Explain the following statements with underline in the program and write out the

Question

Explain the following statements with underline in the program and write out the result of the program. (hint: using pointers in c++ to ans)
(a)
int x = 30;
int* p = &x; (this statement is underline)
cout << "&x is " << &x << endl;   (this statement is underline)
cout << "*p is " << *p << endl;    (this statement is underline)
cout << "p= " << p << endl;   (this statement is underline)
cout << "&p= " << &p << endl;  (this statement is underline)
int y = 10;
p = &y;   (this statement is underline)
cout << "&y is " << &y << endl;
cout << "*p is " << *p << endl;
cout << "p= " << p << endl;
cout << "&p= " << &p << endl;
(b)
double x = 3.5;
double* p1 = &x;   (this statement is underline)
cout << "&x is " << &x << endl;   (this statement is underline)
cout << "*p1 is " << *p1 << endl;  (this statement is underline)
cout << "p1= " << p1 << endl;   (this statement is underline)
cout << "&p1= " << &p1 << endl;   (this statement is underline)
double y = 4.5;
double* p2 = &y;
cout << "&y is " << &y << endl;
cout << "*p2 is " << *p2 << endl;
cout << "p2= " << p2 << endl;
cout << "&p2= " << &p2 << endl;
(c)
string s = "abcd";   (this statement is underline)
string* p = &s;   (this statement is underline)
cout << "&s is " << &s << endl;  (this statement is underline)
cout << "*p is " << *p << endl;    (this statement is underline)
cout << (*p)[0] << endl;    (this statement is underline)
cout << "p= " << p << endl;   (this statement is underline)
cout << "&p= " << &p << endl;   (this statement is underline)

Explanation / Answer

Hi,

Below are the solutions to your questions:

C++ allows us to perform either one of these steps independently on a variable with the & and * operators:

1. &x evaluates to the address of x in memory.

2.*( &x ) takes the address of x and dereferences it – it retrieves the value at that location in memory. *( &x ) thus evaluates to the same thing as x.

Question 1:int x = 30;
int* p = &x; (this statement is underline)
cout << "&x is " << &x << endl;   (this statement is underline)
cout << "*p is " << *p << endl;    (this statement is underline)
cout << "p= " << p << endl;   (this statement is underline)
cout << "&p= " << &p << endl;  (this statement is underline)
int y = 10;
p = &y;   (this statement is underline)
cout << "&y is " << &y << endl;
cout << "*p is " << *p << endl;
cout << "p= " << p << endl;
cout << "&p= " << &p << endl;

Answer:

1.cout << "&x is " << &x << endl;   (this statement is underline)

This statement gives the address of  memory location where the variable 'x' is stored.

output:&x is 0xbfebd5c0(say)

2.cout << "*p is " << *p << endl;    (this statement is underline)

This statement int *p declares the pointer to an integer value, which we are initializing to the address of x. The value stored in a pointer p can be accessed through the dereferencing operator “*”.

Output:*p is 30

3.cout << "p= " << p << endl;   (this statement is underline)

This statement prints the address stored in pointer 'p' variable

output:p= 0xbfc601ac (say)

4.cout << "&p= " << &p << endl;  (this statement is underline)

This statement gives the address of pointer p.

Output:&p=13(say)

5.p = &y;   (this statement is underline)

Currently the contents of y=10 and &y gives the address of variable y which may be say 0xbfebd5b6 so now statement p=&y assigns the address of varaible y to the variable p

Statements:

cout << "&y is " << &y << endl;

cout << "*p is " << *p << endl;

cout << "p= " << p << endl;
cout << "&p= " << &p << endl;

Ouput:

1.&y is 0xbfebd5b77(say)

2.*p is 10(* dereferences the value of varaible y)

3.p=0xbfebd5b77 (address of y which is 0xbfebd5b77)

4.&p=0xbef77(address of pointer p)

Question 2:

double x = 3.5;
double* p1 = &x;   (this statement is underline)
cout << "&x is " << &x << endl;   (this statement is underline)
cout << "*p1 is " << *p1 << endl;  (this statement is underline)
cout << "p1= " << p1 << endl;   (this statement is underline)
cout << "&p1= " << &p1 << endl;   (this statement is underline)
double y = 4.5;
double* p2 = &y;
cout << "&y is " << &y << endl;
cout << "*p2 is " << *p2 << endl;
cout << "p2= " << p2 << endl;
cout << "&p2= " << &p2 << endl;

Answer:

1.double* p1 = &x;   (this statement is underline)

This statement declares a varaible as a pointer and assignes values to it the address of varaiable x.Suppose the address of x is 0xbfebd5c0 then this statement stores this address.

Output of the statements:

Statement:1.cout << "&x is " << &x << endl;   (this statement is underline)

Explaination:This statement gives the address of memeory location where the variable 'x' of type double is stored

output:&x is  0xbfebd5b77(say)

2.cout << "*p1 is " << *p1 << endl;  (this statement is underline)

This statement int *p1 declares the pointer to an double value, which we are initializing to the address of x. The value stored in a pointer p can be accessed through the dereferencing operator “*”.

3.cout << "p1= " << p1 << endl;   (this statement is underline)

This statement prints the address stored in pointer 'p1' variable

output:0xbfc601ac (say)

4.cout << "&p1= " << &p1 << endl;   (this statement is underline)

This statement gives the address of pointer p.

Output:&p1=0x13(say)

Now the statements:

double y = 4.5;
double* p2 = &y;
cout << "&y is " << &y << endl;
cout << "*p2 is " << *p2 << endl;
cout << "p2= " << p2 << endl;
cout << "&p2= " << &p2 << endl;

output:

1.&y is 0xabef77

2.*p2 is 10(* dereferences the value of varaible y)

3.p2=0xbfebd5b77 (address of y which is 0xbfebd5b77)

4.&p2=0xbef77(address of pointer p)

Question 3:

1.string s = "abcd";   (this statement is underline)

A string is a character array that is ‘’ terminated,thus the string s =abcd

string* p = &s;   (this statement is underline)

This statement declares a pointer with name p of type string and then assigns the address of the string s to the pointer p.

Statement:

1.cout << "&s is " << &s << endl;  (this statement is underline)

Explaination:This statement gives the address of the string s in the memory

Output:&s is 0xadcd555(say)

2.cout << "*p is " << *p << endl;    (this statement is underline)

This statement  string *p declares the pointer to an string value, which we are initializing to the address of string s. The value stored in a pointer p can be accessed through the dereferencing operator “*”.

Output:*p is abcd

3.cout << (*p)[0] << endl;    (this statement is underline)

This statement gives the character 0 of the string s which is 'a' and outputs this value.

Output: a

4.cout << "p= " << p << endl;   (this statement is underline)

This statement prints the address stored in pointer 'p' variable

Output: p= 0xbfc601ac (say)

5.cout << "&p= " << &p << endl;   (this statement is underline)

This statement

This statement gives the address of pointer p.

Output: &p=0x13(say)

Hope that helps...HAPPY ANSWERING!!!!!

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