GATE CSE 2016 | Set 1 | Question: 12

Question

Consider the following "C" program.

void f(int, short);
void main()
{
    int i = 100;
    short s = 12;
    short *p = &s;
    ____________;  // call to f()
}

 Which one of the following expressions , when placed in the blank above, will NOT result in a type checking error?

• A. $f(s, *s)$
• B. $i = f(i,s)$
• C. $f(i, *s)$
• D. $f(i, *p)$

Comments

.

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the function f wants integer and short value not any address.

option a and c is wrong as s is a short variable not a pointer.

return type of b is wrong

option  d is right as it returns the content at location p which is a short data type.

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A is incorrect because *s will produce segmentation fault, since *s is pointing at some random address.

B is incorrect because the return type of f is void not int.

C is incorrect, same reason as A.

D is correct bcoz i is an integer value and p is short pointer that's why *p is the short value.

#GOCLASSESOP

Answer 1

• A. $f(s, *s) - 1^{\text{st}}$ argument is short whereas the function expects an int. But in C language short gets implicitly converted to int during a function call and so this won’t be a type error. But second argument is a pointer (can be $64$ bits on a $\textsf{x64}$ machine) where as the function expects a short (can be even $16$ bits). So this will generate a type error. So, WRONG.
• B. $i = f(i,s)$ - return type is not void. So, WRONG.
• C. $f(i, \ast s) - 1^{\text{st}}$ argument is int, second is again syntax error. So, WRONG
• D. $f(i, \ast p)$ - Both the arguments and return type match. $p$ is a pointer to short, so $\ast p$ is value of short. So, D is ANSWER.

Comments

in C it must be semantic error , as it is a type mismatch. but syntax is correct there.

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@Pranavpurkar yess correct.

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The variable s is a standard short integer, not a pointer. You can only use the dereference operator (*) on a pointer variable. Attempting to dereference a non-pointer variable causes a compiler error.

Answer 2

By option elimination method:
Options A & C are wrong because *s is an invalid argument as s is not a pointer

Option B is incorrect because the f function return type is void while in option it shows of type int implicitly.

Option D is correct because f(100,12) is a valid function call for f(int, short).

Comments

What if it would be (i,&s) ??

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Then it will return f(100,2000) as per previous comment assumptions. Therefore it would give error as short is a variable holding a value not address.

Correct me if I'm wrong

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explain something in  Option B which is incorrect because of the f function return type is void while in option it shown of type int implicitly.

Answer 3

Let's check each option.

A. f(s, *s)

• s is a short, which can be promoted to int for the first parameter.
• *s is invalid because s is not a pointer.
• Type error.

B. i = f(i, s)

• f(i,s) is valid as a function call, but f returns void.
• Assigning a void value to i is illegal.
• Type error.

C. f(i, *s)

• i matches int.
• *s is invalid because s is not a pointer.
• Type error.

D. f(i, *p)

• i is int → matches first parameter.
• p is short *, so *p is a short → matches second parameter.
• No type checking error.

Answer: D. f(i, *p)

Answer 4

So option d is correct.