Lecture 5: Macros and Global Variables #

What is the output of the following program?

#include<iostream>
using namespace std;

int x = 1;

void print()
{
    int x = 2;
    {
        int x = 3;
        cout << x << endl;
    }
}

int main()
{
    print();
    return 0;
}

Options

a. 1
b. 2
c. 3
d. Error

c. 3
We declared variables at the innermost scope as the variables in the outer scope(s), hence the outer ones were 'shadwoed',
hence only variable in the nearest scope was printed. 3 in this case, inside a code-block.

What is the output of the following program?

#include <iostream>
using namespace std;

#define MULTIPLY(a, b) a*b

int main()
{
    cout << MULTIPLY(2+3, 3+5);
    return 0;
}

Answer

____

2 + 3*3 + 5 = 16
"MULTIPLY(a, b)" will be replaced by "a*b" in code. So, MULTIPLY(2+3, 3+5) will be replaced by 2+3*3+5. And according to operator precedence, multiply operator (*) has higher precedence than plus operator (+). So, 3*3 will be evaluated first. Hence expression will become : 2+9+5 = 16

What is the output of the following program?

#include <iostream>
using namespace std;

#define SQUARE(x) x*x

int main()
{
    int x = 36 / SQUARE(6);
    cout << x;
    return 0;
}

Answer

____

After the #define directibve is completed.
We get int x = 36/6*6 = 36 / 6 * 6 = (36/6)*6 = 36 // paranthesisation is done by the compiler.
So 36 is printed.

Lecture 6: Inline and Default Arguments #

Inline functions are useful when ____

Options

a. Function is large with many nested loops
b. Function has many static variables
c. Function is small and we want to avoid function call overhead.
d. None of the above

c. Function is small and we want to avoid function call overhead.
Obvious.

What is the output of the following program ?

#include<iostream>
using namespace std;

int getValue(int x = 0, int y = 0, int z)
{
    return (x + y + z);
}

int main()
{
   cout << getValue(10);
   return 0;
}

Options

a. 10
b. 0
c. 20
d. Compilation Error

c. Function is small and we want to avoid function call overhead.
Obvious.

What is the output of the following program ?

Which of the following statement is correct?

Options

a. Only one parameter of a function can be a default parameter.
b. Minimum one parameter of a function must be a default parameter.
c. All the parameters of a function can be default parameters.
d. No parameter of a function can be default.

c. All the parameters of a function can be default parameters.
There's no restriction unless specified.

Lecture 6: Constant Variables #

What is the output of the following program ?

#include <iostream>
using namespace std;

int main()
{
    int  const  p = 5;
    cout << ++p;
    return 0;
}

Options

a. 5
b. 6
c. Error
d. Garbage

c. Error.
Basics. Non change at the address in symbol table.

What is the output of the following program ?

#include <iostream>
using namespace std;

int main()
{
    int p = 5;
    int const *q = &p;
    p++;
    cout << p << endl;
    return 0;
}

Options

a. Error
b. 5
c. 6
d. None

c. 6
Rule 1: Memories cannot be tagged const.
Hence we can make a change through other paths(which just the old variable here).

Which statement(s) will give an error for the following code -

#include <iostream>
using namespace std;

int main()
{
    int p = 5;
    int const *q = &p;
}

Options

a. p++;
b. q++;
c. (*q)++;

c. (*q)++;
a. and b. will run without errors. Please read the declaration from right to left, we are protecting a int(through q), not the value we store in our pointer q. No restriction on the value of q.

p is anyway indepenedent.

Which statement(s) will give an error for the following code -

#include <iostream>
using namespace std;

int main()
{
    int p = 5;
    int * const q = &p;
}

Options

a. p++;
b. q++;
c. (*q)++;

b. q++
When the declaration is read, then we can see that we are protecting q from change, so memory at the address is changeable.

p is anyway independent.

Which statement(s) will give an error for the following code -

#include <iostream>
using namespace std;

int main()
{
    int p = 5;
    int const * const q = &p;
}

Options

a. p++;
b. q++;
c. (*q)++;

b. (*q)++ and c. q++
Reading the declaration, both the CHANGE OF q and the VALUE AT q are const.

p is anyway independent.

Which statement(s) will give an error for the following code -

#include <iostream>
using namespace std;

int main()
{
    const int p = 5;
    int const *q = &p;
}

Options

a. p++;
b. q++;
c. (*q)++;

a. p++ and c. (*q)++
Value at p is const. So p++ will give and error.

Value pointed by q is also const, so changing value through q is also an error.

But changing q's value is still allowed.