C program for pattern matching, pattern matching in c
Pattern matching in c: C programming code to check if a given string is
present in another string, For example the string "programming" is present
in "c programming". If the string is present then it's location (i.e. at
which position it is present) is printed. We create a function match which
receives two character pointers and return the position if matching occurs
otherwise returns -1. We are implementing naive string search algorithm in
our c program
C programming code
#include <stdio.h>
#include <string.h>
int match(char [], char []);
int main() {
char a[100], b[100];
int position;
printf("Enter some text\n");
gets(a);
printf("Enter a string to find\n");
gets(b);
position = match(a, b);
if(position != -1) {
printf("Found at location %d\n", position + 1);
}
else {
printf("Not found.\n");
}
return 0;
}
int match(char text[], char pattern[]) {
int c, d, e, text_length, pattern_length, position = -1;
text_length = strlen(text);
pattern_length = strlen(pattern);
if (pattern_length > text_length) {
return -1;
}
for (c = 0; c <= text_length - pattern_length; c++) {
position = e = c;
for (d = 0; d < pattern_length; d++) {
if (pattern[d] == text[e]) {
e++;
}
else {
break;
}
}
if (d == pattern_length) {
return position;
}
}
return -1;
}
Download Pattern matching program.
Output of program:
C program for pattern matching using pointers
#include<stdio.h>
int match(char*, char*);
main()
{
char a[100], b[100];
int position;
printf("Enter some text\n");
gets(a);
printf("Enter a string to find\n");
gets(b);
position = match(a, b);
if(position!=-1)
printf("Found at location %d\n", position+1);
else
printf("Not found.\n");
return 0;
}
int match(char *a, char *b)
{
int c;
int position = 0;
char *x, *y;
x = a;
y = b;
while(*a)
{
while(*x==*y)
{
x++;
y++;
if(*x=='\0'||*y=='\0')
break;
}
if(*y=='\0')
break;
a++;
position++;
x = a;
y = b;
}
if(*a)
return position;
else
return -1;
}
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