如何让子在C与和strchr [英] how to get substring in c with strchr

问题描述

我想取一个字符串的一部分。结果
我有以下的code：

I'm trying to fetch part of a string.
I have the following code:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char mystring[]="The quick brown fox jumps over the lazy dog";
char word1[]="The";
char * posb,pose;
char * word2;
int b,e,n;

n=memcmp(mystring, word1, sizeof(word1)-1);
if (n==0) printf("the first word is found!\n");
posb=strchr(mystring,' ');   // posb will be a pointer to the first character
b=posb-mystring+1;
printf("posb -> %d\n",posb);
printf("b -> %d\n",b);
pose=strchr(mystring+b+1,' ');   // pose will be a pointer to the second character
printf("calc e\n");
e=pose-sizeof(mystring)+1;
printf("pose -> %s\n",pose);
printf("e -> %d\n",e);
word2 = (char*) malloc (sizeof(char)*(e-b));
memcpy(word2, posb, sizeof(word2));
printf("%s:%i:%i:%i\n",word2, b, e, e-b);
free (word2);

的问题越来越第二个字并将其存储在WORD2。为此，我尝试使用和strchr 来定位的空间。但是我用的是第二次和strchr 我需要一个偏移找到第二空间。我试过如下：

The problem is getting the second word and store it in word2. For this I try using strchr to locate the spaces. But the second time I use strchr I need an offset to find the second space. I tried the following:

pose=strchr(mystring+b+1,' ');
pose=strchr(&mystring[b+1],' ');

变量 B 和电子应包含 mystring中空格字符的位置。 字词2 应包含快速最后。结果
另一种解决方案是使用一个循环的字符串'穿行'，但是这将是欺骗和strchr 功能。

The variables b and e should contain the positions of the space character in mystring. word2 should contain quick eventually.
Another solution would be to 'walk through' the string using a loop, but that would be cheating the strchr function.

推荐答案

为了获得第二个字，请在此

In order to get the second word, check this

char mystring[] = "The quick brown fox jumps over the lazy dog";
char word1[] = "The";
char * posb, pose;
char * word2;
int b, e, n;

n = memcmp(mystring, word1, sizeof(word1)-1);
if (n == 0) printf("the first word is found!\n");
posb = strchr(mystring, ' ');   // posb will be a pointer to the first character

b = posb - mystring + 1;

printf("posb -> %d\n", posb);
printf("b -> %d\n", b);
posb = strchr(posb + 1, ' ');   // pose will be a pointer to the second character
printf("calc e\n");

e = posb - mystring + 1;
printf("e -> %d\n", e);

word2 = (char*)malloc(sizeof(char)*(e - b));
memcpy(word2, mystring + b, e - b);
word2[e-b-1] = '\0';
printf("%s:%i:%i:%i\n", word2, b, e, e - b);
free(word2);

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