# Implicit initialization of variables with 0 or 1 in C

In C programming language, the variables should be declared before a value is assigned to it.
For Example:

```   // declaration of variable a and
// initializing it with 0.
int a = 0;

// declaring array arr and initializing
// all the values of arr as 0.
int arr[5] = {0};
```

However, variables can be assigned with 0 or 1 without even declaring them. Let us see an example to see how it can be done:

 `#include ` `#include ` ` `  `// implicit initialization of variables ` `a, b, arr[3]; ` ` `  `// value of i is initialized to 1 ` `int` `main(i) ` `{ ` `    ``printf``(``"a = %d, b = %d\n\n"``, a, b); ` ` `  `    ``printf``(``"arr[0] = %d, \narr[1] = %d, \narr[2] = %d,"` `                ``"\n\n"``, arr[0], arr[1], arr[2]); ` ` `  `    ``printf``(``"i = %d\n"``, i); ` ` `  `    ``return` `0; ` `} `

Output:

```a = 0, b = 0

arr[0] = 0,
arr[1] = 0,
arr[2] = 0,

i = 1
```

In an array, if fewer elements are used than the specified size of the array, then the remaining elements will be set by default to 0.
Let us see another example to illustrate this.

 `#include ` `#include ` ` `  `int` `main() ` `{ ` `    ``// size of the array is 5, but only array[0], ` `    ``// array[1] and array[2] are initialized ` `    ``int` `arr[5] = { 1, 2, 3 }; ` ` `  `    ``// printing all the elements of the array ` `    ``int` `i; ` `    ``for` `(i = 0; i < 5; i++) { ` `        ``printf``(``"arr[%d] = %d\n"``, i, arr[i]); ` `    ``} ` ` `  `    ``return` `0; ` `} `

Output:

```arr[0] = 1
arr[1] = 2
arr[2] = 3
arr[3] = 0
arr[4] = 0
```

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