Question

How to initialize a 2D array with pointer.

int *mapTable[] = { {1,10} , {2,20} , {3,30} , {4,40} };  // It's giving error

Here int *mapTable is giving an error.

How can I declare them properly?

Answer 1

int *mapTable[] is not a 2D array: it is a 1D array of pointers.

But then you go and use a 2D array initialiser { {1,10} , {2,20} , {3,30} , {4,40} }.

That’s why it’s “giving error”.

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The 2D array way

Try:

int mapTable[][2] = { {1,10} , {2,20} , {3,30} , {4,40} };

And, yes, you do need to specify the size of that second dimension.

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The 1D array of pointers way

This is a little more involved, and is usually too complex to be worth it.

It also usually requires dynamic allocation, causing a total mess with object lifetime:

int *mapTable[] = { new int[2], new int[2], new int[2], new int[2] };

int main() {
   mapTable[0][0] = 1; mapTable[0][1] = 10;
   mapTable[1][0] = 2; mapTable[1][1] = 20;
   mapTable[2][0] = 3; mapTable[2][1] = 30;
   mapTable[3][0] = 4; mapTable[3][1] = 40;

   // then, at the end of your program:
   for (size_t i = 0; i < 4; i++)
      delete[] mapTable[i];
}

As you can see, this is not ideal.

You can avoid dynamic allocation:

int mapTable0[] = {1,10};
int mapTable1[] = {2,20};
int mapTable2[] = {3,30};
int mapTable3[] = {4,40};
int *mapTable[] = { &mapTable0[0], &mapTable1[0], &mapTable2[0], &mapTable3[0] };

But I’m not sure why you’d want to go down this avenue.