这段代码实现了搜索工作的逻辑(其中一种方式)，而不使用任何现成的函数:

public int findSubString(char[] original, char[] searchString)
{
    int returnCode = 0; //0-not found, -1 -error in imput, 1-found
    int counter = 0;
    int ctr = 0;
    if (original.Length < 1 || (original.Length)<searchString.Length || searchString.Length<1)
    {
        returnCode = -1;
    }

    while (ctr <= (original.Length - searchString.Length) && searchString.Length > 0)
    {
        if ((original[ctr]) == searchString[0])
        {
            counter = 0;
            for (int count = ctr; count < (ctr + searchString.Length); count++)
            {
                if (original[count] == searchString[counter])
                {
                    counter++;
                }
                else
                {
                    counter = 0;
                    break;
                }
            }
            if (counter == (searchString.Length))
            {
                returnCode = 1;
            }
        }
        ctr++;
    }
    return returnCode;
}

下面是如何报告找到的子字符串的第一个字符的位置:

在上面的代码中替换这一行:

printf("%s",substring,"\n");

:

printf("substring %s was found at position %d \n", substring,((int) (substring - mainstring)));

#include <stdio.h>
#include <string.h>

int findSubstr(char *inpText, char *pattern);
int main()
{
    printf("Hello, World!\n");
    char *Text = "This is my sample program";
    char *pattern = "sample";
    int pos = findSubstr(Text, pattern);
    if (pos > -1) {
        printf("Found the substring at position %d \n", pos);
    }
    else
        printf("No match found \n");

    return 0;
}

int findSubstr(char *inpText, char *pattern) {
    int inplen = strlen(inpText);
    while (inpText != NULL) {

        char *remTxt = inpText;
        char *remPat = pattern;

        if (strlen(remTxt) < strlen(remPat)) {
            /* printf ("length issue remTxt %s \nremPath %s \n", remTxt, remPat); */
            return -1;
        }

        while (*remTxt++ == *remPat++) {
            printf("remTxt %s \nremPath %s \n", remTxt, remPat);
            if (*remPat == '\0') {
                printf ("match found \n");
                return inplen - strlen(inpText+1);
            }
            if (remTxt == NULL) {
                return -1;
            }
        }
        remPat = pattern;

        inpText++;
    }
}

if (strstr(sent, word) != NULL) {
    /* ... */
}

注意，如果找到单词word, strstr将返回一个指向在sent中单词开头的指针。

使用strstr。

https://cplusplus.com/reference/cstring/strstr

你可以这样写。

char *sent = "this is my sample example";
char *word = "sample";

char *pch = strstr(sent, word);

if(pch)
{
    ...
}

My code to find out if substring is exist in string or not 
// input ( first line -->> string , 2nd lin ->>> no. of queries for substring
following n lines -->> string to check if substring or not..

#include <stdio.h>
int len,len1;
int isSubstring(char *s, char *sub,int i,int j)
{

        int ans =0;
         for(;i<len,j<len1;i++,j++)
        {
                if(s[i] != sub[j])
                {
                    ans =1;
                    break;
                }
        }
        if(j == len1 && ans ==0)
        {
            return 1;
        }
        else if(ans==1)
            return 0;
return 0;
}
int main(){
    char s[100001];
    char sub[100001];
    scanf("%s", &s);// Reading input from STDIN
    int no;
    scanf("%d",&no);
    int i ,j;
    i=0;
    j=0;
    int ans =0;
    len = strlen(s);
    while(no--)
    {
        i=0;
        j=0;
        ans=0;
        scanf("%s",&sub);
        len1=strlen(sub);
        int value;
        for(i=0;i<len;i++)
        {
                if(s[i]==sub[j])
                {
                    value = isSubstring(s,sub,i,j);
                    if(value)
                    {
                        printf("Yes\n");
                        ans = 1;
                        break;
                    }
                }
        }
        if(ans==0)
            printf("No\n");

    }
}