用c需要设计程序截取字符串中第m个开始的n个字符 C语言输入一串字符串从第n个字符开始截取m个字符组成一个新的...
C\u8bed\u8a00\u8f93\u5165\u4e00\u4e32\u5b57\u7b26\u4e32\u4ece\u7b2cn\u4e2a\u5b57\u7b26\u5f00\u59cb\u622a\u53d6m\u4e2a\u5b57\u7b26\u7ec4\u6210\u4e00\u4e2a\u65b0\u7684\u5b57\u7b26\u4e32\u8f93\u51fa\u30021\u3001\u6211\u4eec\u7f16\u8f91\u5982\u56fe\u4e2d\u8f93\u5165\u7684\u4ee3\u7801\u3002
2\u3001\u8fdb\u5165\u6e90\u6587\u4ef6\u7684\u76ee\u5f55\u4e0b\uff0c\u5728\u7ec8\u7aef\u8f93\u5165\u547d\u4ee4gcc -o t test.cpp\u3002
3\u3001\u8f93\u5165./t\u547d\u4ee4\uff0c\u8fd0\u884c\u751f\u6210\u7684\u53ef\u6267\u884c\u6587\u4ef6\u3002
4\u3001\u8fd0\u884c\u7ed3\u679c\u4e3abaidu\u3002
5\u3001strncpy \u7684\u51fd\u6570\u539f\u578bchar *strncpy(char *dest,char *src,size_t n)\uff0c\u6211\u4eec\u5728\u590d\u5236src\u7684\u524dn\u4e2a\u5b57\u7b26\u5230dst\u4e2d\u3002
6\u3001\u5982\u679c\u8d85\u8fc7src\u7684\u5b57\u7b26\u4e32\u957f\u5ea6\uff0c\u81ea\u52a8\u590d\u5236\u5b8c\u5168\u90e8\u7684\u5b57\u7b26\u4e32 \u3002
#include
#include
int main()
{char s[300];
int n,m;
printf("Please enter a string:\n");
gets(s);
printf("n m =?");
scanf("%d%d",&n,&m);
if(n>strlen(s)||m<1)
return 0;
s[n+m-1]='\0';
printf("%s\n",s+n-1);
return 0;
}
方法很多
我用指针写了一种
#include<stdio.h>
#include<string.h>
#include<malloc.h>
int main()
{
int m,n;
char str[]="abcdffdsfds",*p;
printf("原字符串：%s
",str);
printf("输入m,n值:");
scanf("%d%d",&m,&n);
p=&str[m-1];
printf("从%d个字符开始截取%d个字符：",m,n);
while(n-->0)
printf("%c",*p++);
return 0;
}
void cut_str(char *str, char *dst, int m, int n)
{
int i;
for(i = 0; i < n; i ++)
dst[i]=str[i+m];
dst[n]=0;
}
其实一句话就可以搞定.
strncpy(dst, str+m,n);
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绛旓細濡倄='a',瑕佹悶娓呮 瀛楃 鍜 瀛楃涓 鐨勬蹇 鑷充簬浣犺鐨勬彁鍙栧瓧绗︼紝鍙涓嬫柟娉 for(i=0; i<n; i++)//n鏄瓧绗︿覆鍙橀噺y鐨勫瓧绗︿釜鏁 { y[i]=x[i+3];} 杩欐牱灏卞彲浠ヤ簡锛屽氨鏄畝鍗曠殑寰幆璇彞璧嬪硷紝鎶妜[]鏁扮粍绗笁涓瓧绗︹榙鈥欙紙绗0涓瓧绗︽椂'a'锛,浠樼粰鏁扮粍y[0]....
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