please wait
how to find the output of the following segment?
Oct 12, 2012 at 9:15am
May I know what variable u want to output?
Because u did output for both variables that is in the code
Because u did output for both variables that is in the code
Last edited on Oct 12, 2012 at 9:22am
Oct 12, 2012 at 9:21am| how to find the output of the following segment? |
Run the program, and see what the output is.
Oct 12, 2012 at 9:24amWill result in compile errors. The for loop takes only three portions(or any other suitable word) of statements. Syntax:
for(initialization;condition;increment/decrement);Also, if you wanna specify more than one statement in the "portions", then use the comma(',') operator.
eg:
for (x=1, y=5; x+y<=10; x++,y++)@vichu8888
| So I took out my suggestion |
Yeah no problem!
Aceix.
Last edited on Oct 12, 2012 at 10:42am
Oct 12, 2012 at 9:38am
@Aceix
Is your reply for me?
Because I suggested him the code
But later when I tried in my lappy, I examined the error. So I took out my suggestion.
Sorry for that
Is your reply for me?
Because I suggested him the code
for (x=1; y=5; x+y<=10; x++;y++)But later when I tried in my lappy, I examined the error. So I took out my suggestion.
Sorry for that
Last edited on Oct 12, 2012 at 9:39am
Oct 12, 2012 at 2:51pm
for (x=1, y=5; x+y<=10; x++)
{
cout <<x<< “ “<<y<<endl;
y++;
}
Output:
1 5
2 6
3 7
{
cout <<x<< “ “<<y<<endl;
y++;
}
Output:
1 5
2 6
3 7
Oct 12, 2012 at 3:15pm
It's not that hard to work it out in your head.
Both x and y are incremented by 1 at the end of each iteration of the loop. That means (x+y) goes up by 2 each time. Thus (x+y) is 6, 8 10, meaning three passes through the loop. The values of x and y output each time are straightforward.
Both x and y are incremented by 1 at the end of each iteration of the loop. That means (x+y) goes up by 2 each time. Thus (x+y) is 6, 8 10, meaning three passes through the loop. The values of x and y output each time are straightforward.
Oct 13, 2012 at 5:14am
No replies from the OP - could be a typical trolling question?
If not. let's see some feedback / thanks from the OP.
If not. let's see some feedback / thanks from the OP.
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