c++ - 打印指向指针的指针

Question

这可能很简单，但我很困惑!从指针到指针提取信息时出现段错误。请参阅 main() 中的 cout 部分。任何帮助将不胜感激。
谢谢..

森

#include <stdlib.h>
#include <iostream>

typedef struct{
  int hour;
  int minute;
} Time;

Time* GetNextTime(void)
{
 Time *p_time = new Time;
 return p_time;  
}

void GetTime( Time **sometime )
{
 int length = 10;
 sometime = new Time*[length];
 for(int i=0; i<length; i++)
 {
    sometime[i] = GetNextTime();
    sometime[i]->hour = rand()%24 ;
    sometime[i]->minute = rand()%60;
    std::cout << "Entered times " << sometime[i]->hour << " hour " << sometime[i]->minute << " minutes " << std::endl;
  }
}


int main()
{
  Time** _time;
  GetTime( _time );

 //here is the question
 // I cant print them from original _time
  for( int i=0; i<10; i++)
  std::cout <<  " Print times " << (*_time)[i].hour << " hour " << (*_time)[i].minute << " minutes " << std::endl;

}

Answer 1

您有时按值而不是按引用传递 ，因此它保持未初始化状态。将 GetTime 更改为以下内容:

void GetTime( Time ** &sometime ) //& means pass by reference

因为您正在创建一个指针数组，所以您也可以在打印期间使用数组符号来访问它们。

std::cout <<  " Print times " << _time[i]->hour << " hour "
          << _time[i]->minute << " minutes " << std::endl;