Create an array based on the input size

Question

I am pretty new to C and I am trying to read the user input (some sentence or string in general) and then I want to create an array based on the input lenght. Is there a reasonable way to do it? Thanks for answers

Answer 1

malloc/free, in the case of strings a strlen will get you it's length.

Answer 2

Just for an overview of why all the answers are suggesting pointers instead of arrays:

When I was learning C one thing that helped was to understand arrays and pointers and how similar they are.

For the most part, they can have the same syntax, you can use * syntax with either or you can use [] syntax with either.

The differences are:

1) Arrays have memory allocated for them by the system and pointers don't, you have to "set" a pointer to some memory that you have allocated.

2) I don't think arrays can change where arrays point, they always point at their pre-allocated spot.

Since arrays are pre-allocated and can't be repointed, you want a pointer. You can treat it exactly as an array (You can use [] syntax) but you have to allocate memory for it first.

So for example, if a array with and p is a pointer, a[0]=1, *a=1, p[0]=1 and *p=1 are all identical functions, and while *++p=1 is valid, I don't think *++a=1 is valid because you can't change where a points.

So the short version would be, you need a pointer, not an array, and to change how much is allocated, you allocate the new size (With malloc or something similar), copy what you want to retain over and free the old space (Or you might be able to increase the size of the first one--realloc?, not sure, my C is decades old)

Answer 3

You can use malloc to allocate new memory, Note that since C's memory isn't managed (contrary to Java, Python or any other high level language), you will have to free the memory once you are done using it.

int arr_size = 0;
int* arr;
printf("Please enter a size to the array:");
scanf("%d", &arr_size);
arr = malloc(arr_size * sizeof(int))
// Use array
free(arr);

void *malloc(size_t size);

The malloc() function allocates size bytes and returns a pointer to the allocated memory. The memory is not initialized. If size is 0, then malloc() returns either NULL, or a unique pointer value that can later be successfully passed to free().

Answer 4

It depends on the standard you're compiling against and your compiler. You can only rely on variable length arrays in C99

The only way to be certain is to use malloc, though you need to ensure you free the memory afterwards:

int length;

// Do something to set the size

// Allocates a contiguous block of memory that is
// (length * size of a char primitive) in length
char *array = (char *)malloc(length * sizeof(char));

// Do whatever you need do to with the array

free(array);

In C, declaring a variable as a pointer (char *a) and as an array (char a[3]) allows you to use that variable in exactly the same way. The only difference is that with a pointer you need to allocate and free the memory yourself, while with the array that block of memory is given to you automatically and it is freed when it goes out of scope.

With the code above, you can still access each individual character via an index like so:

array[0] = 'f';
array[1] = 'o';
array[3] = 'o';