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 <stdio.h>
#include <stdlib.h>
a, b, arr[3];
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;
}
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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 <stdio.h>
#include <stdlib.h>
int main()
{
int arr[5] = { 1, 2, 3 };
int i;
for (i = 0; i < 5; i++) {
printf ( "arr[%d] = %d\n" , i, arr[i]);
}
return 0;
}
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Output:
arr[0] = 1
arr[1] = 2
arr[2] = 3
arr[3] = 0
arr[4] = 0
Last Updated :
19 Jun, 2018
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