我相信我有最简单的答案。在这个程序中不需要string.h库，也不需要stdbol .h库。简单地使用指针和指针算术将帮助您成为一个更好的C程序员。

如果是False，返回0(没有找到子字符串)，如果是True，返回1(是的，在整个字符串“str”中找到了子字符串“sub”):

#include <stdlib.h>

int is_substr(char *str, char *sub)
{
  int num_matches = 0;
  int sub_size = 0;
  // If there are as many matches as there are characters in sub, then a substring exists.
  while (*sub != '\0') {
    sub_size++;
    sub++;
  }

  sub = sub - sub_size;  // Reset pointer to original place.
  while (*str != '\0') {
    while (*sub == *str && *sub != '\0') {
      num_matches++;
      sub++;
      str++;
    }
    if (num_matches == sub_size) {
      return 1;
    }
    num_matches = 0;  // Reset counter to 0 whenever a difference is found. 
    str++;
  }
  return 0;
}

使用C -没有内置函数

String_contains()完成所有繁重的工作，并返回基于1的索引。其余是驱动程序和辅助程序代码。

指定指向主字符串和子字符串的指针，匹配时增加子字符串指针，当子字符串指针等于子字符串长度时停止循环。

read_line()—一个额外的代码，用于读取用户输入，而不需要预先定义用户应该提供的输入大小。

#include <stdio.h>
#include <stdlib.h>

int string_len(char * string){
  int len = 0;
  while(*string!='\0'){
    len++;
    string++;
  }
  return len;
}

int string_contains(char *string, char *substring){
  int start_index = 0;
  int string_index=0, substring_index=0;
  int substring_len =string_len(substring);
  int s_len = string_len(string);
  while(substring_index<substring_len && string_index<s_len){
    if(*(string+string_index)==*(substring+substring_index)){
      substring_index++;
    }
    string_index++;
    if(substring_index==substring_len){
      return string_index-substring_len+1;
    }
  }

  return 0;

}

#define INPUT_BUFFER 64
char *read_line(){
  int buffer_len = INPUT_BUFFER;
  char *input = malloc(buffer_len*sizeof(char));
  int c, count=0;

  while(1){
    c = getchar();

    if(c==EOF||c=='\n'){
      input[count]='\0';
      return input;
    }else{
      input[count]=c;
      count++;
    }

    if(count==buffer_len){
      buffer_len+=INPUT_BUFFER;
      input = realloc(input, buffer_len*sizeof(char));
    }

  }
}

int main(void) {
  while(1){
    printf("\nEnter the string: ");
    char *string = read_line();
    printf("Enter the sub-string: ");
    char *substring = read_line(); 
    int position = string_contains(string,substring);
    if(position){ 
      printf("Found at position: %d\n", position);
    }else{
      printf("Not Found\n");
    }
  }
  return 0;
}

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");

    }
}

下面这段简单的代码也能达到同样的效果:

int main(void)
{

    char mainstring[]="The quick brown fox jumps over the lazy dog";
    char substring[20];
    int i=0;
    puts("enter the sub string to find");
    fgets(substring, sizeof(substring), stdin);
    substring[strlen(substring)-1]='\0';
    if (strstr(mainstring,substring))
    {
            printf("substring is present\t");
    }
    printf("and the sub string is:::");
    printf("%s",substring,"\n");
   return 0;
}

但棘手的部分是报告子字符串从原始字符串的哪个位置开始…

我相信我有最简单的答案。在这个程序中不需要string.h库，也不需要stdbol .h库。简单地使用指针和指针算术将帮助您成为一个更好的C程序员。

如果是False，返回0(没有找到子字符串)，如果是True，返回1(是的，在整个字符串“str”中找到了子字符串“sub”):

#include <stdlib.h>

int is_substr(char *str, char *sub)
{
  int num_matches = 0;
  int sub_size = 0;
  // If there are as many matches as there are characters in sub, then a substring exists.
  while (*sub != '\0') {
    sub_size++;
    sub++;
  }

  sub = sub - sub_size;  // Reset pointer to original place.
  while (*str != '\0') {
    while (*sub == *str && *sub != '\0') {
      num_matches++;
      sub++;
      str++;
    }
    if (num_matches == sub_size) {
      return 1;
    }
    num_matches = 0;  // Reset counter to 0 whenever a difference is found. 
    str++;
  }
  return 0;
}

尝试使用指针…

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

int main()
{

  char str[] = "String1 subString1 Strinstrnd subStr ing1subString";
  char sub[] = "subString";

  char *p1, *p2, *p3;
  int i=0,j=0,flag=0;

  p1 = str;
  p2 = sub;

  for(i = 0; i<strlen(str); i++)
  {
    if(*p1 == *p2)
      {
          p3 = p1;
          for(j = 0;j<strlen(sub);j++)
          {
            if(*p3 == *p2)
            {
              p3++;p2++;
            } 
            else
              break;
          }
          p2 = sub;
          if(j == strlen(sub))
          {
             flag = 1;
            printf("\nSubstring found at index : %d\n",i);
          }
      }
    p1++; 
  }
  if(flag==0)
  {
       printf("Substring NOT found");
  }
return (0);
}