C语言(不是C++)中如何从指定字符串中提取指定开始位置到指定结束位置的一个子串给数组？ C++如何从String的指定位置搜索一个子串？
c/c++\u4e2d\u600e\u6837\u4ece\u6307\u5b9a\u7684\u4f4d\u7f6e\u62f7\u8d1d\u5b57\u7b26\u4e32\u7684\u4e00\u90e8\u5206C++ :
string str2 = "how are you"\uff1b
string str1=str2.substr(4,3);
C:
char str1[4];
for(int i=0;i<3;++i)
str1[i]=str2[4+i];
str1[3]='\0';
\u53ef\u4ee5\u7528find()\u51fd\u6570\uff0c\u5177\u4f53\u7684\u8bed\u6cd5\u5982\u4e0b\u6240\u793a\uff1a
(1) size_type find(E c, size_type pos = npos) const;\u7528\u6765\u67e5\u627e\u5355\u4e2a\u5b57\u7b26\u5728\u5b57\u7b26\u4e32\u4e2d\u51fa\u73b0\u7684\u4f4d\u7f6e\u5e76\u8fd4\u56de\u4f4d\u7f6e\u57fa\u4e8e0\u7684\u7d22\u5f15\u503c\u3002
(2) size_type find(const E *s, size_type pos = npos) const;\u7528\u6765\u67e5\u627e\u4ee5\u7a7a\u5b57\u7b26\u7ed3\u5c3e\u7684\u5b57\u7b26\u6570\u7ec4\u5728\u5b57\u7b26\u4e32\u4e2d\u51fa\u73b0\u7684\u4f4d\u7f6e\u5e76\u8fd4\u56de\u8be5\u4f4d\u7f6e\u57fa\u4e8e0\u7d22\u5f15\u503c\u3002
(3) size_type find(const E *s, size_type pos, size_type n = npos) const;\u7528\u6765\u67e5\u627e\u4ee5\u7a7a\u5b57\u7b26\u7ed3\u5c3e\u7684\u5b57\u7b26\u6570\u7ec4\u5728\u5b57\u7b26\u4e32\u4e2d\u51fa\u73b0\u7684\u4f4d\u7f6e\u5e76\u8fd4\u56de\u8be5\u4f4d\u7f6e\u57fa\u4e8e0\u7d22\u5f15\u503c\uff0c\u5b83\u662f\u4ecenpos\u5f00\u59cb\u67e5\u627e\u7684\u3002
(4) size_type find(const basic_string &str, size_type pos = npos) const;\u7528\u6765\u67e5\u627e\u5b57\u7b26\u4e32\u5e76\u4e14\u8fd4\u56de\u8be5\u641c\u7d22\u7684\u5b57\u7b26\u4e32\u7684\u57fa\u4e8e0\u7d22\u5f15\u503c\u7684\u4f4d\u7f6e\u503c\uff0c\u5b83\u662f\u4ecenpos\u5f00\u59cb\u67e5\u627e\u7684\u3002
学程序，关键是学方法！
void substr(char *str1,char *str2,int start,int end)
{
int ii;
for (ii=start;ii<=end;ii++)
str1[ii-start]=str2[start-1];
str1[ii-start]='\0';
}
main()
{
char str1[30]="dwew13lkoilesiwokdwok";
char str2[50]="e22dww_dwoiewolksdjfmmm"
char str[50];
substr(str,str1,1,10);
printf("1--%s\n",str);
substr(str,str2,3,18);
printf("2--%s\n",str);
}
1 #include <stdio.h>
2
3 char *sub_str(int start,int end,char *str)
4 {
5 char *sub = NULL;
6 int i = 0;
7 int j = 0;
8 sub = (char *)malloc(end-start);
9 if(NULL == str)
10 return NULL;
11 for(i = start;i<end;i++)
12 {
13 sub[j++] = str[i];
14 }
15
16 sub[end-start]='\0';
17
18 return sub;
19
20 }
21
22 int main ()
23 {
24 char str1[30]="dwew13lkoilesiwokdwok";
25
26 char *sub = NULL;
27
28 sub = sub_str(0,10,str1);
29
30 printf("%s\n",sub);
31
32 return 0;
33 }
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