Double array through pointers in C - c

Question

I want to scan a 2D array with the help of pointers and have written this code, could you tell me why the compiler gives errors?
#include<stdio.h>
#include<stdlib.h>
int main(void) {
int i,j,n,a,b;
int (*(*p)[])[];
printf("\n\tEnter the size of the matrix in the form aXb\t\n");
scanf("%dX%d",&a,&b);
p=(int (*(*p)[b])[a])malloc(b*sizeof(int (*p)[a]));
for(i=0;i<b;i++) {
p[i]=(int (*p)[a])malloc(a*sizeof(int));
printf("\t\bEnter Column %d\t\n");
for(j=0;j<a;j++)
scanf("%d",&p[i][j]);
}
return 0;
}

Answer 1

This statement has several problems:
p=(int (*(*p)[b])[a])malloc(b*sizeof(int (*p)[a]));
First, malloc returns a void*. You are casting that pointer using (int (*(*p)[b])[a]) which yields a value, not a data type. That isn't a valid cast, so that's one reason that the compiler is yelling at you. At this point, p hasn't been initialized so the de-referencing taking place here can crash your program if this statement was executed.
Inside your malloc call, you are using sizeof(int (*p)[a]). The statement int (*p)[a] isn't a valid C statement.
It seems that you are making this a bit more complex that it needs to be. There are two ways of building a 2D array. You can build an array using malloc(a * b * sizeof(int)) as Reinderien explains. You can also build a 1D array of pointers, each pointing to an array of type int. From your code, it seems you are trying to do the latter.
The easier way to do this would be something like this:
int **p;
... get input from user ...
// Declare an array of int pointers of length b
p = malloc(b * sizeof(int*));
// For each int* in 'p' ...
for (i = 0; i < b; ++i) {
// ... allocate an int array of length 'a' and store a pointer in 'p[i]' ..
p[i] = malloc(a * sizeof(int));
// ... and fill in that array using data from the user
printf("\t\bEnter Column %d\t\n");
for(j = 0; j < a; j++)
scanf("%d", &p[i][j]);
}
Using this method of building a 2D array allows you to use the syntax p[x][y]. Since p is a pointer-to-pointer, p[x] is a pointer to an array and p[x][y] is an item in the pointed-to array.

Answer 2

That's some pretty contorted syntax. Usually when you make a 2D array:
The declaration is simply int *p;
The allocation is simply p = malloc(a*b*sizeof(int));
You cannot write p[i][j]. You must do one of several things - either make a secondary array int **q that contains row pointers to be able to write q[i][j] (better performance and legibility), or write p[b*i + j] (fewer steps).
Additionally, note that:
Your printf will spew garbage due to the missing %d parameter.
Since C is not typesafe, using scanf will hide any errors in indirection that you may make.
About the closest thing I could think of that remotely resembles what you were trying to do:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i, j;
const int a = 3, b = 4;
int m[4][3];
int (*p[4])[3];
for (i = 0; i < b; i++)
{
p[i] = &m[i];
printf("\t\bEnter Column %d\t\n", i);
for (j = 0; j < a; j++)
{
int x;
scanf("%d", &x);
(*p[i])[j] = x;
}
}
return 0;
}
It compiles and functions as expected, but it's pointlessly complicated. p is an array of pointers to arrays.