i have a question of integer array 's enter

March 28, 2023

int num = 0, k = 0;

scanf("%d %d", &num, &k);

int* A = &num;

    for (int i = 0; i < num; i++)
    {
        scanf("%d", &A[i]);//scanf("%d", A+i);
    }

int that code, i want
4 5 1 3 2 integers put in the array A

input is

5 7

4 5 1 3 2

but int debug mode,there are

      *A    4           int
  *(A+1)    5           int
  *(A+2)    1           int
  *(A+3)    3           int
  *(A+4)    -858993460  int
  *(A+5)    -858993460  int

why is the *(A+4) not 2??????

i tried the num+1, but that is not ultimate solution.

for (int i = 0; i < num+1; i++)
{
    scanf("%d", &A[i]);
}

>Solution :

To define an array, you’d do

int A[num];

Your variable A is a pointer to an integer, not an array. You can treat it as an array, as arrays are basically pointers in C. But the memory you’re writing from scanf() is not ‘allocated’.