Quote:
problem,it's not counting all the strings
The first action is to add code to track what is doing your code:
...
while(j<=(strlen(a)/2))
{
if(a[j]!=a[strlen(a)-j-1])
{
c=0;
}
j++;
}
if (c==1)
{
printf("%s\n",a);
}
return c;
}
Then see which palindromes are found and which are not.
If the hint of the missing ones is enough, go to correct.
Otherwise, jump to debugger and see why it is missing a palindrome.
There is a tool that allow you to see what your code is doing, its name is
debugger. It is also a great learning tool because it show you reality and you can see which expectation match reality.
When you don't understand what your code is doing or why it does what it does, the answer is
debugger.
Use the debugger to see what your code is doing. Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute.
Debugger - Wikipedia, the free encyclopedia[
^]
Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[
^]
Basic Debugging with Visual Studio 2010 - YouTube[
^]
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't find bugs, it just help you to. When the code don't do what is expected, you are close to a bug.
[update]
By the way, your algorithm is very naive, and very inefficient. The workload grow with the square of length of
s
, it is O(n²).
A little analyze show that any large palindrome is built around a smaller one until size 2 or 3. So if you don't have a small palindrome around a given center, you will not have a larger one around same center.
This means that with 'abcdefghabcdefghabcdefghabcdefghabcdefghabcdefgh' you don't need to check large palindromes because you don't have small ones, this optimization is O(n).