我试图检查一个字符串是否包含C中的子字符串,如:

char *sent = "this is my sample example";
char *word = "sample";
if (/* sentence contains word */) {
    /* .. */
}

在c++中可以用什么来代替string:: ?


当前回答

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)
{
    ...
}

其他回答

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;
}

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;
}

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;
}

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)
{
    ...
}