C programming

Question

#include <stdio.h>
int f(int *a,int b) {
  b = b-1;
  if(b == 0) return 1;
  else {
    *a = *a+1;
    
  return  *a + f(a,b)  ;
  }
}
int main() {
  int X = 5;
  printf("%d\n",f(&X,X));
}

Predict the output

If the return statement was

return  f(a,b)+ *a ;

What would have been the output here ?  Explain why there is change in output (if any)

Comments

my approach,,

work till the last function ,which will be return (10+1)=11 so 11 will written to the adress ,so 11+11=22

then 22+22

then 44+44 then 88+88=176

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I think, it should be  6+7+8+9 + 1. ??

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Roughly im getting 37 But,@uddipto can u pls show how exactly this implementation would be ?

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I am getting 31.
@pC you will get the output as 37 if the return statement is "return(fun(a,b)+a);"

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Yes, i made mistake,also getting 31...

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return(fun(a,b)+*a);

Here it is 31

return(*a+fun(a,b));

Here it is 37 . But why ?  What I thought was the value a in *a+ was resolved while returning back

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The value of a is calculated while going into the depth of the tree.

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@GateSet,

#include <stdio.h>
int foo(static int n) 
{
	static int r;
	if(n<=1)
		return 1;
	r=n+r;
	return(r+foo(n-2));
} 
int main () {
   printf("value : %d",foo(5));
}

Like you said if we are computing a while going into depth , What would be the output here :
Moreover in this code there is no change in output even if the return statement is

return(foo(n-2)+r);

Why ?

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You all have solved previous GATE papers? A similar question was asked in 2013 or 2012 and explanation is given here only. Sad to see that no one knows it :(

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Yes @Arjun sir, it had * instead of +. Same type.

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@pC

The program posted by you is using static variable.

When the program has "return(r+foo(n-2));", we are calculating r and then calling (r+foo(n-2)).

When the program has "return(foo(n-2)+r);", when we backtrack, the local opy of r at each procedure is added to foo(n-2).

In the program posted by uddipto, return(a+fun(a,b)) has same effet as your case 1.

But during return(fun(a,b)+a)), while we backtrack, we try to add the local copy of a, but its already modified by the last call, because it is call by reference.

I have tried to explain what is happening, maybe I am not that clear!!

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The difference is made by call by reference actually. While call by reference changes the value permanently at the address, static does nothing except to retain the value of the variable till the last procedure call. When you rollback, you get the same local copy at each function, which you get while going into the depth.

But while you rollback during call by reference, variable is now changed permanently.

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I think I have missed that Gate Question while solving previous questions :(
Can someone share the link to it

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https://gateoverflow.in/60/gate2013_42

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Undefined behaviour !! :(

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@GateSet

• When the program has "return(r+foo(n-2));", we are calculating r and then calling (r+foo(n-2)).
• When the program has "return(foo(n-2)+r);", when we backtrack, the local opy of r at each procedure is added to foo(n-2). ie  foo(n-2) calcuclated first then r is added

foo(n-2)+r

foo r+(n-2)

Do you mean to say this ? 

In both base output will be 17 when I run on machine

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It completely depends upon the language as to how the operation '+' is implemented.

i.e. if the operation is (a+b) then it depends upon the language whether to compute 'a' first or 'b' first.

 

Source: Dennis-Ritchie

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Ok, so I think my explanation for the call by reference case is wrong!!

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@Sushant_Gokhale  Why don't you convert this into answer  (with some reference/ atleast pic of the portion of statement from Dennis Riche for users who read this thread )

Both case it is undefined behaviour

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@pC we should calculate the value of a variable in these type of cases while coming back right?

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@Pc. Will try to get the picture of the same

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@GateSet , No What @Sushant_Gokhale said is absolutely right !

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Have posted the answer.

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So, Order of evaluation is undefined. That's it. And no other error ? right ? sushant ?

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yes here order of evaluation is not maintained. But give some result.rt?

Answer 1

See this article from Dennis Ritchie:

REF: Right below the precedence table of the ANCI book here

Comments

Understood!!! Thanks!!