Question the increment and decrement in this code.

[papagym177]

Parts of this program are missing. The last few lines are confusing, since the variable 'a' gets incremented then decremented. But there are no loops. I understand that the value of 'a' is passed to 'c' before 'a' is changed in both cases.

But where, and when, do the changes take place? Is the decrement ever processed? Is there a better way to write these lines?

 main(){ int a = 21;int b = 10;int c ;    
   c = a++;    
cout << "Line 6 - Value of c is :" << c << endl ;
   c = a--;
   cout << "Line 7 - Value of c is  :" << c << endl ;
   return 0;}

[MutantJohn]

Keep in mind, these operators are really just doing

a = a + 1;

// or...

a = a - 1;

Out of curiousity, what's your output and is it as you expect it?

[Matticus]

I understand that the value of 'a' is passed to 'c' before 'a' is changed in both cases.

It would be more clear to say the value of 'a' is assigned to 'c' before 'a' is updated.

But where, and when, do the changes take place?

Think of it as a two-step process:

(1) The value of 'a' is assigned to 'c'
(2) 'a' is incremented (or decremented)

Is the decrement ever processed?

Have you tried printing out the value of 'a' to see what it is?

[Elysia]

int c = a--;
<==>
int c = a;
--a;

[papagym177]

Thank you all for the replies. I'm aware of what all of you have written above. I guess I was not clear when I asked my question. It is hard to put this into words.

The code is written so the increment happens after c = a++; and the decrement happens after c = a--;. Therefore Line 3 prints out 21 and Line 5 prints out 22. (exactly as I expected).

I hope this makes sense. Where exactly do these changes (increment & decrement) take effect? Does the code do this automatically after each cout lines (number 3 & 5)? Or does the compiler make the change?

[King Mir]

Thank you all for the replies. I'm aware of what all of you have written above. I guess I was not clear when I asked my question. It is hard to put this into words.

The code is written so the increment happens after c = a++; and the decrement happens after c = a--;. Therefore Line 3 prints out 21 and Line 5 prints out 22. (exactly as I expected).

I hope this makes sense. Where exactly do these changes (increment & decrement) take effect? Does the code do this automatically after each cout lines (number 3 & 5)? Or does the compiler make the change?

They happen after the value is read from, after the previous line of code, and before the next line of code. Nothing else is guaranteed. Because of this, it is, in the general case, undefined behavior to read and write from the same memory location within the same statement.

C++ defines virtual model of how a program runs. But that model isn't what actually happens in hardware. The compiler ensures that generated machine code obeys the constraints of the C++ programing model. In doing so the compiler is free to reorder code arbitrarily, so long as the externally visible behavior of the program is the same. For instance, in this example, a compiler could make a and c not exists in memory at all, instead using the constants 20 and 21 directly.