Writing and Calling Functions Part A True or False A functio

Writing and Calling Functions

Part A

True or False?

A function has exactly one return statement.

A function has at least one return statement.

A function has at most once return value.

A procedure (with return value void) never has a return statement.

When executing a return statement, the functions exists immediately.

A function without parameters always has sideeffects.

A procedure (with a return value void) always has a side effect.

A function without side effects always returns the same value when called with the same parameter values.

Part B

Consider these functions:

Double F(double x) { return g(x) + sqrt(h(x)); }

Double G(double x) { return 4 * h(x); }

Double H(double x) { return x * x + k(x) – 1; }

Double K(double x) { return2 * (x + 1); }

Without actually compiling and running a program, determine the results of the following function calls:

Double x1 = F(2);

Double x2 = G(H(2));

Double x3 = K(G(2) + H(2));

Double x4 = F(0) + F(1) + F(2);

Double x5 = F(-1) + G(-1) + H(-1) + K(-1);

Part C

Write a procedure sort3(int& a, int& b, int& c) that swaps its three inputs to arrange them in sorted order.

For example:

int v = 3;

int w = 4;

int x = 1;

sort3(v, w, x); //V is now 1, w is not 3, x is now 4

Solution

PART A:

1. A function has exactly only one return statement.

Above statement is false

2. A function has at least one return statement.

Above statement is false;

3. A function has at most return one value.

Above statement is true;

4. A procedure (with return value void) never has a return statement.

Above statement is false;

5. When executing return statement function exits immediately.

Above statement is true;

6. A function without parameters always has side-effects.

Above statement is false

7. A procedure with return value of void always has side-effect.

Above statement is false

8. A function without side effects always returns same value with same parameter values.

Above statement is true

PART B:

Consider these functions:

double F(double x) {return G(x) + sqrt (H(x)); }

double G(double x) {return 4*H(x); }

double H(double x) {return x*x+ K(x)-1; }

double K(double x) {return 2*(x+1); }

1. double x1 = F(2) ;

= G(2) + sqrt (H(2))

= 4*(H(2) + sqrt(H(2)

= 4*9+sqrt(9)

= 36+3

x1 = 39

Where H(2) = 2*2+K(2)-1

= 4+2(2+1)-1

= 4+4+2-1 = 9

2. double x2 = G(H(2));

= G(9)

= 4*H(9)

= 4*100

x2= 400

Where H(9) = 9*9+K(9)-1

= 81+2(9+1)-1

= 81+2*9+1

= 81+18+1 = 100

3. double x3= K(G(2)+H(2));

G(2) = 4*H(2)

= 4*9 = 36

H(2) = 9

K(36+9) = K(45)

= 2*(45+1)

= 2*46

x3= 92

4. double x4 = F(0) + F(1) + F(2);

F(0) = G(0) + sqrt(H(0))

= 4*H(0) + sqrt(H(0))

= 4*(0+1)^2 + 0+1

= 4*1+1 = 5

F(1) = G(1) + sqrt(H(1))

= 4*H(1) + sqrt(H(1))

= 4*(1+1)^2 + 1+1

= 4*4+2 = 18

F(2) = G(2) + sqrt(H(2))

= 4*H(2) + sqrt(H(2))

= 4*(2+1)^2 + 2+1

= 4*9+3 = 39

x4= 5+18+39 = 62

5. double x5 = F(-1) + G(-1) + H(-1) + K(-1);

F(0) = G(-1) + sqrt(H(-1))

= 4*H(-1) + sqrt(H(-1))

= 4*(-1+1)^2 + -1+1

= 4*0+0 = 0

G(-1) = 4*H(-1)

= 4*(-1+1)^2 = 4*0 = 0

H(-1) = (-1+1)^2= 0

K(-1) = 2*(-1+1) = 0

x5 = 0+0+0+0 = 0

PART C:

void sort3(int& a,int& b, int& c)

{

if(a>b) { int temp =a; a=b; b= temp; }

if(b>c) { int temp =b; b=c; c= temp; }

if(a>b) { int temp =a; a=b; b= temp; }

}