The function pal
int pal(char p[],int i, int j)
{
if (i > j)
return 1;
if (p[i] != p[j])
{
return 0;
}
pal(p, i++, j--);
}
has undefined behaviour because it returns nothing in case when not i > j and not p[i] !+ p[j].
You have to write
int pal(char p[],int i, int j)
{
if (i > j)
return 1;
if (p[i] != p[j])
{
return 0;
}
return pal(p, ++i, --j);
}
Also pay attention to that you have to use the pre-increment and pre-decrement operators.
return pal(p, ++i, --j);
Otherwise you are passing to a next call of the function pal
the same values of i
and j
.
Also the first parameter of the function should have the qualifier const
.
The function can be defined much simpler with using only two parameters.
Here is your program with the updated function definition and its calls.
#include <stdio.h>
#include <string.h>
int pal( const char *s, size_t n )
{
return n < 2 ? 1 : s[0] == s[n-1] && pal( s + 1, n - 2 );
}
void palTest( void )
{
char p1[] = "hello";
char p2[] = "elle";
int x;
x = pal( p1, strlen( p1 ));
if (x == 0)
printf("p1 is not a palendrom\n");
else
printf("p1 is a palendrom\n");
x = pal( p2, strlen( p2 ) );
if (x == 0)
printf("p2 is not a palendrom\n");
else
printf("p2 is a palendrom\n");
}
int main(void)
{
palTest();
return 0;
}
Instead of the conditional operator in the return statement of the function you could use a logical expression like
int pal( const char *s, size_t n )
{
return ( n < 2 ) || ( s[0] == s[n-1] && pal( s + 1, n - 2 ) );
}
Bear in mind that according to the C Standard the function main without parameters shall be declared like
int main( void ).