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;
}
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…
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。
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;
}
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;
}
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);)
