一维数组在函数中被视为二维数组 - C

Question

Please tell me if the title is not entirely correct.

I first initialize a pointer to array using malloc in the main, then I pass it to a function which receives a pointer to pointer to int like this:

void readfile(int**);

int main(void)
{
  int (*obsF) = malloc(sizeof(int)*16);
  readfile(&obsF);
  free(obsF);
}

void readfile(int **obsF)
{
  //I can do this without errors:
  obsF[0][1] = 1;

}

I don't understand why I have to treat it as a 2D array, because if I do obsF[1] = 1 I get an error. Could someone explain me why this happens?

Answer 1

Change your function signature from void readfile(int**) to void readfile(int*) and from void readfile(int **obsF) to void readfile(int *obsF) .

When you put the function signature as int**, the compiler treats it as a pointer to a pointer, and thus it is treated as a 2D array, and not a 1D array.