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:: ?
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
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;
}
其他回答
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;
}
#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++;
}
}
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);
}
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;
}
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)));