0
#include<stdio.h>
#include<conio.h>

int f(int & ,int );//function prototype

main()
{
  int x,p=5; //p is initialized to 5
  x=f(p,p);
  printf("\n Value is : %d",x);//print the value of x
  getch();
}

int f (int & x, int c)
{
c=c-1;
if (c==0) return 1;
x=x+1;
return f(x,c) * x; //recursion
}

output : 6561

can anyone explain me the flow the program This question is from gate i couldn't understand it. It seems that the function is called with value of p = 5. It is catched in the function f by int &x the problem is here. Is the value i.e 5 is stored in x or in address of x.

Massimiliano
  • 7,842
  • 2
  • 47
  • 62
Swapnil
  • 1,004
  • 1
  • 13
  • 21

6 Answers6

7

This code uses a C++ reference, which is what the int & syntax means.

A reference is, basically, syntactic sugar for a pointer. So when you call f(p, p), the function argument x is a reference to p in main(), while c is merely a copy of the value at the time of the call.

This means f can change the value of p in main(), through the reference. Since f calls itself recursively, passing the same reference toitself, it's always a reference to the p in main().

To track the recusion, I would suggest adding logging print-outs at strategic places inside f().

unwind
  • 391,730
  • 64
  • 469
  • 606
0

The first creates a reference which is more like a cleaner syntax instead of a pointer.

You can use it to change contents of variables among calls to function.

Although you must note, the code you have posted is not valid C++. C++ does not allow main without a return type.

Aniket Inge
  • 25,375
  • 5
  • 50
  • 78
0

More simple:

int f(int &a){
  a = 5;
}
int x = 0;
f(x);
//now x equals 5

int f2(int b){
 b = 5;
}
int y = 0;
f2(y);
//y still equals 0
Marc-Emmanuel Ramage
  • 290
  • 1
  • 4
0

Actaully it is like this the reference part in the question was to distract u it is just x pointed to the address of the the p but the 'x' itself contained the value 5 at the first time but the recurrsion ran for 4 times so in the stack it was like ----[x when value 6(bottom of the stack), xwhen value 7, xwhen value was 8 , x when value is 9(top of the stack)] ...at the top it was 9 but as we know that in recurrsion first loop execution takes place and then the evaluation so when the value of x changed to 9 then all the four block of the stack changed to 9 since they pointed to the same 'x'....so it came out to be 9*9*9*9=6561 ..as simple as that.

garveet parmar
  • 1
0
  • & is reference: it means it must be an object's reference (address);
  • * is a pointer: it points to a object.

The difference is that "&x" can not be "NULL", but &x can be NULL.

These two prototype are different.

int f(int & ,int );
int f(int * ,int ); //are different function prototypes

This is a good question.

Sterconium
  • 559
  • 4
  • 20
user12319994
  • 1
-3

The code is calculating (2P-1)^(P-1) Try various values for P to find the pattern. Here, P=5; so it yields 9^4 = 6561.

پویا پاکاریان

پویاپاکاریان

پویا پاکاریان
  • 1
  • 1