### Simple if statement question?

Hi,

I am new to C++ and are teaching myself from a book and have come across this small and simple program(Below) and I understand 90% of it but just need someone to clarify the if statement.

Before, you read the code the program is: a program that searches an array of ten integers for duplicates values. Have the program display each duplicate found.

The code that I'm a little stuck on is: if(nums[i] == nums[j]) and the j for loop now I try to work it out for myself but after I go through the program myself once I get lost. I know this program is very simple but sometimes the book I am teaching myself from doesn't explain things very well.

Here is the whole program:

 12345678910111213 #include using namespace std; int main() { int nums[] = {1, 1, 2, 3, 4, 4, 4, 5, 6, 6 }; for(int i=0; i < 10; i++) for(int j=i+1; j<10; j++) if(nums[i] == nums[j]) cout << "Duplicate: " << nums[i] << "\n"; return 0; }

Thanks,

Luke.
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Okay, let's focus on this:

 123456 int n[] = { 1, 2, 2, 3 }; for(int i = 0; i < 3; i++) for(int j = i + 1; j < 4; j++) if(n[i] == n[j]) cout << n[i] << endl;

For clarity's sake, I have reduced the size of the array.

Let's picture this array as follows:

http://oi42.tinypic.com/29di7t2.jpg

Let i be red and j be blue. Then, this is the effect the loops achieve (highlighted squares denote a found duplicate):

http://oi40.tinypic.com/2qlev5i.jpg

Sorry for the links, but this forum apparently doesn't implement image tags.

Just for fun, notice that if n is the size of the array, then this algorithm performs a total of n * (n - 1) / 2 comparisons, thus rendering it an O(n2) algorithm.

There are better alternatives. For example, if loop through the array and use a hash table to both store its elements and check for duplicates, then we'd end up with an O(n) algorithm.
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Yes I understand arrays. Its just the for loop for the int value 'j': for(int j=i+1; j<10; j++) and the if: if(nums[i] == nums[j]) .

I understand the rest of the program.

Sorry, I've been told the book I have isn't very good for beginner's even though it is a beginner's guide.

Thanks you for your time. :)
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The first parameter of the while loop is the initial value, so for the code you quoted, j starts at whatever i is plus 1. It is still just an integer. nums[j]==nums[i] doesn't mean j must be equal to i, just that those 2 elements have the same value.
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Erhm... I'm sure manudude has the right idea, but I'll explain it in my own words.

There are three things going on in a for loop's statement:
 12345 for( int j = I + 1; //Create a variable j and initialize it with I + 1. j will act as a counter. j < 10; //Keep running iterations (keep looping) while j is less than 10 j++ //After each iteration (loop), increment j by 1 )

Then, the if statement:
 1234567 /* The == operation checks if the two values are equal to each other. Here it tests if the integer at index I is equal to the integer at index j. If they are equal, print a statement. */ if(num[I] == nums[j]) cout << "Duplicate: " << nums[I] << "\n";

Overall, this code iterates through the array, and then for each value in the array, iterate again through the array--starting one position forward--to find any matches.
Arrhh, see my book doesn't explain it like that. In fact its nothing like the way you explain it. Yeah I couldn't se how the for loop for the j worked first time round.

You all really helped me out, cheers!

Thanks for describing this program in a better way.
I have edited my first reply above.
Thank you
 Josue Molina
.

I understand it now, cheers.

Luke.
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