CodePudding user response:
Char * func (char * STR, int the begin, int num)
{
Char * result=(char *) malloc (sizeof (char) * (num + 1));
Memcpy (result, STR + the begin, sizeof (char) * num);
The result (num)='\ 0';
return result;
}
CodePudding user response:
//the string "abcdefghijklmn", read in two Numbers, such as the begin, num, assuming the begin to 1, num is 5,
//the application memory, dynamic storage substring "bcdef", finally outputs the substring,
//simplicity, assume that the begin, num are within the effective range, design the main function, at the same time to reveal a string and dynamic memory address
#include
#include
#include
#include
Int main () {
Char * s="abcdefghijklmn";
Int the begin, num;
Char * p;
The scanf (" % d % d ", & amp; The begin, & amp; Num);
P=(char *) malloc (num + 1);
{if (NULL==p)
Printf (" Can not malloc (% d bytes)! \ n ", num + 1);
return 1;
}
Strncpy (p, s + the begin, num);
P (num)=0;
Printf (" % s \ n ", p);
Printf (" 0 x % \ n p ", p);
free(p);
return 0;
}
//enter
//1 5
//output
//bcdef
//0 x012ddc38
//
