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Implicit initialization of variables with 0 or 1 in C
We know that we need to declare variables before using it in our code. However, we variables can be assigned with 0 or 1 without declaration. In the following example we can see this.
Example
#include <stdio.h>
#include <stdlib.h>
x, y, array[3]; // implicit initialization of some variables
int main(i) {
//The argument i will hold 1
int index;
printf("x = %d, y = %d
", x, y);
for(index = 0; index < 3; index++)
printf("Array[%d] = %d
", i, array[i]);
printf("The value of i : %d", i);
}Output
x = 0, y = 0 Array[0] = 0 Array[1] = 0 Array[2] = 0 The value of i : 1
Sometimes, if some array is initialized with few values, then rest of the values will hold 0.
#include <stdio.h>
#include <stdlib.h>
int main() {
//The argument i will hold 1
int index;
int array[10] = {1, 2, 3, 4, 5, 6};
for(index = 0; index < 10; index++)
printf("Array[%d] = %d
", index, array[index]);
}Output
Array[0] = 1 Array[1] = 2 Array[2] = 3 Array[3] = 4 Array[4] = 5 Array[5] = 6 Array[6] = 0 Array[7] = 0 Array[8] = 0 Array[9] = 0
Updated on: 30-Jul-2019
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