hello! I am having problems with a palindrome program. I have to input an integer between 1 and 9. and then the output has to be:

for example if I input 4:
output: 1234321

if I input 5:
output: 123454321

if 1
output: 1

hello! I am having problems with a palindrome program. I have to input an integer between 1 and 9. and then the output has to be:

for example if I input 4:
output: 1234321

if I input 5:
output: 123454321

if 1
output: 1

the problem is that I don't know how to do it. I tried but I don't get it. can somebody help me, my mind is going to blow!!!

ANd the problem is?

that I have been trying but I don't get it. my mind is going to explode!!!

Well, what to you have so far? Code wise?

#include <stdio.h>
int main()
{
int n
printf ("Enter digit: ");
scanf ("%d",&n);

if ((n < 0) || (n > 9))
{
printf("number not in range\n");
}
else
printf("n-1...n...n-1")

Ok. Consider using two for loops. One to count up to a number and one to count down. Print the looping variable.

Ok. Consider using two for loops. One to count up to a number and one to count down. Print the looping variable.

something like:
for (count = 1; count <=n; count = count + 1)
for(count=1; count>=n; count = count - 1)
printf(“%d”, count);

?

Sorta. But you don't want to nest the loops. printf() after each for loop.
for ( counting up )
printf();

for ( counting down )
printf()
The second loop will be going from the upper limit to 1, remember. So limits of count=n; count>=1; count-- would make more sense.

Sorta. But you don't want to nest the loops. printf() after each for loop.
for ( counting up )
printf();

for ( counting down )
printf()
The second loop will be going from the upper limit to 1, remember. So limits of count=n; count>=1; count-- would make more sense.

like:
#include <stdio.h>
int main()
{
int n
printf ("Enter digit: ");
scanf ("%d",&n);

if ((n < 0) || (n > 9))
{
printf("number not in range\n");
}
else


for (count = 1; count <=n; count = count + 1);
pintf("%d",count);
for(count=n; count>=1; count = count - 1)
printf(“%d”, count);

return 0;
}

?

Does that work?
I think the limits of the loops need work, and that you need to consider the input=1 case, but it looks fine. ANd yer missing a ; in places. And you don't need the else. And you have a ; where you don't need one (end of the first for line).

Edited 7 Years Ago by twomers: n/a

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