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BUG in VC++ ? :()

Posted on 1998-11-10
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well, here goes the C++ question

  Why does the following code produce 211 as the answer

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

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Question by:umzilber
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jhance earned 800 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.
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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.
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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.
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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.
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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.
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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!
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