this function run a series of numbers.
if n=3
out put is:
1,1,1
1,1,2
1,1,3
1,2,1
1,2,2
1,2,3
1,3,1
1,3,2
1,3,3
Please help me understand how the increasing process in this function works. I cann't see how the array's elemt increase their values.
Code:void print_num(int n) { int times; times=pow(n,2); while (times--) { for(i=0;i<n-1;i++) { printf ("%d,", arr[i]); } printf ("%d\n",arr[i]); if (++arr[n-1]>n) { arr[n-1]=1; for(j=i-1;j>=0;j++) if (++arr[j]>n) arr[j]=1; else break; } } }
Last edited by ronenk; 07-21-2004 at 01:21 AM.
if (++arr[n-1]>n)
++ is the prefix operator for addition
before arr[n-1]>n is evaluated ++arr[n-1] is executed and it increases arr[n-1] with 1.
Take a look at this link
http://crasseux.com/books/ctutorial/...---and---.html
This I know:
...and still don't get it.++ is the prefix operator for addition
From what I see in debugger I can tell the increment process happens in this loop:
[code]
for(i=0;i<n-1;i++)
{
printf ("%d,", arr[i]);
}
printf ("%d\n",arr[i]);
/[code]
on the other hand when put this in comment:
if (++arr[n-1]>n)
increment process doesn’t happen at all with output of:
1,1,1
1,1,1
1,1,1
1,1,1
1,1,1
1,1,1
1,1,1
1,1,1
1,1,1
please try to clarify it.
I'm not sure what you don't understand but..
if (++arr[n-1]>n)
++ has higher precedance then > so it get executed first.
You can see it like this
arr[n-1] = arr[n-1] + 1;
if(arr[n-1] >n )
