C Programming - Pointers - Discussion

for(i=0; i<n; i++)2, 4, 6, 8, 10};
*(b+1) = *(b+i)+5;
In this for loop,only (b+1) location value is updated but not entire array
i.e 1st iteration (b+1)=(b+0)+5<=>2+5=7
2nd iteration (b+1)=(b+1)+5<=>4+5=9
after 5 iterations, the final value that is in (b+1) is
*(b+1)=*(b+4)+5<=>10+5=15;
change() will return an array *b whose *(b+1) location is updated by 15 Hence array a[] will hold values 2,15,6,8,10. since array a[] is passed into array b[](using pass-by-address) hence any changes made in b[] will reflect back in array a[].

Can anyone explain me the concept clearly. I could not understand. What is the value of b?

Only a[1] value is changed. Reaming arrays value are same,

*(b+1) = *(b+i)+5;

b is base address +1 means assign the value to b[1];

Every time, each row is increment by 5 and assign to b[1].

At last value increment with five and finally overwrite value 10+5=15.

Answer is [B]. 2, 15, 6, 8, 10.

Because in the above code there are few mistakes.

1) There is no link between main program and function. i.e there is no prototype.

2) Return type is void.

*(a+i)=a[i]. that should give the address of a[i] right? why is it giving the value?

Answer must be option a function definition is perform and return the value to the function call so it should be a option (A).

for(i=0; i<n; i++)
*(b+1) = *(b+i)+5; /*( b[1]=b[i]+5 ) . */

End time loop.
b[1] = b[5]+5.
b[1] = 10+5.
b[1] = 15.

Ans = 2, 15, 6, 8, 10.

Change the position of a[i] with respect to the position of the second element of the array.

*[b+1] = *[b+i]+5;
*[a+1] = *[b+0]+5;
a[1] = b[0]+5;
a[1] = 2+5;
a[1] = 7;

Now it will continuous till the b[i] = 10.
So a[1] = 15;

For other it will remain same.
2 15 6 8 10.
a[0] a[1] a[2] a[3] a[4].

#include<stdio.h>
int main(){
int a=5,b=10,c=15;
int *arr[]={&a,&b,&c};
printf("%d",*arr[1]);
return 0;

What is the o/p of above program?