我正在尝试检查字符串是否包含C中的子字符串，如：
char *sent = "this is my sample example";
char *word = "sample";
if (/* sentence contains word */) {
/* .. */
}
在C ++中使用什么代替string::find？
答案 0 :(得分：228)
if(strstr(sent, word) != NULL) {
/* ... */
}
请注意，如果找到单词strstr，sent会返回指向word中单词开头的指针。
答案 1 :(得分：28)
请使用strstr。
http://www.cplusplus.com/reference/clibrary/cstring/strstr/
所以，你会把它写成......
char *sent = "this is my sample example";
char *word = "sample";
char *pch = strstr(sent, word);
if(pch)
{
...
}
答案 2 :(得分：10)
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);
}
答案 3 :(得分：4)
您可以尝试使用此选项来查找子字符串的存在并提取并打印它：
#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;
}
答案 4 :(得分：4)
此代码实现了搜索的工作原理（其中一种方法），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;
}
答案 5 :(得分：2)
以下是如何报告找到的子字符串中第一个字符的位置：
在上面的代码中替换此行：
printf("%s",substring,"\n");
使用：
printf("substring %s was found at position %d \n", substring,((int) (substring - mainstring)));
答案 6 :(得分：1)
我自己的谦虚（区分大小写）解决方案：
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（avr8-gnu-toolchain-3.5.4.1709）上进行了测试;）
答案 7 :(得分：0)
使用这个更简单的代码也可以实现同样的目的：为什么要使用这些：
int main(void)
{
char mainstring[]="The quick brown fox jumps over the lazy dog";
char substring[20];
int i=0;
puts("enter the sub stirng 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串开始在原始字符串中的哪个位置......
答案 8 :(得分：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");
}
}
答案 9 :(得分：0)
我相信我的回答最简单。您不需要此程序中的string.h库，也不需要stdbool.h库。简单地使用指针和指针算术将帮助您成为更好的C程序员。
对于False（未找到子字符串），仅返回0；对于True（对，在整个字符串“ str”中找到子字符串“ sub”），仅返回1：
#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;
}
答案 10 :(得分：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。剩下的就是驾驶员和助手代码。
为主字符串和子字符串分配一个指针，匹配时递增子字符串指针，当子字符串指针等于子字符串长度时停止循环。
read_line（）-应该提供一些用于读取用户输入而无需预先定义用户输入大小的奖励代码。
#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;
}
答案 11 :(得分：-1)
#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++;
}
}