Question

Let’s say that any C function has a pointer already declared, but not assigned any value yet. We will int for our examples.

int *ptr;  

The goal of the function is not to assign ptr any dynamic memory on the heap, so no malloc call. Instead, we want to have it point to an array of fixed size n. I know I could accomplish this like so:

int arr[n];
ptr = arr;

However, the code could get very messy and hard to read if we need to do this many times in a function, ie, a struct of many pointer fields all need to point to an array of fixed length. Is there a better way to accomplish this in one line? I was thinking of something similar to below, but it looks too ambiguous and uncompilable:

int *ptr;
// Many other things happen in between...
ptr[n];

***EDIT***
Here, the below additional information may help guide some more answers (not saying that the current answers are not fine). In my use case, the pointers are declared in a struct and, in a function, I am assigning the pointers to an array. I want to know if there is a simpler way to accomplish this than in the below code (all pointers to point to fixed-length array):

struct foo {
    int* a;
    short* b;
    char* c;
    ...
};

void func(void) {
    struct foo f;
    int n = ...;
    int tempArr1[n];
    f.a = tempArr1;
    short tempArr2[n];
    f.b = tempArr2;
    char tempArr3[n];
    f.c = tempArr3;
    ...
}

Answer 1

You cannot declare an array and assign it to an existing pointer in a single declaration. However, you can assign an array pointer to a newly declared pointer, like this:

int arr[n], *ptr = arr;

If you insist on staying within a single line, you could use an ugly macro, like this:

#define DECL_ASSIGN_INT_ARRAY(name,size,pointer) int name[(size)]; pointer = name;

The clarity of this one-liner is far lower than that of a two-line version from your post, so I would keep your initial version.

EDIT (in response to the edit of the question)

Another option is to create an unused pointer variable in a declaration, and assign your pointer in an initializer, like this:

void func(void) {
    struct foo f;
    int n = ...;
    int tempArr1[n], *tempPtr1 = f.a = tempArr1;
    short tempArr2[n], *tempPtr2 = f.b = tempArr2;
    char tempArr3[n], *tempPtr3 = f.c = tempArr3;
    ...
}