Scanf array limit in C?

Question

I have some little error in C:

Error: expression must have a constant value

I know, that's mean that my limit must have a constant value, but how i can to solve that when i have this situation?

printf("Type limit: ");
scanf("%i",&limit);
int arr[limit];

Thanks.

EDIT:

Ok guys, another problem, sorry if i spam.

    int num,limit,i;
    printf("Type limit: ");
    scanf("%i",&limit);
    int *arr = (int*)malloc(limit*sizeof(int));
    for(i=0;i<limit;i++)
    {
        printf("Type num %i: ",i);
        arr[i] = scanf("%i",&num);
    }
    system("pause");
    return 0;

error 4 error c2109 subscript requires array or pointer type

Answer 1

You should use malloc:

printf("Type limit: ");
scanf("%i",&limit);
int *arr = malloc(sizeof(int) * limit);

Answer 2

Variable-length arrays with automatic storage duration are allowed since C99. In C89, it is not possible to allocate an array with automatic storage duration, size of which is not known at the compile time. Use malloc to allocate it dynamically:

printf("Type limit: ");
scanf("%i", &limit);

int* arr = malloc(limit * sizeof(int));

and don't forget to call free(arr) to deallocate this memory once you don't need it anymore.

---

To your question about initializing this array with values being read from stdin in a loop:

for(i = 0; i < limit; ++i)
    arr[i] = scanf("%i", &num);

reads each value, stores it into num variable and then 1 is assigned into arr[i] since scanf returns "number of input items successfully matched and assigned" (which is 1 in this case). You can read into array elements directly:

for(i = 0; i < limit; ++i)
    scanf("%i", &arr[i]);

Answer 3

C89 and earlier versions of C didin't support run-time sizing of arrays. You need to turn on C99 (or newer) support in your compiler.

If you are using Linux you can either type:

gcc -std=c99

or

c99

to compile code written for c99.

Setting std=c99 flag in GCC

Answer 4

int *arr=malloc( limit*sizeof(int) );

This will allocated sufficient memory in the heap for your array of limit int’s. But this array will be “dynamical” (the size is set at run-time), and it will be your responsibility to “free” this memory when you not longer need it. Your variable arr will be just a pointer to that memory. int arr1[10]; on the other hand separate a memory space for 10 int in the stack, and your variable arr1 is that memory. The compiler need to know the size. If you past it to a function taking int* it will “decay" to int*, that is, a pointer to the first element arr1[0].