Question

I NEED HELP WITH C PROGRAMMING. Answer each of the following.

Assume that unsigned integers are stored in 4 bytes and that the starting address of the array is at location 200 600 in memory.

i)What address is referenced by vPtr + 5? What value is stored at that location?

j) Assuming vPtr points to values[6] , what address is referenced by vPtr -= 4? What value is stored at that location?

Also .....

For each of the following, write a single statement that performs the indicated task. Assume that short integer variables value1 and value2 have been defined and that value1 has been initialized to 80000.

a) Define the variable lPtr to be a pointer to an object of type short.

b) Assign the address of variable value1 to pointer variable lPtr.

c)Print the value of the object pointed to by lPtr.

d) Assign the value of the object pointed to by lPtr to variable value2.

e)Print the value of value2.

f)Print the address of value1.

g)Print the address stored in lPtr. Is the value printed the same as the address of value1?

My code is below. I keep getting the error message " Q2.c: In function ‘main’: Q2.c:6:1: warning: overflow in implicit constant conversion [-Woverflow] short int value1=80000,value2; ^ /tmp/ccSYVDKb.o: In function `main':

/u/jforrow/C291-Spring1-2019/exercise/hwk4/Q2.c:21: undefined reference to `clrscr'

/u/jforrow/C291-Spring1-2019/exercise/hwk4/Q2.c:42: undefined reference to `stdscr'

/u/jforrow/C291-Spring1-2019/exercise/hwk4/Q2.c:42: undefined reference to `wgetch'

collect2: error: ld returned 1 exit status"

#include
#include

void main()
{
short int value1=80000,value2;

//short range is -32,768 to 32,767
// 80000 is initialized to value1 but it can not value
//beacuse of range
// so the out put is not 80000
// 80000-32767=47233
//47233-32767=14466
//two times zeros we need subtract
//14466-2=14464
//14464 is out put

//a)
short *lPtr;
clrscr();//clear the previous output

//b)
lPtr=&value1;

//c)
printf("\n The value of the object pointed by lPtr is %d\n",*lPtr);

// \n for new line
//d)
value2=*lPtr;

//e)
printf("\n The value of value2=%d\n",value2);

//f)
printf("\n The address of value1 is %d\n",&value1);

//g)
printf("\n The address stored in lPtr is %d, and the address of value1 is %d",lPtr,&value1);

getch();
}

Answer 1

Assume that unsigned integers are stored in 4 bytes and that the starting address of the array is at location 200 600 in memory.

hence values array is located at starting address 200600

assume array ---> values=[0,1,2,3,4,5,6]

-----------------------------------------------------------------------------

i)What address is referenced by vPtr + 5? What value is stored at that location?

if vPtr stores starting address of values array :

unsigned int *vPtr=values;

hence printf ("%u",*vPtr) ---> 0

hence printf("%u", *(vPtr+5)) -----> 5

hece vPtr will store the address (starting_address + (index * sizeof(unsigned int =4 )) ) =200600+ (0*4)= 200600

hece vPtr+1 will store the address (starting_address + (index * sizeof(unsigned int =4 )) ) =200600+ (1*4)= 200604

hece vPtr+2 will store the address (starting_address + (index * sizeof(unsigned int =4 )) ) =200600+ (1*4)= 200608

hece vPtr+5 will store the address (starting_address + (index * sizeof(unsigned int =4 )) ) =200600+ (5*4)= 200620

and so on ....

about values :

*vPtr=values[0]=0

*(vPtr+1)=values[1]= 1

*(vPtr+2)=values[2]= 2

and hence  * (vPtr + 5) =values[5] = 5 is value stored

REFERENCE OUTPUT

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Question j) Assuming vPtr points to values[6] , what address is referenced by vPtr -= 4? What value is stored at that location?

if vPtr points to values[6] then vPtr stores the address 200600+ (6*4)= 200624

and values stored will be values[6]=6

******* after vPtr - = 4 ******

vPtr stores the address = 200624 - (4*4) =2006008 which will hold the address of values[2]

and values stored will be *(vPtr) = values[2]=2

------------------------------------------------------------------------

Question ) For each of the following, write a single statement that performs the indicated task. Assume that short integer variables value1 and value2 have been defined and that value1 has been initialized to 80000.

Your code works fine -->just gives the warning

main.c:4:18: warning: overflow in implicit constant conversion [-Woverflow]

short int value1=80000,value2;

This is because the range of short range is -32,768 to 32,767

So either Ignore the warning or else assign the value1 and value2 in between -32,768 to 32,767 to avoid warning otherwise your code is fine

1) with warning OUTPUT

2)without warning OUTPUT