C – How to Initialize All Elements of an Array Separately

arraysc++initialization

when we write something like this

int arr[5] = 0; or int arr[5] = {0};

there is no problem

but when we do something like this

int arr[5];
arr[5] = {0};

an error occurs. Any explanation for this ?

Best Answer

It is simply a part of the language definition that arrays can be initialised, but not directly assigned. You can do what you want in C99 using memcpy() with a compound literal:

int arr[5];

/* ... */

memcpy(&arr, &(int [5]){ 0 }, sizeof arr);

With GCC's typeof extension, you can add a little more safety:

memcpy(&arr, &(typeof(arr)){ 0 }, sizeof arr);

In C89 you must give the source array a name:

{ static const int zero[5] = { 0 }; memcpy(&arr, &zero, sizeof arr); }

Related Solutions

C Arrays – Declaring and Initializing Arrays in C

In C99 you can do it using a compound literal in combination with memcpy

memcpy(myarray, (int[]) { 1, 2, 3, 4 }, sizeof myarray);

(assuming that the size of the source and the size of the target is the same).

In C89/90 you can emulate that by declaring an additional "source" array

const int SOURCE[SIZE] = { 1, 2, 3, 4 }; /* maybe `static`? */
int myArray[SIZE];
...
memcpy(myarray, SOURCE, sizeof myarray);

C – Possible to Initialize an Array After Declaration in C?

You cannot assign to arrays so basically you cannot do what you propose but in C99 you can do this:

CGFloat *components;
components = (CGFloat [8]) {
    0.0, 0.0, 0.0, 0.0,
    0.0, 0.0, 0.0, 0.15
};

the ( ){ } operator is called the compound literal operator. It is a C99 feature.

Note that in this example components is declared as a pointer and not as an array.

Best Answer

Related Solutions

C Arrays – Declaring and Initializing Arrays in C

C – Possible to Initialize an Array After Declaration in C?

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