Consider the following code:

function modify(a,b)
{
integer c, d = 2
c = a*d + b
return c
}

function calculate( )
{
integer a = 5, b = 20, c
integer d = 10
c = modify(a, b);
c = c + d
print c
}

Assume that a and b were passed by value. What will be the output of the function calculate( ) ?

• at first it goes to function c = modify(a,b);
  passing a=5 and b=20;
  c=a*d+b; //where as d is a local variable which is located in modify function.
  c=5*2+20;
  c=30
  and it returns to caluclate() function
  and c=c+d;//here d is declared as d=10;
  c=30+10;
  c=40;
  
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• 40
  the function follows the value in the given value in respective function
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• In claculate function there is another function i.e modify(a,b).....so it goes to modify( ) i.e....c=a*d+b;
  c=5*2+20=30
  c=20..it returns c value
  here again c=c+d..i.e
  c=30+10=40
  therefore ans ..is 40...
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• modify(5,20)
  c = 5*2+20 = 30
  c = 30 + 10
  c = 40
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• Since c is declared inside the function. It's scope will end at the function exists so the value of c would become zero ans is 10
  
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• output of function calculate will be 40
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• statement return c returns value 30. then c=30 and d=10. final output of funcction calculate() is 40
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• 40 is the correct answer
  
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• C value=40
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• c=40
  bcz when function modify is called the value of c=5*2+20=30 and its value is returned to calling function in main.then we calculate c by takng d=10 bcz in main the value of d=10 due to local scope and calculate c=30+10=40
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• 40 i guess
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• c=modify(5,20)
  modify(5,10)
  {
  
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• ans will be 40
  
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• c=40
  when a=5 and b=20 pass through modify function,c will get calculated as (5*2)+20
  and c=30 is returned
  which is added with the value of local function calculate c=30+10
  i.e. c=40
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• value of c 40
  
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• Ans will be 40
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• ans is 40
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• ans is 40
  
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