为什么getchar（）在2d数组中存储重复值？

所以基本上我试图在每一行的第6个位置存储一个字符，我已经成功地使用了它，但是我得到了相同的值，在期望的位置和它的后续位置存储了两次

我不知道出了什么问题

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• 我尝试过使用while（（getchar（））！='\n'） 使用calloc而不是malloc。这样内存将被初始化为零
  请不要忘记将分配大小（即6）和索引计算更新为相同的数字。您已分配6，但使用了&p[i*5+5]
  粘贴了两个更改

  char *p=(char*)calloc(limit*6, sizeof(char)); 
  scanf("%c",&p[i*6+5]);  
  

  ---

  &p[i*5+5]

  在分配之外。它仍然在存储重复的值请记住

  malloc

  并没有将内存设置为零。使用

  calloc

  。为什么同时使用

  getchar

  和

  scanf

  ？欢迎使用堆栈溢出！您的代码不完整；特别是，它似乎缺少一个

  main（）

  函数和至少一个

  #include

  。请输入您的代码，使其成为您问题的一部分（包括任何必要的输入，但最好不需要任何输入），然后我们可以尝试复制并解决它。您还应该阅读。那么我不必使用getchar（）？我已经在使用calloc而不是malloc时发布了输出

  Value at row 0 is = 
  Value at row 0 is = 
  Value at row 0 is = 
  Value at row 0 is = 
  Value at row 0 is = 
  Value at row 0 is = 1
  Value at row 1 is = 1
  Value at row 1 is = 
  Value at row 1 is = 
  Value at row 1 is = 
  Value at row 1 is = 
  Value at row 1 is = 2
  Value at row 2 is = 2
  Value at row 2 is = 
  Value at row 2 is = 
  Value at row 2 is = 
  Value at row 2 is = 
  Value at row 2 is = 3
  Input Verified
  
  

  Value at row 0 is = 0
  Value at row 0 is = 
  Value at row 0 is = 
  Value at row 0 is = 
  Value at row 0 is = 
  Value at row 0 is = 1
  Value at row 1 is = 
  Value at row 1 is = 0
  Value at row 1 is = 
  Value at row 1 is = 
  Value at row 1 is = 
  Value at row 1 is = 2
  Value at row 2 is = 
  Value at row 2 is = 
  Value at row 2 is = 0
  Value at row 2 is = 
  Value at row 2 is = 
  Value at row 2 is = 3
  

      char *arr_input(int limit){
          char *p=(char*)calloc(limit*6,sizeof(char));
          int i=0;
          while(i<limit){
              scanf(" %c",&p[i*6+5]);             
              i++;
          }
          zeroIn(p,limit);
          arr_display(p,limit);
          return p;
      }
      void *zeroIn(char *p ,int limit){
          int i=0,j=0;
          for(i=0;i<limit;i++){
              for(j=0;j<5;j++)
                  p[i*6+j]='0';
          }       
              //arr_display(p,limit); 
      }
      char *arr_display(char *disp,int size){
          int i=0,j=0;
          for (i = 0; i <size ;i++)
          {
              for(j=0;j<6;j++){
           printf("Value at row %d is = %c\n",i,disp[i*6+j]);
              }
          }
      }
  

  Value at row 0 is = 0
  Value at row 0 is = 0
  Value at row 0 is = 0
  Value at row 0 is = 0
  Value at row 0 is = 0
  Value at row 0 is = 1
  Value at row 1 is = 0
  Value at row 1 is = 0
  Value at row 1 is = 0
  Value at row 1 is = 0
  Value at row 1 is = 0
  Value at row 1 is = 2
  Value at row 2 is = 0
  Value at row 2 is = 0
  Value at row 2 is = 0
  Value at row 2 is = 0
  Value at row 2 is = 0
  Value at row 2 is = 3
  
  

  char *p=(char*)calloc(limit*6, sizeof(char)); 
  scanf("%c",&p[i*6+5]);