What will be the output of the program?
#include<stdio.h>
int main()
{
int arr[1] = {10};
printf("%d", 0[arr]);
return 0;
}
#include<stdio.h>
int main()
{
int arr[1] = {10};
printf("%d", 0[arr]);
return 0;
}A. 1
B. 0
C. 10
D. 6
E. None of these
Answer: Option C
Solution(By Examveda Team)
>> int arr[1]={10}; The variable arr[1] is declared as an integer array with size '2' i.e. arr[0] and arr[1] and it's first element is initialized to value '10'(means arr[0]=10)
and arr[1] = garbage value or zero
>> printf("%d", 0[arr]); It prints the first element value of the variable arr.
Hence the output of the program is 10.
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Comments ( 2 )
Related Questions on Arrays and Strings
#include
int main()
{
int arr[1] = {10}; // ={10,0);
printf("%d", 0[arr]); // *(arr+0)=arr[0]=10;
return 0;
}
output: 10
#include
int main()
{
int arr[1] = {10};
printf("%d", 1[arr]); // if *(arr+1)=arr[1]=0
return 0;
}
output: 0
how can the size be 2?
even if size is 2 then elements are arr[0]=0
arr[1]=10
so first element should be 0