c - 为什么 getchar() 在二维数组中存储重复值？

Question

所以基本上我试图在每一行的第六个位置存储一个字符，并且我已经成功了，但是我在所需位置及其后续位置中存储了两次相同的值。

我不知道出了什么问题。

• 我尝试过使用 while ((getchar()) != '\n'); <-- 没有帮助
• 将 getchar() 放在 scanf 之前
• 最后，scanf之后的getchar()至少允许我输入值，上面两个只是打印空数组

编辑1- 将列从 5 更改为 6，因为“chux”表示先前的值导致外部分配。编辑2- 向下滚动查看工作代码和最后一个错误

    char *arr_input(int limit){
        char *p=(char*)malloc(limit*6*sizeof(char)); // Edit 1
        int i=0,j;
        //while ((getchar()) != '\n');
        while(i<limit){
            getchar();
            scanf("%c",&p[i*5+5]);    

            i++;
        }

        arr_display(p,limit);
        return p;
    }

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

使用 calloc() 声明时的输出

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

除了 calloc() 之外的工作代码没有初始化 0 值，所以我手动初始化了

    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

Answer 1

使用calloc代替malloc。这样内存就会被初始化为零。

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。您分配了 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]);