如何在搜索用C字符数组的字符串？ [英] How to search a string in a char array in C?

问题描述

例如我有：

 字符的buff [1000];
 

我想如果搜索字符串hassasin是该字符数组。这里是什么我都试过了。

 字符字[8] =hassasin;
字符缓冲区[1000] = sdfhksfhkasd /./。fjka（hassasin）hdkjfakjsdfhkksjdfhkjh ....等
INT K = 0;
INT T = 0;
INT LEN = 0;
INT SOR = 0;
对于（K = 0; K＆LT; 1000; k ++）{
    对于（t = 0; T＆LT; 8;吨++）{
        如果（缓冲[K] ==字[T]）LEN ++;
        如果（LEN == 8），它开创0.9.1
    }
}
 

解决方案

如果chararray包含stringend或者不以\\ 0结束了，你可以使用这些code，因为的strstr都会对这些的刹车：

 的#include＆LT;＆stdio.h中GT;
诠释的main（）
{
    炭c_to_search [5] =ASDF;    字符文本[68] =你好，我的名字是\\ 0有它背后\\ n \\ 0 ASDF其他一些字符串;    INT pos_search = 0;
    INT pos_text = 0;
    INT len_search = 4;
    INT len_text = 67;
    为（pos_text = 0; pos_text＆下; len_text  -  len_search ++ pos_text）
    {
        如果（文[pos_text] ==​​ c_to_search [pos_search]）
        {
            ++ pos_search;
            如果（pos_search == len_search）
            {
                // 匹配
                的printf（从％d个匹配％d个\\ N，pos_text-len_search，pos_text）;
                返回;
            }
        }
        其他
        {
           pos_text  -  = pos_search;
           pos_search = 0;
        }
    }
    //不匹配
    的printf（无匹配\\ n）;
   返回0;
}
 

http://ideone.com/2In3mr

For example I have:

char buff[1000];

I want to search if the string "hassasin" is in that char array. Here is what I have tried.

char word[8] = "hassasin";
char Buffer[1000]=sdfhksfhkasd/./.fjka(hassasin)hdkjfakjsdfhkksjdfhkjh....etc 
int k=0;
int t=0; 
int len=0; 
int sor=0; 
for (k=0; k<1000; k++){ 
    for (t=0; t<8; t++){ 
        if (Buffer[k]==word[t]) len++; 
        if (len==8) "it founds 0.9.1" 
    } 
}

解决方案

if the chararray contains stringend or do not end with \0 you can use these code, because strstr will brake on these ones:

#include <stdio.h>
int main()
{
    char c_to_search[5] = "asdf";

    char text[68] = "hello my name is \0 there is some other string behind it \n\0 asdf";

    int pos_search = 0;
    int pos_text = 0;
    int len_search = 4;
    int len_text = 67;
    for (pos_text = 0; pos_text < len_text - len_search;++pos_text)
    {
        if(text[pos_text] == c_to_search[pos_search])
        {
            ++pos_search;
            if(pos_search == len_search)
            {
                // match
                printf("match from %d to %d\n",pos_text-len_search,pos_text);
                return;
            }
        }
        else
        {
           pos_text -=pos_search;
           pos_search = 0;
        }
    }
    // no match
    printf("no match\n");
   return 0;
}

http://ideone.com/2In3mr

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