C Programming: Initialize a 2D array of with numbers 1, 2, 3…etc

问题

I am having trouble creating a 2D Array of a size defined by the user, with numbers 1, 2, 3.etc.

If the user chooses for example: a = 2 and b = 2, the program produces:

3 4

3 4

instead of:

1  2

3  4

My program looks like:

#include <stdio.h>

int main()
{
    int a = 0;
    int b = 0;
    int Array[a][b];
    int row, column;
    int count = 1;

/*User Input */
    printf("enter a and b \n");
    scanf("%d %d", &a, &b);

/* Create Array */
    for(row = 0; row < a; row++)
    {
        for(column = 0; column <b; column++)
        {
            Array[row][column] = count;
            count++;
        }
    }

/* Print Array*/
    for(row = 0; row<a; row++)
    {
        for(column = 0; column<b; column++)
        {
            printf("%d ", Array[row][column]);
        }
        printf("\n");
    }

    return 0;
}

回答1:

int a, b;

variables a and b are uninitialized and their value is undetermined by C language

int Array[a][b];

You declare an array which has [a,b] size. The problem is that a and b are undetermined and using them at this point is undefined behavior.

scanf("%d %d", &a, &b);

you get a and b values -- but the Array remains the same!

Simplest solution: try to put Array declaration after scanf. Your compiler may allow it (I think C99 is required to do so).

回答2:

Variable length array is not supported in c89 standard.

int Array[a][b]; is meaningless. Because values of a and b is unknown at that time. so change it to Array[2][2].

回答3:

Since your array size is not known at compile-time, you'll need to dynamically allocate the array after a and b are known. like code as follows:

int **allocate_2D_array(int rows, int columns)
{
    int k = 0;
    int **array = malloc(rows * sizeof (int *) );

    array[0] = malloc(columns * rows * sizeof (int) );
    for (k=1; k < rows; k++)
    {
        array[k] = array[0] + columns*k;
        bzero(array[k], columns * sizeof (int) );
    }

    bzero(array[0], columns * sizeof (int) );

    return array;
}

回答4:

Since your array size is not known at compile-time, you'll need to dynamically allocate the array after a and b are known.

Here's a link that describes how you can allocate the multi-dimensional array (really an array of arrays): http://www.eskimo.com/~scs/cclass/int/sx9b.html

Applying the example code from that link, you might do this:

int **Array; /* Instead of int Array[a][b] */

...

/* Create Array */
Array = malloc(a * sizeof(int *));
for(row = 0; row < a; row++)
{
    Array[row] = malloc(b * sizeof(int));
    for(column = 0; column <b; column++)
    {
        Array[row][column] = count;
        count++;
    }
}

来源：https://stackoverflow.com/questions/17563075/c-programming-initialize-a-2d-array-of-with-numbers-1-2-3-etc

标签

multidimensional-array

user-input