。
[此贴子已经被作者于2020-3-14 13:52编辑过]
for (p = &str[m - 1]; p <= &str[n - 1]; p++)
str1.push_back(*p);
程序代码:
void Partial_replication(string str)
{
int StringLength(string str);
string str1 = str;
int n, m, i = 0;
char *p, *p1;
n = StringLength(str);
cout << "请输入被复制的字符个数m,(m<" << n << "):";
cin >> m;
for (p = &str[m - 1], p1 = &str1[0]; p <= &str[n - 1];)
*p1++ = *p++;
while(*p1)*p1++ = 0;
cout << "被复制的部分字符串为:" << str1 << endl;
}
[此贴子已经被作者于2020-3-14 01:52编辑过]