C Programming - Arrays - Discussion

arr[1]={10};
Is the same as
arr[3]={5,69};
The first example is saved in memory like this;

arr[0]=10 and arr[1]=0 why equal to zero this is referred to the basics of the array you initialize the array in the same line of declaration the non initialized elements will be initialized by def to zero

The second example will be saved in memory like this;
arr[0]=5, arr[1]=69 and arr[2]= 0
And also to print contents of any array using for loop as a example
for (int i=0;i<3;i++)
printf("%d\n", arr[i]); //or printf("%d\n", i[arr]);

I hope you understand my simple explanation

Here, arr[1]=10;

It means that the value 10 is initialized to an array arr[] with index 1. The values arr[0] and arr[2] is not initialized if we want to access those elements then it will give the garbage values.

for(i=o;i<n;i++)
printf("%d",a[i]);
and
for(i=0;i<n;i++)
printf("d",i[a]);
both gives the same value...
Someone please clarify this problem with how address calculation happens?

@MOON. It will be 6 if you will give 7 then it will print garbage value.

int main()
{
int arr[7] = {1,2,3,4,5,6,7};
printf("%d\n",6[arr]);
return(0);
}

Output: 7

How is this possible?

#include<stdio.h>

int main()
{
int arr[1]={10};
printf("%d\n", 0[arr]);
return 0;
}


Someone plz clarify.

It is the same as declaring an array of a number of variables. Here it is an array of size 1. The usual syntax is used for initialising the array.

#include<stdio.h>
int main()
{
int arr[7] = {1,2,3,4,5,6,7};
printf("%d\n",7[arr]);
return(0);
}

why it return's 2?

arr[i] = i[arr].
So it is printing 10
Here it is given arr[1] = {10} that means arr[1] = {10,0}
So arr[0] = 10 and arr[1] = 0;

The first thing is that array is of size 1.

arr[i]=*(arr+i)

i[arr]=*(i+arr)

So, 0[arr]=arr[0]

That's why answer is 10.

arr[1]=10 means {0,10},
or arr[1]=10 means {10,0};
which one is true?
and arr[1] has length 1 or 2;