Yeah, but also confused about what types are. I was trying to get him to understand the distinction between integer and char, or integer and int*.PeterDonis said:
Read page 492 to 495 of Gaddis bookPeterDonis said:
I am trying to distinguish the address of the variable with the content of the variable.Jarvis323 said:
The & operator acts like a function, it takes myFloat as an argument and returns something. Both the argument myFloat and the returned value each have a C++ type. The type of the argument is (float). The return type is (float*). Those, with their exact syntax are valid types that the compiler can recognize and parse. The compiler flagged an error because (int*) is not the same type as (float*), and it has been decided by the C++ committee that this type mismatch should throw an error. Why? Because it can be problematic. As in Mark's post, getting the value from a pointer (de referencing) has to interpret the bits as something. Those bits could be encoding a float or an int, the only way to know is if the compiler keeps track of what is supposed to be there. If you really want to convert a value of type (float*), into a type (float*), then you can do type casting as Mark shows, where you ask the compiler to convert/consider the value of type (float *) as a type (int*), and maybe that means that the bits which were encoding 3.75 will now be encoding 1081081856 as far as the compiler knows.yungman said:
yungman said:
yungman said:
float *. So the correct way of coding the original snippet of yours that started this subthread, assuming your purpose was to store a pointer to a float variable, would be:float myFloat;
float *pFloat = &myFloat;// address of float variable
#include <iostream>
using namespace std;
int main()
{
float test;
int* ptr;
cout << "the address of float test is " << &test << "\n\n";
return 0;}yungman said:
yungman said:
yungman said:
float *pFloat to make the pointer variable name more descriptive).double val = 3.75;
double * dPtr, * dPtr2;
dPtr = &val;
dPtr2 = dPtr + 1;
cout << "dPtr: " << dPtr << "\n";
cout << "dPtr2: " << dPtr2 << "\n";dPtr: 00EFFEC8
dPtr2: 00EFFED0Sorry I did not see that when I was busy responding. Now I am happy. that there is a float *ptr for &myFloat.PeterDonis said:
#include <iostream>
using namespace std;
int main()
{
char testC; char* ptC; ptC = &testC;
cout << "the address of char testC is = " << &testC << " The content of ptC is = " << ptC << "\n\n";
float test; float* ptr; ptr = &test;
cout << "the address of float test is = " << &test << " The content of ptr is = " << ptr << "\n\n";
double* ptD; double testD; ptD = &testD;
cout << "the address of double testD is = " << &testD << " The content of ptD is = " << ptD << "\n\n";
const int* ptCS; const int testCS=7; ptCS = &testCS;
cout << "the address of const int testCS is = " << &testCS << " The content of ptCS is = " << ptCS << "\n\n";
return 0;
}The cout class has member functions called operator <<. One of them takes a char * as an argument, and considers it as a c-string (null terminated array of char), so it tries to output the characters pointed to by the pointer rather than the pointer itself. They chose this behaviour so it's convenient to print c-strings.yungman said:
Didn't think of "*" is a character! So how do you define a char pointer? I flip through half the chapter on pointers, no mention on any of these. and this is the best book I have so far.Jarvis323 said:
In the context of your example, * is not a character. The type char * is a pointer to a character.yungman said:
char testC = 'a';
char* ptC = &testC;
cout << "the address of char testC is = " << (void *)ptC << " The character pointed to is = " << *ptC << "\n\n";yungman said:
yungman said:
PeterDonis said:
Yes, I was confused because the way the book is written for this chapter. I decided to stop talking and went through most of chapter 9. It is NOT a well written chapter. It just gave the example that pointing to a float is an error and left it at that. The book later just talk about each type one at a time and mostly with example instead of listing pointers of different type from the beginning. That cause my confusion and concern. Like it spread these through 20 to 30 pages and never have a summation of all the pointers and their reason of why to have different types.PeterDonis said:
My advice is to definitely study Ch. 13. The concepts presented in that chapter start to cover what makes C++ an object-oriented language. Without it, what you've been doing is pretty much C plus the C++ take on I/O (i.e., using the cout and cin stream objects). The fact that your grandson's class covered only the first 11 chapters shouldn't enter into your decision. After all, his class was constrained by time, with either 10 weeks for a quarter of 15 weeks or so for a semester. No doubt his class was in Intro to Programming with C++. You don't have those constraints.yungman said:
Not very much. You'll need another book for data structures and algorithms, but what you learn from the book you're working in will be helpful, as it will likely show examples in C++ (unless you get a book that is Java-based). The data structures will include things like singly- and doubly-linked lists, binary trees, and other structures, plus algorithms for searching through the various structures or sorting them. A fair amount of the presentation on algorithms will be focused on how efficient various algorithms are.yungman said:
Ch. 13 and at least the parts in Ch. 14 that are about writing classes will definitely help.yungman said:
That reminded me to check. Yes, the obsfucated C contest is still active.jedishrfu said:
Thank you for the advice. I will study cpt 13 and at least the part you suggested on cpt 14. What is the next logical thing I should study after that? I suspect I will finish those by Christmas.Mark44 said:
//Function return pointers
#include <iostream>
using namespace std;
void getRNum(int*, int);//return a pointer from the functionint main()
{
int size = 5;
int *number;//initiate a pointer number
getRNum(number, size);//get the pointer to array of 5 random numbers
for (int count = 0; count < 5; count++)// display the numbers
cout << *(number + count) << "\n\n";
delete[] number;//free the memory of 5 element array number[5]
return 0;
}
void getRNum(int *arr, int num)//receive an integer 5 to num, return a pointer to array arr.
{
arr = new int[num]; //request 5 elements of int of memory from computer
for (int count = 0; count < num; count++)// input 5 integers
{
cout << " Enter number " << count + 1 << " = ";
cin >> arr[count]; cout << "\n\n";
cout << " In memory location: " << (arr + count) << "\n\n";
}
}yungman said:
main function like this:int** pNumber;void getRNum(int** pArr, int num);getRNum function you would have to do this to pass the address of the array of 5 numbers back to the main function:*pArr = arr;yungman said:
yungman said:
Thanks for the reply.PeterDonis said:
yungman said:
yungman said:
yungman said:
ThankPeterDonis said:
yungman said:
yungman said:
// pseudo code example
my_function(...) {
data = my_alloc()
.. do something with data...
data.my_free(); // alternatively my_free(data);
}Thanksjedishrfu said:
#include <iostream>
#include <cstdlib>//for rand and srand
#include <ctime>//for time function
using namespace std;
int *getRNum(int);//return a pointer from the functionint main()
{
int *number; // to point to number
number = getRNum(5);//get the pointer to array of 5 random numbers
for (int count = 0; count < 5; count++)// display the numbers
cout << number[count] << "\n\n";
delete[] number;//free the memory of 5 element array number[5]
number = 0;//set pointer to 0
return 0;
}
int *getRNum(int num)//receive an integer 5 to num, return a pointer to array arr.
{
int *arr;// initialize point to array arr to hold 5 elements, num = 5
if (num <= 0)
return NULL;
arr = new int[num]; //request 5 elements of int of memory from computer
srand(time(0)); // Seed the random number
for (int count = 0; count < num; count++)
arr[count] = rand();
return arr;//return the starting address of arr.
}I tried what you suggested, this is the program and the error message:PeterDonis said:
#include <iostream>
using namespace std;
void getRNum(int**, int);//return a pointer from the functionint main()
{
int size = 5;
int **pNumber;//initiate a pointer number
getRNum(pNumber, size);//get the pointer to array of 5 random numbers
for (int count = 0; count < 5; count++)// display the numbers
cout << *(pNumber + count) << "\n\n";
delete[] pNumber;//free the memory of 5 element array number[5]
return 0;
}
void getRNum(int **pArr, int num)//receive an integer 5 to num, return a pointer to array arr.
{
int *Arr = new int[num]; //request 5 elements of int of memory from computer
*pArr = Arr;
for (int count = 0; count < num; count++)// input 5 integers
{
cout << " Enter number " << count + 1 << " = ";
cin >> *pArr[count]; cout << "\n\n";
cout << " In memory location: " << (pArr + count) << "\n\n";
}
}yungman said:
pNumber variable needs to be initialized. The obvious way to initialize it is with a null pointer: int** pNumber = NULL;.yungman said:
pNumber, and you are passing the address of that variable to the getRNum function so the function can assign a value to it, namely the address of the array you allocate with new.Thanks, I did not know about assigning NULL to pointer. Yes, that eliminates the error. But it still did not return the pointer back to main.PeterDonis said:
//Function return pointers
#include <iostream>
using namespace std;
void getRNum(int*, int);//return a pointer from the functionint main()
{
int size = 5;
int* pNumber = NULL;//initiate a pointer number
getRNum(pNumber, size);//get the pointer to array of 5 random numbers
cout << " Back in main from getRNum.\n\n";
for (int count = 0; count < 5; count++)// display the numbers
cout << *(pNumber + count) << "\n\n";
delete[] pNumber;//free the memory of 5 element array number[5]
return 0;
}
void getRNum(int* pArr, int num)//receive an integer 5 to num, return a pointer to array arr.
{
pArr = new int[num]; //request 5 elements of int of memory from computer
for (int count = 0; count < num; count++)// input 5 integers
{
cout << " Enter number " << count + 1 << " = ";
cin >> pArr[count]; cout << "\n\n";
cout << " In memory location: " << (pArr + count) << "\n\n";
}
}yungman said:
pArr in the function signature for getRNum is still wrong; it needs to be int**, not int*. Then you pass the address of pNumber to the function using that parameter.PeterDonis said:
int" to be a null pointer.new.int" to be a null pointer.new to the variable at the address that was passed to it.new--i.e., the pointer to the array of ints that was initialized in the function--in the variable whose address was passed.int main()
{
int size = 5;
int* pNumber = NULL;//initiate a pointer number
getRNum(pNumber, size);//get the pointer to array of 5 random numbers
cout << " Back in main from getRNum.\n\n";
for (int count = 0; count < 5; count++)// display the numbers
cout << *(pNumber + count) << "\n\n";
delete[] pNumber;//free the memory of 5 element array number[5]
return 0;
}// Pointer to a pointer
#include <iostream>
using namespace std;
void changePtVal(int**);
int Gvar = 42;
int main()
{
int var = 23;
int* ptr = &var;
cout << " Before calling changePtVal, *ptr = var = " << *ptr << " ptr = " << ptr << "\n\n";// the content in address ptr = var =23.
changePtVal(&ptr);//**pptr = &ptr.
cout << " After calling changePtVal, *ptr = " << *ptr << "\n\n";
cout << " address of var = " << &var << " ptr = " << ptr << " *ptr = " << *ptr << "\n\n";
return 0;
}
void changePtVal(int** pptr)
{
*pptr = &Gvar;
}
//The result is:
//Before calling changePtVal, * ptr = var = 23 ptr = 012FFE48
//After calling changePtVal, *ptr = 42
//address of var = 012FFE48 ptr = 00EEC008 * ptr = 42#include <iostream>
using namespace std;
void getRNum(int**, int);//return a pointer from the functionint main()
{
int size = 5;
int*pNumber = NULL;//initiate a pointer number
getRNum(&pNumber, size);//pass address of pNumber
cout << " Back to main, pNumber = " << pNumber << "\n\n";
cout << " The array pointed by pNumber = {";
for (int count = 0; count < size-1; count++)// display the numbers
cout << *(pNumber + count) << " ";
cout << *(pNumber + size-1) << "}\n\n";
delete[] pNumber;//free the memory of 5 element array number[5]
return 0;
}
void getRNum(int** pArr, int num)//receive &pNumber the addresses of pNumber.
{
int* Arr = NULL; // set pointer Arr to Null
Arr = new int[num]; //request 5 elements of int of memory from computer pointed by Arr.
*pArr = Arr;// *pArr = address of Arr[0].
for (int count = 0; count < num; count++)// input 5 integers to Arr[0..4].
{
cout << " Enter number " << count + 1 << " = ";
cin >> *(*pArr + count);//*pArr = address of Arr[0], pArr + elements = address of Arr[1...]
cout << " In memory location of Arr: " << (Arr + count) << "\n\n";
}
}
/*
Enter number 1 = 2
Enter number 1 = 2
In memory location of Arr: 012B4D50
Enter number 2 = 3
In memory location of Arr: 012B4D54
Enter number 3 = 4
In memory location of Arr: 012B4D58
Enter number 4 = 5
In memory location of Arr: 012B4D5C
Enter number 5 = 6
In memory location of Arr: 012B4D60
Back to main, pNumber = 012B4D50
The array pointed by pNumber = {2 3 4 5 6}
*/yungman said:
new returns to the variable whose address is contained in the function parameter pArr. Since the caller passed the address of the pointer variable pNumber in that function parameter, that variable is what the function writes to.Ha ha, I can barely bent my mind to this so far, I need to play with this more to get a better feel of this. I think I have enough for today. Believe me, I have been working on this. It's hard to to twist around the address point to an address and all that. This is the first time in all 9 chapters that is difficult to me, or somehow, I have a mental block on this.PeterDonis said:
yungman said:
#include <iostream>
using namespace std;
int main()
{
int var = 23;
int* ptr= &var;
int* pptr = ptr;
cout << " var = " << var << " *ptr = " << *ptr << " **pptr = " << *pptr << "\n\n";
return 0;
}#include <iostream>
using namespace std;
int main()
{
int var = 23;
int* ptr =&var;
int* pptr;
*pptr = ptr;// Error
cout << " var = " << var << " *ptr = " << *ptr << " **pptr = " << *pptr << "\n\n";
return 0;
}yungman said:
I posted the error message right under the program " cannot convert int* to int."PeterDonis said:
yungman said:
No, this is the same mistake you made before.yungman said:
No.yungman said:
