C传递2D数组以起作用，打印结果

Question

I am trying to print a 2D array by passing it to a function, but I got weird results. Here is my code.

#include <stdio.h>

int main()
{
    int b[2][3] = {{1,2,3},{4,5,6}};
    printArray(b);
    return 0;
}

void printArray(int (*ptr)[3])
{
    int i, j;
    for(i = 0; i < 2; i++)
    {
        for(j = 0; j < 3; j++)
        {
            printf("%d\t", *((*ptr+i)+j));
        }
        printf("\n");
    }
}

However, the output is

1  2   3

2  3   4

I think it is something to do with my 'j' variable but I can't seem to pinpoint it. Please help.

Answer 1

您需要将i乘以3才能加上j ...

printf("%d\t", *((*ptr+(i*3))+j));

Answer 2

You need to apply the addition operator before using the dereference operator. You need to use parenthesis as the dereference operator( * ) has higher precedence than the addition operator( + ).

So, this

printf("%d\t", *((*ptr+i)+j));

should be

printf("%d\t", *((*(ptr+i))+j));

or better

printf("%d\t", ptr[i][j]);

---

Also, you need to move the function printArray before main or provide a function prototype before main like:

void printArray(int (*ptr)[3]);

Answer 3

Abandoning indexes in favor of pointers, assuming matrix is stored by rows (whic is normally the case):

#include <stdio.h>

void printArray(int *);

int main()
{
 int b[2][3] = {{1,2,3},{4,5,6}};
 printArray((int*)b);
 return 0;
}

void printArray(int *ptr)
{
int i, j;
 for(i = 0; i < 2; i++)
 {
    for(j = 0; j < 3; j++)
    {
     printf("%d\t", *ptr++);
    }
  printf("\n");
 }
}

Outputs:

1   2   3   
4   5   6

Answer 4

printArray can take nb_column nb_line param and deal with all sort of 2D array

#include <stdio.h> 

static void printArray(int *ptr,  int nb_columns, int nb_lines) 
{ 
    int i, j; 
    for(i = 0; i < nb_lines; i++) 
    { 
        for(j = 0; j < nb_columns; j++) 
        { 
            printf("%d\t", *(ptr++)); 
        } 
        printf("\n"); 
    } 
} 
int main() 
{ 
    int b[2][3] = {{1,2,3},{4,5,6}}; 
    printArray((int *)b, 3, 2); 
    return 0; 
}