我想从函数返回一个字符数组。然后我想打印出来 main。我怎样才能重新获得角色阵列 main 功能？
#include<stdio.h>
#include<string.h>
int main()
{
int i=0,j=2;
char s[]="String";
char *test;
test=substring(i,j,*s);
printf("%s",test);
return 0;
}
char *substring(int i,int j,char *ch)
{
int m,n,k=0;
char *ch1;
ch1=(char*)malloc((j-i+1)*1);
n=j-i+1;
while(k<n)
{
ch1[k]=ch[i];
i++;k++;
}
return (char *)ch1;
}
请告诉我，我做错了什么？
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
char *substring(int i,int j,char *ch)
{
int n,k=0;
char *ch1;
ch1=(char*)malloc((j-i+1)*1);
n=j-i+1;
while(k<n)
{
ch1[k]=ch[i];
i++;k++;
}
return (char *)ch1;
}
int main()
{
int i=0,j=2;
char s[]="String";
char *test;
test=substring(i,j,s);
printf("%s",test);
free(test); //free the test
return 0;
}
这将编译正常，没有任何警告
#include stdlib.h test=substring(i,j,s); m 因为它未被使用 char substring(int i,int j,char *ch) 或者在main之前定义它 丹尼尔是对的： http://ideone.com/kgbo1C#view_edit_box
更改
test=substring(i,j,*s);
至
test=substring(i,j,s);
此外，您需要转发声明子字符串：
char *substring(int i,int j,char *ch);
int main // ...
评论中的懒惰笔记。
#include <stdio.h>
// for malloc
#include <stdlib.h>
// you need the prototype
char *substring(int i,int j,char *ch);
int main(void /* std compliance */)
{
int i=0,j=2;
char s[]="String";
char *test;
// s points to the first char, S
// *s "is" the first char, S
test=substring(i,j,s); // so s only is ok
// if test == NULL, failed, give up
printf("%s",test);
free(test); // you should free it
return 0;
}
char *substring(int i,int j,char *ch)
{
int k=0;
// avoid calc same things several time
int n = j-i+1;
char *ch1;
// you can omit casting - and sizeof(char) := 1
ch1=malloc(n*sizeof(char));
// if (!ch1) error...; return NULL;
// any kind of check missing:
// are i, j ok?
// is n > 0... ch[i] is "inside" the string?...
while(k<n)
{
ch1[k]=ch[i];
i++;k++;
}
return ch1;
}