怎样可以让ch1也是12345呢？

Question

用strncpy时字符数组和字符指针有着怎样的区别？#include <iostream>using std::cout;using std::endl;int main(){char ch1[5];char* ch2;char* ch3 = "12345678";strncpy(ch1,ch3,5);//ch1输出结果：12345&^% (注：后面几个是乱码)strncpy(ch3,ch2,5);//ch3输出结果：12345cout << ch1 <<endl; cout << ch3 <<endl; return 0;}能解释一下strncpy的运行结果么？？int main(){char ch1[5],ch5[5];char ch2[11],ch6[11];char* ch3 = new char[6];char* ch7 = new char[6];char* ch4 = new char[11];char* ch8 = new char[11];char* ch9 = new char[6];char* ch = "12345678";//strcpy(ch1,ch); cout << 1 << ch1 << endl;//strcpy(ch2,ch); cout << 2 << ch2 << endl;//strcpy(ch3,ch); cout << 3 << ch3 << endl;//strcpy(ch4,ch); cout << 4 << ch4 << endl; //上面四行结果均是12345678strncpy(ch5,ch,5); cout << 5 << ch5 << endl; //512345后面乱码strncpy(ch6,ch,5); cout << 6 << ch6 << endl; //612345后面乱码strncpy(ch7,ch,5); cout << 7 << ch7 << endl; //712345后乱码strncpy(ch7,ch,6); cout << 7 << ch7 << endl; //7123456后乱码strncpy(ch8,ch,5); cout << 8 << ch8 << endl; //812345后面乱码strncpy(ch9,ch,5); cout << 9 << ch9 << endl; //912345后面乱码system("pause");return 0;}

Answer 1

操作字串时，不要忘了给最后的'\0'分配空间，不然输出时就会带着乱码，正确的做法是：
char *ch2 = new char[6];
strncpy(ch2, ch3, 5);
ch2[5] = '\0';
cout << ch2 << endl;
不能少了这句
delete[] ch2;
之所以出现乱码，是因为执行strncpy的过程中并没有将字串结束符'\0'赋值给目标，目标字串没有结束符输出时自然就不知道应该什么时候停止输出，除非遇到'\0'字符,因此你会看到乱码，就这样。

Answer 2

分析下strncpy的源代码应该就明白了 ，库函数并没有对dest检查，有可能dest不是以'\0'结尾， 输出的时候就会出错了 
char * __cdecl strncpy ( 
char * dest, 
const char * source, 
size_t count 
) 
{ 
char *start = dest; 

while (count && (*dest++ = *source++)) /* copy string */ 
count--; 

if (count) /* pad out with zeroes */ 
while (--count) 
*dest++ = '\0'; 

return(start); 
}