Finding a substring, using pointers - c

Question

As the title mention, I want to check if a substring is found or not into another string.
#include <stdio.h>
#include <stdlib.h>
int isIncluded(char *text, char* pattern);
int main()
{
char text[30];
char pattern[30]; int result;
printf(" Please introduce your text \n");
scanf("%s", &text);
printf(" Please introduce the pattern you are looking for \n");
scanf("%s", &pattern);
result = isIncluded( &text, &pattern);
if ( result == 1)
{
printf(" Your pattern has been found in your text \n " ) ;
}
if ( result == 0)
{
printf(" no substring found \n " ) ;
}
}
int isIncluded(char *text, char* pattern)
{
int ct = 0;
int numberofcharacters = 0;
while ( *pattern != '\0')
{
pattern++;
numberofcharacters++;
}
while ( *text != '\0' && pattern != '\0')
{
if ( *pattern == *text)
{
pattern++;
ct++;
text++;
}
else
{
text++;
}
}
if ( ct == numberofcharacters )
{
return(1);
}
else
{
return(0);
}
}
The idea is to compare the first character of the text variable with the pattern variable, lets take an example:
Suppose we have "TEXT" in text variable and "EX" in pattern:
I start to compare T with E, in this case, no match.
I point at E and compare again, there is a match.
Because of the match, I point at X in pattern and do the same in text, and I do another test.
2nd match,therefore, the number of characters in the pattern variable will be the same as the ct variable, which counts only when there is a match.
Therefore the return should be equal to 1.
The code returns always zero. I dont understand why ?

Answer 1

#include <stdio.h>
#include <stdlib.h>
int isIncluded(char *text, char* pattern);
int main()
{
char text[30];
char pattern[30]; int result;
printf(" Please introduce your text \n");
scanf("%s", text);
printf(" Please introduce the pattern you are looking for \n");
scanf("%s", pattern);
result = isIncluded( text, pattern);
if ( result == 1)
{
printf(" Your pattern has been found in your text \n " ) ;
}
if ( result == 0)
{
printf(" no substring found \n " ) ;
}
}
int isIncluded(char *text, char* pattern)
{
char *tempPattern = pattern;
int ct = 0;
int numberofcharacters = 0;
while ( *tempPattern != '\0')
{
tempPattern++;
numberofcharacters++;
}
while ( *text != '\0' && pattern != '\0')
{
if ( *pattern == *text)
{
pattern++;
ct++;
text++;
}
else
{
text++;
}
}
if ( ct == numberofcharacters )
{
return(1);
}
else
{
return(0);
}
}
It was the first loop that was problematic, because, as mentionned in the comment section, it completely changed the variable pattern to nothing, therefore there was nothing to compare to. This code works fine.

Answer 2

assuming this is a homework problem or you are learning consider this revision for starters:
int isIncluded ( char *text, char *pattern ); /* this is ok */
/*
can also write it this way
int isIncluded ( char text[], char pattern[] );
*/
int main ( void )
{
char text[30];
char pattern[30];
int result;
printf(" Please introduce your text \n");
scanf("%s", text); /* don't use &text here */
printf(" Please introduce the pattern you are looking for \n");
scanf("%s", pattern); /* don't use &pattern here */
/* don't pass a pointer to a pointer, the strings text and pattern are already pointers, you passing &text would mean isIncluded( char **text ) */
result = isIncluded( text, pattern );
if ( result == 1 )
{
printf(" Your pattern has been found in your text \n " ) ;
}
else if ( result == 0 )
{
printf(" no substring found \n " ) ;
}
else
{
/* don't overlook possibilities of missing simple stuff by not making use of else with if statements */
printf(" Error: value of result is %d\n", result );
}
}
I haven't looked into why isIncluded is always returning zero, but passing a pointer to a pointer to it was not helping.
If you are nice maybe I'll write partial code to get u started,
but what you are wanting to do is already completely accomplished by strstr() provided by # include <string.h>
also, defining char text[30] is the same a defining a pointer to type char called text but in addition it reserves space in memory to hold [n] characters which you said to be 30. It may be helpful in this case to understand the difference between defining and declaring in addition to what is actually happening when doing text[3] versus *(text+3) both of which are valid, unambiguous, and do the same thing.

Answer 3

In your isIncluded function first for calculating the length of pattern you change pattern and then was not the first one, you can calculate length with some other way like strlen but if you want to calculate with this way you should create a new variable as temp and change temp variable.
Another problem in this function in condition of end of pattern in second while loop should use *pattern != '\0' instead of pattern != '\0'
and when you want to call a function whit char* argument and you have array of character you should send array and &text type is char**
and in end here what you want:
#include <stdio.h>
#include <stdlib.h>
int isIncluded(char *text, char* pattern);
int main()
{
char text[30];
char pattern[30]; int result;
printf(" Please introduce your text \n");
scanf("%s", &text);
printf(" Please introduce the pattern you are looking for \n");
scanf("%s", &pattern);
result = isIncluded(text, pattern);
if (result == 1)
{
printf(" Your pattern has been found in your text \n ");
}
if (result == 0)
{
printf(" no substring found \n ");
}
}
int isIncluded(char *text, char* pattern)
{
int ct = 0;
int numberofcharacters = 0;
char *temp = pattern; //<- define new variable
while (*temp != '\0')
{
temp++;
numberofcharacters++;
}
while (*text != '\0' && *pattern != '\0')
{
if (*pattern == *text)
{
pattern++;
ct++;
text++;
}
else
{
text++;
}
}
if (ct == numberofcharacters)
{
return(1);
}
else
{
return(0);
}
}