c++ please help
Write a short C program that declares and initializes (to any value you like) a char, an int, and a double. Next, declare and initialize a pointer to each of the three variables. Your program should then display the address of, value stored in, and the memory size (in bytes) of each of the six variables as well as the dereferenced pointers for a total of nine results. Use the address-of operator when displaying the memory address of a variable (not the value of a variable). Remember, if a variable is defined as a pointer, then the value of this variable is a memory address. When displaying the value of such a pointer variable, a memory location will be displayed, so it is not necessary to use the address-of operator. Use the size of function to determine the memory size allocated for a variable (including a pointer variable). You should see memory addresses that look something like this: \"0times0842a018\". The initial characters \"Ox\" tell you that hexadecimal notation is being used; the remainder of the digits give the memory address itself.
Solution
#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
char c = \'a\';
int i=10;
double d=15;
int * iptr = &i;
char * cptr = &c;
double * dptr = &d;
cout<<\"The address of integer value is \"<<iptr<<endl;
cout<<\"The address of character value is \"<<cptr<<endl;
cout<<\"The address of double value is \"<<dptr<<endl;
cout<<\"The memory size of int is \"<<sizeof(i);<<endl;
cout<<\"The memory size of double is \"<<sizeof(d);<<endl;
cout<<\"The memory size of char is \"<<sizeof(c);<<endl;
cout<<\"The memory size of char pointer is \"<<sizeof(cptr);<<endl;
cout<<\"The memory size of integer pointer is \"<<sizeof(iptr);<<endl;
cout<<\"The memory size of double pointer is \"<<sizeof(dptr);<<endl;
getch();
}
.
Basic Civil Engineering first year Notes- Chapter 4 Building.pptx
c++ please help Write a short C program that declares and initializes.docx
1. c++ please help
Write a short C program that declares and initializes (to any value you like) a char, an int, and a
double. Next, declare and initialize a pointer to each of the three variables. Your program should
then display the address of, value stored in, and the memory size (in bytes) of each of the six
variables as well as the dereferenced pointers for a total of nine results. Use the address-of
operator when displaying the memory address of a variable (not the value of a variable).
Remember, if a variable is defined as a pointer, then the value of this variable is a memory
address. When displaying the value of such a pointer variable, a memory location will be
displayed, so it is not necessary to use the address-of operator. Use the size of function to
determine the memory size allocated for a variable (including a pointer variable). You should see
memory addresses that look something like this: "0times0842a018". The initial characters
"Ox" tell you that hexadecimal notation is being used; the remainder of the digits give the
memory address itself.
Solution
#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
char c = 'a';
int i=10;
double d=15;
int * iptr = &i;
char * cptr = &c;
2. double * dptr = &d;
cout<<"The address of integer value is "<<iptr<<endl;
cout<<"The address of character value is "<<cptr<<endl;
cout<<"The address of double value is "<<dptr<<endl;
cout<<"The memory size of int is "<<sizeof(i);<<endl;
cout<<"The memory size of double is "<<sizeof(d);<<endl;
cout<<"The memory size of char is "<<sizeof(c);<<endl;
cout<<"The memory size of char pointer is "<<sizeof(cptr);<<endl;
cout<<"The memory size of integer pointer is "<<sizeof(iptr);<<endl;
cout<<"The memory size of double pointer is "<<sizeof(dptr);<<endl;
getch();
}
