CODEWITHSUNDEEP
c
Aung
Aung

Reputation: 263

Finding substring in string without using library function

Below is the code template and under /* write your code here */ is my own code. The template should be correct but there is sth wrong with my code.

My algorithm is to iterate through str until finding the null character. Then compare each character, if they are the same then iterate through both str and sub, otherwise set continue to iterate through str and reset to the first character of substr.

#include <stdio.h>
int findSubstring(char *str, char *substring);
int main()
{
    char str[40], substr[40];
    printf("Enter the string: ");
    gets(str);
    printf("Enter the substring: ");
    gets(substr);
    printf("findSubstring(): %d\n", findSubstring(str, substr));
    return 0;
}
int findSubstring(char *str, char *substr)
{
    /* write your code here */
    int i = 0, j = 0;
    while ((str[j] != '\0')||(substr[i] != '\0')) {
        if (substr[i] != str[j]) {
            j++;
            i = 0;
        }
        else {
            i++;
            j++;
        }
    }
    if (substr[i] == '\0')
        return 1;
    else
        return -1;
}

Upvotes: 3

Views: 46919

Answers (3)

Rajeev
Rajeev

Reputation: 167

/*--------------------------One more simple example-----------------------------

Find the words from a set of words containing a given substring?

Input: Set of Words: [blackcat, blackdog, blackrat, whitetiger, blueelephant],

Substring: black

Output:[blackcat, blackdog, blackrat]

-----------------------------------------------------------------------------------------*/

#include <iostream>
#include <cstring>
int substring(char* sub,char* string);
int main()
{
    const char* Names[] { "blackcat", "blackdog", "blackrat", "whitetiger", "blueelephant" };
    char substr[]{ "black" };
    int found{ -1 };
    for (auto strings: Names)
    {
        found = substring(substr, const_cast<char*>(strings));
        if (found != -1) {
            std::cout << strings  << " ";
        }
    }
    std::cout << std::endl;
    return 0;
}

int substring(char* sub, char* string)
{
    int i{};
    int j{};
    while ((string[i] != '\0') && (sub[j] != '\0'))
    {
        if (string[i] != sub[j]) {
            j++;
            i = 0;
        }
        else {
            i++;
            j++;
        }
    }
    if (sub[j] == '\0' && i != 0) {
        return 1;
    }
    else {
        return -1;
    }
}

Upvotes: 0

MikeCAT
MikeCAT

Reputation: 75062

  • Do not use gets(), which has unavoidable risk of buffer overrun.
  • The condition of the loop is wrong. The loop should exited if one of *(str + j) or *(substr + i) is a (terminating) null character.

Fixed code:

#include <stdio.h>
int findSubstring(char *str, char *substring);
void safer_gets(char *str, size_t max);
int main(void)
{
    char str[40], substr[40];
    printf("Enter the string: ");
    safer_gets(str, sizeof(str));
    printf("Enter the substring: ");
    safer_gets(substr, sizeof(str));
    printf("findSubstring(): %d\n", findSubstring(str, substr));
    return 0;
}
int findSubstring(char *str, char *substr)
{
    int i = 0, j = 0;
    while ((*(str + j) != '\0')&&(*(substr + i) != '\0')) {
        if (*(substr + i) != *(str + j)) {
            j++;
            i = 0;
        }
        else {
            i++;
            j++;
        }
    }
    if (*(substr + i) == '\0')
        return 1;
    else
        return -1;
}
void safer_gets(char *str, size_t max)
{
    int i;
    fgets(str, max, stdin);
    for (i = 0; *(str + i) != '\0'; i++) {
        if (*(str + i) == '\n') {
            *(str + i) = '\0';
            break;
        }
    }
}

Upvotes: 7

Mahesh Asalkar
Mahesh Asalkar

Reputation: 1

#include<stdio.h>
#include<conio.h>
void main()
{
    char s[100],sub[50];
    int i,j,c=0;
    clrscr();
    printf("enter string and substring\n");
    gets(s);
    printf("\n");
    gets(sub);
    printf("\n");
    i=0;
    j=0;
    while(s[i]!='\0')
    {
        if(s[i]!=sub[j])
            i++;
        else if(s[i]==sub[j])
        {
            while(sub[j]!='\0')
            {
                if(s[i]==sub[j])
                {
                    i++;
                    j++;
                    c++;
                }
                else
                {
                    c=0;
                    break;
                }
            }
        }
    }
     if(c!=0)
     printf("\nsubstring  is present \n ");
     else
     printf("\nsubstring  is absent \n ");
    getch();
}

Upvotes: -1

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