Solved

BUG in VC++ ? :()

Posted on 1998-11-10
6
178 Views
Last Modified: 2010-04-02
well, here goes the C++ question

  Why does the following code produce 211 as the answer

  int i= 1;
  cout << i << i++ << i << endl;

0
Comment
Question by:umzilber
  • 3
  • 3
6 Comments
 
LVL 32

Accepted Solution

by:
jhance earned 200 total points
ID: 1177467
The order of evaluation of the arguments to cout is not defined by the C++ language standard.  In other words, you cannot be assured that just because you have placed a variable in the left most position in the statement, that it will get evaluated in any particular order.
0
 
LVL 32

Expert Comment

by:jhance
ID: 1177468
BTW, it appears in your example above that the order of evaluation is from right to left.  This makes the right most i = 1, the middle one is also = 1, it is then incremented and then the left most i is now = 2.
0
 
LVL 32

Expert Comment

by:jhance
ID: 1177469
Just another thought.  This is much like the C statement:

printf("%d%d%d\n", i, i++, i);

What will this statement print out?  It might be 211, it might not be, it's not defined.
0
Free Tool: Path Explorer

An intuitive utility to help find the CSS path to UI elements on a webpage. These paths are used frequently in a variety of front-end development and QA automation tasks.

One of a set of tools we're offering as a way of saying thank you for being a part of the community.

 
LVL 22

Expert Comment

by:nietod
ID: 1177470
>>order of evaluation of the arguments to cout is not defined by the C++ language standard
This has nothing to do with the cout.  it has to do with the order of evaluation of operands in an expression and it IS defined by the standard.  This will work the same on all C++ compilers (assuming they don't have a bug).  This will work with the >> operator in all cases.
0
 
LVL 22

Expert Comment

by:nietod
ID: 1177471
I take that back.

The order in which subexpressions are evaluated is implimentation defined (except with "," ,"&&", and "||").  You are guarranteed that the << operators are called in left to right order, but the values that are passed to them might not be.
0
 
LVL 22

Expert Comment

by:nietod
ID: 1177472
To make that clearer.  in

cout << i << i++

You are guaranteed that the order of evaluation of the << operators is

(cout <<  i) << i++

but you are not guarantted the order in which the i and i++ subexpressions are calculated.  The only guarantee that you get is that the second call to operator << will have the value of i before it  was ++ed there is no guarantee that the ++ is done after the call to the 2nd operator <<, what's mroe there is no guarantee that is done after the call to the 1st!
0

Featured Post

Free Tool: Site Down Detector

Helpful to verify reports of your own downtime, or to double check a downed website you are trying to access.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Introduction This article is the first in a series of articles about the C/C++ Visual Studio Express debugger.  It provides a quick start guide in using the debugger. Part 2 focuses on additional topics in breakpoints.  Lastly, Part 3 focuses on th…
This article shows you how to optimize memory allocations in C++ using placement new. Applicable especially to usecases dealing with creation of large number of objects. A brief on problem: Lets take example problem for simplicity: - I have a G…
The viewer will learn how to pass data into a function in C++. This is one step further in using functions. Instead of only printing text onto the console, the function will be able to perform calculations with argumentents given by the user.
The viewer will learn additional member functions of the vector class. Specifically, the capacity and swap member functions will be introduced.

821 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question