Question

1. Which exemplifies C's weak type checking? (a) the ability to assign pointers of any type to/from a (void *) variable without warning (b) the requirement that all variables be declared before being used (c) the static nature of variable typese.g., once declared an int, always an int (d) the fact that data type widths (e.g., for ints) can be dependent on the platform 2. Consider the following macro definition and variable declaration #define F00(x) (2 * x-x) int val 10; 5)? What is the value of the expression F00 (val (a) 15 (b) 20 (c) 25 (d) 35 3. Consider the following variable and function definitions int g = 10; int q30 static int g = 5; return ++g; int q40 [ extern int g; return ++g; int q5O int g = 1; return ++g; What is the value of the expression q3O (a) 32 (b) 34 (c) 38 (d) 40 30 q4) q4) q5 q5)?

Answer 1

`Hey,

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1) What exemplifies C's Weak type checking?

The compiler cannot statically check most of the uses of void/* . void* may change to a pointer of any sort of type without any cast (Which applies on to C, And not C++) which is another weakness. C is taken into account to be typed fragile. As a result of which you will be able to convert any type to the other type with the cast and without a compile error.

So, Option A is the answer. i.e, The ability to assign pointers of any type to/form a (void *) variable without warning.

2) What is the value of the expression F00(val + 5)?

Considered the following macro definition and variable declaration : -

#define F00(x) (2 * x - x)

int val=10;

Answer

First option is in the given question let us test with 15 first. So x is 15,

F00(val + 5) = F00(15)

therefore x=15

So,

F00(15) = (2 * 15 - 15) = 15

Condition satisfies and the answer is Option A.

The condition does not satisfy for rest of the options.

3) What is the value of the expression q3() + q3() + q4() + q4() + q5() + q5() ?

Here, q3() is 6

q4() is 11

q5() is 2

Therefore, q3() + q3() + q4() + q4() + q5() + q5()

6+6+11+11+2+2 = 38.

Option C = 38 is the answer.

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