this-ambiguity – Mat Feb 07 '15 at 22:51
@Mat won't `b` (not `f(b)`) be evaluated before `a` ? – Quentin Feb 07 '15 at 23:00

score 1 · Answer 2 · answered Feb 07 '15 at 23:02

In the last line of your code, you make several function calls (hidden, but each << operator is actually a function call). The evaluation of the parameters to these function calls is unspecified. So it is unspecified which "pop() function is called first.

std::cout statements evaluation order

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