Syntax error when assigning values to a dynamically allocated 2d array of pointers

You have many problems with that code. The first is that only pmat1 is an array of arrays of pointers to int, the others (pmat2 and pmat3) are only arrays of arrays of int (i.e. not pointers).

The second problem is the initialization, you simply can't assign like that, you have to initialize each field separately.

Thirdly, *pmat1[3] doesn't do what you probably expect it to do, besides it's out of bounds, as array indexes start from zero.

If we are going down to a more basic level, if what you want is an array of three arrays of three integers, then you are doing it completely wrong, then it's just enough with

int pmat1[3][3];

The above will cause the compiler to allocate space for a total of nine (3 * 3) int. No need for any dynamic allocation.

You could then initialize it like

pmat1[0][0] = pmat1[0][1] = pmat1[0][2] = 1;
pmat1[1][0] = pmat1[1][1] = pmat1[1][2] = 0;
pmat1[2][0] = pmat1[2][1] = pmat1[2][2] = 2;

I want to directly assign values to a 2d array of pointers for which the memory is allocated dynamically.

You can't do this. For example:

int *p = malloc(sizeof(*p) * 3);
*p = {1, 2, 3};

is completely wrong. You're assigning {1, 2, 3}, an array of type int[3], to *p, an int. If you've to, do it one by one:

int a[] = {1, 2, 3};
for (size_t i = 0u; i < 3u; ++i)
    p[i] = a[i];

The only time you can do a direct assignment is in an automatic array initialization, such as the a in the above snippet, where the unnamed array {1, 2, 3} is , conceptually, copied to the elements of a.

As Joachim noted, make sure you stay within bounds. [0, n − 1] is the range allowed to be dereferenced for an array of size n. Hence a[3] = 4; is illegal, as a is of size 3, although it'll happily compile, you're treading on UB land.

In C's syntax for pointer declaration, the * operator is associated to the identifier and not the type. In int* a, b;, although * is written next to int, the type is still int and not int*. Only a is of type int* and b is still an int.

Can you please look below code, this code show how pointer array stores values of integer type,

#include <stdio.h>

const int MAX = 3;

int main ()
{
   int  var[] = {10, 100, 200};
   int i, *ptr[MAX];

   for ( i = 0; i < MAX; i++)
   {
      ptr[i] = &var[i]; /* assign the address of integer. */
   }
   for ( i = 0; i < MAX; i++)
   {
      printf("Value of var[%d] = %d\n", i, *ptr[i] );
   }
   return 0;
}

And, below code shows how pointer array stores string type values,

#include <stdio.h>

const int MAX = 4;

int main ()
{
   char *names[] = {
                   "Zara Ali",
                   "Hina Ali",
                   "Nuha Ali",
                   "Sara Ali",
   };
   int i = 0;

   for ( i = 0; i < MAX; i++)
   {
      printf("Value of names[%d] = %s\n", i, names[i] );
   }
   return 0;
}