Thread: help!!! code needs fixing

Originally Posted by Salem

> for ( int i = 1; i <= num; i++ )
In old C++, the scope of i was the rest of the block.
In new C++, it's just the for loop.

What?

> if ( ( i == sum ) && ( i != i ) )
This is outside the loop, so strictly speaking, i is out of scope.
The error is telling you that you're ASSUMING old-style C++ for the scope of i in the for loop.

ok what do I do

What doed i != i do anyway?
Isn't that trivially false?
Doesn't that make the whole if statement trivially false?

i != i is so I don't get the same number twice. and I go 1 + 7 and 3 + 5

Originally Posted by Salem

> i != i is so I don't get the same number twice. and I go 1 + 7 and 3 + 5
It doesn't have a memory of all it's previous values.
If you want that, you need to store them yourself, and then check each one.

In old C++, the declaration of i looks something like this

Code:

int i;
for ( i = 0 ; i < N ; i++ ) {
}

// i still exists here

In new C++, the declaration of i looks something like this

Code:

{
  int i;
  for ( i = 0 ; i < N ; i++ ) {
  }
}

// i DOES NOT exist here

Note the extra braces, hiding the scope of i from later code.

Solutions
- declare i outside the for loop
- declare another variable which can carry the value of i you want out of the loop.

This doesn't answer my question at all.

It answers your first question, namely "What?"

You are not a very appreciative person. He gave you a solution to one of your problems, you should be able to use logic to solve to rest.

>> I get the sum how do I get the added pairs to show up?
You have to keep track of the numbers that you try for the target sum. When you reach the target sum, you would output all the numbers you kept track of.

>> This doesn't answer my question at all.
Salem is offering you important advice though. i != i is trivially false, like he said, *and* you use i in places beside the for loop... so you might want to take a look at his post again.

Is this supposed to work with other numbers than eight?

Originally Posted by citizen

>> I get the sum how do I get the added pairs to show up?
You have to keep track of the numbers that you try for the target sum. When you reach the target sum, you would output all the numbers you kept track of.

That's what I've been tring to do this whole time extracting digits, but I don't know how to.

>> This doesn't answer my question at all.
Salem is offering you important advice though. i != i is trivially false, like he said, *and* you use i in places beside the for loop... so you might want to take a look at his post again.

I think I need to explain my myself better lol I just made a new code has no error.

s this supposed to work with other numbers than eight?

Yes it works for all numbers, BUT it doesn't extract the digits. It gives me the sum of the odd numbers, BUT I can't extract the digits which add to the number.