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#include<iostream.h>
void main()
{
    int A=5,B=10;
    for(int I=1;I<=2;I++)
    {
        cout<<"Line1="<<A++<<"&"<<B-2<<endl;
        cout<<"Line2="<<++B<<"&"<<A+B<<endl;
    }
}

The output of this program is 

Line1=5&8
Line2=11&16
Line1=6&9
Line2=12&18

I thought that it will produce 17 and 19 in place of the 16 and 18 in the second and fourth lines of the output. This is because, in the first run of the loop, first the value of A is 5 and the first command prints 5&8 and should increment the value of A by 1, making it 6. In the second command it should print 11&(6+11) which should print 11&17 but the output is not that.

Where is the loophole in my reasoning??

Jabberwocky
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Martund
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  • What is your platform/compiler/IDE/OS etc.? This code should not compile with any C++ compiler less then 15 years old. – Jabberwocky Apr 30 '20 at 12:04
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    Does this answer your question? [Order of evaluation: subexpressions, sequence points and postfix increments in C](https://stackoverflow.com/questions/50608248/order-of-evaluation-subexpressions-sequence-points-and-postfix-increments-in-c) – Jean-Baptiste Yunès Apr 30 '20 at 12:04
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    @Jean-BaptisteYunès Your link applies to C. (Though I'm not aware of whether there are differences between C and C++. I roughly remember that the rules changed a bit with C++17 where some cases are now Implementation Defined Behavior which were Undefined Behavior before.) – Scheff's Cat Apr 30 '20 at 12:07
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    FYI: [cppreference - Order of evaluation](https://en.cppreference.com/w/cpp/language/eval_order) – Scheff's Cat Apr 30 '20 at 12:09
  • Does this answer your question? [Undefined behavior and sequence points](https://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points) – ChrisMM Apr 30 '20 at 12:22
  • @scheff It's just a terrible mistake... But the diagnostic is the right one. – Jean-Baptiste Yunès Apr 30 '20 at 13:06

1 Answers1

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Im not an expert on the subject, but I believe it is because of the order of operations that are performed in the background.

Mainly "<<" is something called an overloaded operator, which basicly means that someone, somewhere wrote what it should do, and how to do it. And if you have a bunch of things written one after the other, like you have:

cout<<"Line2="<<++B<<"&"<<A+B<<endl;

The compiler has to do some fancy tricks to make it work. The way the program runs through such code, is from right to left. So in essence it kinda runs in reverse to the way you would think it does. First it pushes endl, then it does A+B and pushes it, then it pushes &, then it increments B and it also pushes it, finaly it pushes Line2= forming the complete "sentence". These are then taken to the console (or whatever else you might have) to be printed to your screen at once.

As a solution to the issue, try separating cout into 2 lines; something like this:

cout <<"Line2="<<++B<<"&";
cout <<A+B<<endl;

Or ,if allowed to, try swaping ++B and A+B, this should also solve the issue, however your results will also be reversed.

cout<<"Line2="<<A+B<<"&"<<++B<<endl;

tl;dr: A+B happendes before B++, doing them in spearate lines or swaping the positions should solve the problem

Antonio Klasić
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  • Re: "The way the program runs through such code, is from right to left" -- not necessarily. Yes, `A+B` was evaluated before `B++` if that's the result that you see. But that's not required by the language definition. In fact, this code results in undefined behavior (the language definition does not tell you what the program does) because the value of `A` is modified and read in the same statement. Fiddling with the order of the operations won't fix this problem. It might make the code do what you expect, but if you use a different compiler or different settings it can (and will) change. – Pete Becker Apr 30 '20 at 14:12