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
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);)
其他回答
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。
#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++;
}
}
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);)
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");
}
}
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)
{
...
}
