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Hello.
That's exactly what the method you're invoking returns, the height/width referenced to its parent coordinate system.
If you want to get the offset referenced to the screen (1st level of DOM) you can get it by adding all offset values invoking recursively to 'get_offsetParent' method (n-1) times.
Hope it helps.
Regards
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can any one give me a hint on how to write a code in for loop that prints only even numbers with an input of n?
i.e. 0, 2, 4, 6, .....
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What is n supposed to be?
This sounds a lot like a homework question... so I'll just give you a hint... divide by 2.
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What would YOUR code look like without the "only even numbers" constraint?
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"Show me a community that obeys the Ten Commandments and I'll show you a less crowded prison system." - Anonymous
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David: this code is for print n from 0 to n (n is an input number from the user).
int n;
cout << "Enter a number and press ENTER: ";
cin >> n;
for (int i = 0; i <= n; i = i++)
{
cout << i << " ";
}
cout << "\n";
I am suppose to do something to i++ + 2, or n-- - 2, or something.
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Yolande MR wrote: for (int i = 0; i <= n; i = i++)
This is questionable at best. Try:
for (int i = 0; i <= n; i = i + 1)
Yolande MR wrote: I am suppose to do something to i++ + 2, or n-- - 2, or something.
If each iteration of your loop is incrementing i by 1, then why not try incrementing by...
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"Show me a community that obeys the Ten Commandments and I'll show you a less crowded prison system." - Anonymous
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Incrementing by....well, certain number... works, but only if you start from an already even number. Important to note that.
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Albert Holguin wrote: ...but only if you start from an already even number. Important to note that.
It was already noted...zero. My suggestion reflected that.
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"Show me a community that obeys the Ten Commandments and I'll show you a less crowded prison system." - Anonymous
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Chris:
I got i+=2, works the same!
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That doesn't really give you even numbers if i happens to be an odd number. If you want to do a better job, look at using the modulus operator.
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Why would anyone vote this down? It's a fact... care to explain yourself downvoter?
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5 got you 3! oops. 
Unrequited desire is character building. OriginalGriff
I'm sitting here giving you a standing ovation - Len Goodman
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Finally a reply that covers all issues in one operator. Can't go wrong on the modulo.
My 5.
Cheers, AT
Cogito ergo sum
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Here[^] is a useful reference for you.
Unrequited desire is character building. OriginalGriff
I'm sitting here giving you a standing ovation - Len Goodman
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Divide the user input by 2 and check for remainder. If there is an remainder means it is an odd number, vice versa. 
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No kidding?
Unrequited desire is character building. OriginalGriff
I'm sitting here giving you a standing ovation - Len Goodman
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Probably shouldn't have posted this as a reply to Richard... but you are correct.
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They've written a book on even numbers now? 
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Use the modulus operator % to get the remainder
For instance: if (number % 2 == 0) then its even, and if (number % 2 != 0) then its odd.
Code to print only EVEN numbers..
#include <iostream>
using namespace std;
int main()
{
int num;
cout <<"Printing only EVEN numbers..\n\n ";
for(num = 1; num < 210;num++)
{
if ( num % 2 == 0 )
{
cout << num << " is even\t ";
}
}
cout <<"\n ";
return 0;
}
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Or you could use the simple expression
int odd = num & 1;
Unrequited desire is character building. OriginalGriff
I'm sitting here giving you a standing ovation - Len Goodman
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Is there any way to get Network Adapter Name, Link speed and network utilization like Task manager using native api?
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Check out this page[^] in MSDN for information on the network APIs.
Unrequited desire is character building. OriginalGriff
I'm sitting here giving you a standing ovation - Len Goodman
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Now that's a page which is going to get a bookmark!
Cheers, AT
Cogito ergo sum
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It's not always easy to find, but MSDN has the answer to a lot of questions.
Unrequited desire is character building. OriginalGriff
I'm sitting here giving you a standing ovation - Len Goodman
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