Question

2. AC++ program contains the following C++ code (assume an int occupies 2 bytes each in memory): int x[ ] = { 10, 20, 30, 40, 50, 60, 70, 80); int *ptrx ptrx = x; Array x starts at memory location 2140. a. What is the value assigned to ptrx? b. What is the value of (x+2)? c. What is the value of *x? d. What is the value of (*x+2)? e. What is the value of *(x+2)? f. What is the value of ptrx +3? g. What is the value of ptrx after the statement ptrx &(x[4]); Write a loop that will access each member of x using pointer offset notation h. 3. Given the following declarations: char ptrstuff int n Write the one statement that will dynamically allocate a character array of size n. a. b. Write the one statement that would delete the dynamically allocated array

basic c++

Answer 1

int x[] = {10,20,30,40,50,60,70,80};

int *ptrx;

ptrx = x;

pointer ptrx would be equivalent and would have very similar properties. The main difference being that ptrx can be assigned a different address, whereas array x can never be assigned anything, and will always represent the same block of 20 elements of type int.

Here array x starts at memory location 2140. I am taking same addresses of both x and ptrx. So ptrx has same memory address as x .

Answers:

a. .

b. For value of (x+2), x starts form location 2140. Here (x+2) represents the memory address of third     element of array x, because index starts from 0 , 1 ,2 …. And int occupies 2 byte each in memory. So the value will be 2140+2+2 = 2144.

c. *x represent the value stored at first index i.e. 0, that is 10

d. In C part the value of *x is 10 so the value of (*x + 2) = 10+2 = 12

e. *(x+2) represents the value stored at third index. So value of *(x+2) = 30

f. The value of ptrx+3 will be the address of fourth element in x. So per+3 = 2140 + 2 + 2 + 2 = 2146

g. value of ptrx after statement ptrx = &(x[4]); indicates the address of fifth element address of x that will be = 2140 + 2 + 2 + 2 + 2 = 2148

h. loop:-

                int i;

                for(i=0;i<8;i++){

                                cout<< *(ptrx+i) << ‘\n’;                               // It print the value pointed by ptrx on ith index

}

2. char * ptrstuff;

    int n;

                1. For allocate dynamically:

                *ptrstuff = (char*)malloc(sizeof(char) * n);

                2. To delete dynamically allocated array:

                delete[] ptrdtuff;