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incrementing variables in C++

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# 8  
Old 07-27-2009
In short: it sets the variable i to 5, increments it twice while multiplying, outputs the result of the multiplication, and then outputs the value of i.

If you need more information, we need to be sure that it's not homework, or post in this forum (because of rule 6, to which you agreed when registering)
# 9  
Old 07-29-2009
OK here is what happens:
  1. It assigns 5 to i,
  2. Then, i++ increases i to 6 before multiplication,
  3. Now, i with its value being 6, is multiplied by ++i,
  4. ++i means use the value of i, then, increase it, so it will use 6 here, too.
  5. After the multiplication is done, i is incremented to 7, which is the value echoed by cout.

I hope it is clear now.
From my point of view, this is a drill or an exercise to explain precedence.

faizlo
# 10  
Old 07-29-2009
Lightbulb
Hello Milhan,
The post of faizlo will explain this as best someone can, but the result of this will be:
Code:
ubuntu@ubuntu-laptop:~$ ./test
36
7
ubuntu@ubuntu-laptop:~$

Thanks,
Nathan Paulino Campos
# 11  
Old 07-29-2009
The outputs are 36 and 7 respectively.

Associativity is right to left in this statement. Hence
Code:
cout << i++ * ++i << endl;

is actually the same as
Code:
cout << ++i * i++ << endl;

# 12  
Old 08-03-2009
The result of this operation is undefined.
g++-4.2 gives the following error message:
warning: operation on 'i' may be undefined
The reason for this is explained at
Sequence point - Wikipedia, the free encyclopedia
# 13  
Old 08-04-2009
Quote:
Originally Posted by mrtiller
The result of this operation is undefined.
g++-4.2 gives the following error message:
warning: operation on 'i' may be undefined
The reason for this is explained at
Sequence point - Wikipedia, the free encyclopedia
Interesting. g++ 4.4.0 does not give such a warning.
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