为什么memcpy(calendar, cal, sizeof(CALENDAR))这里输出是错误的啊，搞不懂了

  • Z时代
  • 2024-01-10
  • 分类：IT

不知道是什么原因

#include<iostream>
using namespace std;
typedef struct
{
	int year;
	int month;
	int date;
	int hour;
	int minute;
	int second;
	int millisecond;
}CALENDAR;
CALENDAR *getCalendar()
{
	CALENDAR cal ;
	cal.year = 2015;
	cal.month = 8;
	cal.date = 15;
	cal.hour = 14;
	cal.minute = 34;
	cal.second = 23;
	cal.millisecond = 123;
	return &cal;
}
int main()
{
	CALENDAR calendar;
	CALENDAR* cal;
	cal = getCalendar();
	memcpy(&calendar, cal, sizeof(CALENDAR));
	cout << calendar.year << " "
		<< calendar.month << " "
		<< calendar.date << " "
		<< calendar.hour << " "
		<< calendar.minute << " "
		<< calendar.second << " "
		<< calendar.millisecond << " "
		<< sizeof(CALENDAR) << endl;
}

输出是这样的
图片描述

回答：

局部变量CALENDAR cal ;是在栈上定义的，在退出函数作用域时就内存就释放掉了，后续的代码将覆盖这块内存区域，输出就是未确定的值；

可以改用直接返回对象而非指针

CALENDAR getCalendar()
{
CALENDAR cal ;
cal.year = 2015;
cal.month = 8;
cal.date = 15;
cal.hour = 14;
cal.minute = 34;
cal.second = 23;
cal.millisecond = 123;
return cal;
}

也可以改用引用（或指针）传入返回；

void getCalendar(CALENDAR& cal)
{
cal.year = 2015;
cal.month = 8;
cal.date = 15;
cal.hour = 14;
cal.minute = 34;
cal.second = 23;
cal.millisecond = 123;
}
CALENDAR calendar;
getCalendar(calendar);

回答：

CALENDAR *getCalendar()
{
	CALENDAR cal ;
	cal.year = 2015;
	cal.month = 8;
	cal.date = 15;
	cal.hour = 14;
	cal.minute = 34;
	cal.second = 23;
	cal.millisecond = 123;
	return &cal; // 返回局部变量的引用 闹哪样？
}

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