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Lorelei Symphony
Asked: 2022-09-04 14:06:17 +0800 CST 2022-09-04 14:06:17 +0800 CST

Why can't I define the dimensions of a two-dimensional array with int variables? C

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I have the following code in C

int main(int argc, char *argv[]) {
   int row=2, col=3, f,c;
   char array[row][col] = { {'A', 'B', 'C'}, {'a', 'b', 'c'} };

   for(f=0; f<row; f++) {
      for(c=0; c<col; c++)
         printf("%c", a[f][c]);
      printf("\n");

   }

   return 0;
}

Why does this line NOT work like this: char array[row][col] = { {'A', 'B', 'C'}, {'a', 'b', 'c'} };

error: excess elements in array initializer

But why does it DO work like this? char array[2][3] = { {'A', 'B', 'C'}, {'a', 'b', 'c'} };

Errors(0) Warnings(0)

array

1 Answers

  1. Best Answer

    In C and C++ the size of arrays must be known at compile time. That's why you can't use variables to declare them.

    What you can do is define the array and let the compiler infer the size.

    So:

    int arreglo[][] = { { 1, 2, 3 }, { 4, 5, 6 } };

    Or if you need the size to be dynamic, you should use the heap.

    int n = 2, m = 3;
// Reservas memoria
int *arreglo = (int*)malloc(n * m * sizeof(int));
// Puedes "mapear" el arreglo como bidimensional mediante algunos cálculos
int pos1 = 1, pos = 2;
int elemento = arreglo[pos1 * m + pos2];
// Liberas la memoria cuando no la usas más
free(arreglo);
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