我正在尝试在 C 语言中检查一个字符串是否包含子字符串,例如:
char *sent = "this is my sample example";
char *word = "sample";
if (/* sentence contains word */) {
/* .. */
}
在C++中,除了string::find
,还有什么可以使用的东西?
if (strstr(sent, word) != NULL) {
/* ... */
}
请注意,如果在 sent
中找到了单词 word
,strstr
将返回指向该单词开头的指针。
strstr
返回一个指针;当我测试指针时,我喜欢表述得更加明确。 - nneonneostrcasestr
的作用与 strstr
相同,但它忽略大小写。 - amonettstrstr
的实现:https://github.com/lattera/glibc/blob/master/string/strstr.c。它比朴素的实现要优化得多,可能比一个直接定义的函数更快。然而,当有疑问时,请进行基准测试。 - nneonneo使用strstr
函数就可以实现这个功能。
https://cplusplus.com/reference/cstring/strstr
因此,你可以这样写:
char *sent = "this is my sample example";
char *word = "sample";
char *pch = strstr(sent, word);
if(pch)
{
...
}
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);
}
#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;
}
if(strcmp((ret=strstr(mainstring,substring)),substring))
是不正确的:它只匹配substring
如果它是mainstring
的后缀。函数的其余部分是一种复杂的方式来编写printf("and the sub string is:::%s\n", substring);
。 - chqrlie以下是我个人谦虚的(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(使用avr8-gnu-toolchain-3.5.4.1709)上进行了测试;)
该代码实现了搜索工作原理的逻辑(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;
}
Length
的命名来看。 - Foxint 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;
}
但是棘手的部分在于报告子字符串在原始字符串中的起始位置...
printf("%s",substring,"\n");
使用:
printf("substring %s was found at position %d \n", substring,((int) (substring - mainstring)));
#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;
}
char a [3] =“asd”; char b [2] =“as”; is_substr(a,b);
输入字符串没有以NUL结尾,因此您会越界数组。 - alx - recommends codidactchar a[4] = "asd"; char b[3]= "as"; is_substr(a+4, b);
- alx - recommends codidactstrnstr()
存在的原因(至少在 libbsd 上)。 - alx - recommends codidact使用C语言 - 不使用内置函数
string_contains()函数完成了大部分工作,并返回基于1的索引。其余代码是驱动程序和辅助程序。
将指针分配给主字符串和子字符串,当匹配时增加子字符串指针,当子字符串指针等于子字符串长度时停止循环。
read_line() - 一个小型的额外代码,用于读取用户输入而不预定义输入大小。
#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;
}
strchr
混淆了。 - JonH