Question

) The following functions show() and calling() are a part of some class. Assume that the parameter n is greater than 1 when the function is invoked. It returns the value 1 when true otherwise it returns 0. Show the dry run/working.
void calling()
{
int f=2;
show(n,f);
}
int show(int n, int f)
{
if(n==f)
return 1;
if(n%f==0 || n==1)
return 0;
else
return(show(n,f+1));
}
i) What will the function show() return when the value of n is 11? [ 2 ]
ii) What will the function show() return when the value of n is 27? [ 2 ]
iii) In one line state what function show() is doing? [ 1 ]​

Answer 1

usuquzbiw zja zahzahz h

bwxhsxbpsxskxbie xiwz

Explanation:

gnlqNoqxuwvxuevuxeuvxbuexbowonzibaxv xqxkwlnzaz

Answer 2

Answer:

i) n%f==0

11%2==0

1+2=3

ii) (27%2==0)

1 ans

because function show(n,f+1) so 1 is ans

Explanation:

thks