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;
}

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);
}

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)));

我自己的简单解决方案(区分大小写):

uint8_t strContains(char* string, char* toFind)
{
    uint8_t slen = strlen(string);
    uint8_t tFlen = strlen(toFind);
    uint8_t found = 0;

    if( slen >= tFlen )
    {
        for(uint8_t s=0, t=0; s<slen; s++)
        {
            do{

                if( string[s] == toFind[t] )
                {
                    if( ++found == tFlen ) return 1;
                    s++;
                    t++;
                }
                else { s -= found; found=0; t=0; }

              }while(found);
        }
        return 0;
    }
    else return -1;
}

结果

strContains("this is my sample example", "th") // 1
strContains("this is my sample example", "sample") // 1
strContains("this is my sample example", "xam") // 1
strContains("this is my sample example", "ple") // 1
strContains("this is my sample example", "ssample") // 0
strContains("this is my sample example", "samplee") // 0
strContains("this is my sample example", "") // 0
strContains("str", "longer sentence") // -1
strContains("ssssssample", "sample") // 1
strContains("sample", "sample") // 1

在ATmega328P上测试(avr8-gnu-toolchain-3.5.4.1709);)

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;
}

你可以尝试这一个既找到子字符串的存在，并提取和打印它:

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

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

    for(i=0;i<strlen(substring);i++)
    {
            printf("%c",*(ret+i));

    }
    puts("\n");
    return 0;
}