我如何接受逻辑地址数据？ [英] How do I accept logical address data?

问题描述

Hey Guys，

我编写了一段代码来演示操作系统中的分页概念。我假设框架和页面大小为4个字节。

问题是我无法输入逻辑地址值！

请求帮助：

Hey Guys ,
I've coded a little piece of code that demonstrates the paging concept in Operating Systems. I've assumed the frame and page size to be 4 bytes.
The problem is that i'm unable to enter the Logical Address Values!
Request assistance on the same:

#include<iostream>
#include<conio.h>
#include<time.h>
using namespace std;
int main()
{       
    char* logical[16]; 
    char* physical[33];
    int ft[4]; int pg=0; 
    int s[2]; int i,j; int page_move=0; int frame=0;
    int test[8]={0,4,8,12,16,20,24,28}; //Used for the random selection of frame numbers assuming the frame size to be 4 bytes
    time_t t;
    srand((unsigned) time(&t));
    s[0]=test[rand()%8]; i=s[0];  // We need two random frames to be occupied to represent a real scenario better !
    s[1]=test[rand()%8]; j=s[1];  // 's' contains those two frame numbers which will be made 'Full' later.
    cout<<"\n"<<s[0]<<" "<<s[1];
    for(int i=0;i<32;i++)// initialize all addresses to Empty
    { 
      physical[i]="Empty";
      logical[i]="Empty";
    }
    printf("\nThe Logical Address is Now->\n");
    for(int i=0;i<16;i++)
    {
      cout<<"\n"<<i<<" "<<logical[i];
    }
    for(int k=0;k<4;k++) // For filling in two random frames 
    { 
      physical[i]="Full";
      physical[j]="Full";
      ++i;
      ++j;
    }
    cout<<"\nEnter the Logical Address Data->\n";
    for(int i=0;i<16;i++)   // Entering Logical Data ( Doesn't Work!!!)
    {
       if(i%4==0){cout<<"\n---Page "<<page_move++<<"----\n";}
       cin>>logical[i];  
    }
    i=0;page_move=0;
    while(i<=28)   // Constructing The Frame Table
    {
       if(physical[i]=="Empty" && page_move!=4)
       {
         ft[page_move]=frame;
         ++page_move; 
       }
       ++frame;
       i=i+4;
    }
   
    cout<<"\n::::FRAME TABLE::::\n";
    cout<<"\n Page Frame\n";
    for(int i=0;i<4;i++)  // Display The Frame Table Created Above
    { 
      cout<<"\n  "<<i<<"  "<<ft[i];
    }
    printf("\nThe Physical Address is Now->\n");
    for(int i=0;i<32;i++)  //Display Physical Addresses and their Data
    {
      cout<<"\n"<<i<<" "<<physical[i];
    }
    
    getch();    
}

我的尝试：



我已经尝试改变我接受逻辑地址数据的方式。

1.）使用另一个char *字然后将其分配给逻辑[i] ...没有工作

2.）使用char * s =逻辑;然后输入s，然后输入++ s;没有工作

What I have tried:

I've tried changing how i accept the logical address data.
1.) Used another char* word and then assigned it to logical[i]...Didn't work
2.) Used char*s= logical; and then entered s followed by ++s; Did not work

推荐答案

你有以下声明：

You have the following statement:

cin>>logical[i];

但逻辑是一个 char * 所以你不能直接在其中一个单元格中写入一些文本。您需要将数据读入新的字符串，然后将其地址发布到数组项中。你也不应该使用如下表达式：



but logical is an array of char* so you cannot directly write some text into one of its cells. You need to read the data into a new character string and then post its address into the array item. You should also not use expressions like:

if(physical[i]=="Empty"

因为这不是有效的字符串比较你应该使用 strcmp 或其衍生产品之一。

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