I drew this for a better understanding. In the PC's memory there are succesive memory zones with length of 1 byte ( the green ones ). They can be grouped for representing bigger data ( as an int in our example ).
What i want to add to this picture is that the green squares are succesive addresses like 0x000000 followed by 0x000001 etc. Then the addresses from arr are jumping by four like 0x000000 and the next one would be 0x000004 ( because an int in this care is 4bytes ).
The code_1:
int arr[4] = {1,2,3,4};
int *p = arr;
cout << p << endl;
cout << ++p << endl;
cout << ++p << endl;
cout << ++p << endl;
The output_1:
0x69fedc
0x69fee0
0x69fee4
0x69fee8
The code_2:
char arrr[5] = {'1','2','3','4', '\n'};
char *ptr = arrr;
cout << &ptr << endl;
cout << &(++ptr) << endl;
cout << &(++ptr) << endl;
cout << &(++ptr) << endl;
The output_2:
0x69fed0
0x69fed0
0x69fed0
0x69fed0
The problem:I expect in output_2 i expect that the addresses would be 0x69fed0,0x69fed1,0x69fed2,0x69fed4
This is because you are displaying the address of the pointer instead of the address stored in the pointer:
char *ptr = 0;
std::cout << &ptr; // address where the pointer is placed
std::cout << (void*)ptr; // address managed by the pointer = 0
++ptr;
std::cout << &ptr; // this value never changes
std::cout << (void*)ptr; // Now this value should be 1
Related
I am a student in a CompSci intro class and I have a very basic understanding of pointers in C++. I had noticed in attempting to complete an assignment that a character array / c-string uses pointers differently than other data types.
For example, please consider the following code I created:
#include <iostream>
using std::cout, std::endl;
int main()
{
int inta[] = {1,2,3};
int* p1 = inta;
cout << "p1 = " << p1 << endl;
cout << "*p1 = " << *p1 << endl;
cout << "sizeof(p1) = " << sizeof(p1) <<
", sizeof(*p1) = " << sizeof(*p1) << endl;
char stra[] = "Dog";
char* p2 = stra;
cout << "p2 = " << p2 << endl;
cout << "*p2 = " << *p2 << endl;
cout << "sizeof(p2) = " << sizeof(p2) <<
", sizeof(*p2) = " << sizeof(*p2) << endl;
return 0;
}
The output of *p1 and *p2 are both the first value of the array. However, while the output of p1 is the pointer to the first element of inta (which tracks from online research), the output of p2 is the entire word "Dog". The sizes of p1 and p2 are the same, the size of *p1 and *p2 are 4 and 1 respectively. Is there something I am missing?
I am using Visual Studio Community 2022 and created a normal project.
Thank you, and I appreciate your help!
p1 is a pointer to an int. *p1 is thus an int -> size = 4
p2 is a pointer to a char, *p2 is thus a char -> size = 1
passing a char pointer to an output operator (cout<< or printf etc) will keep reading chars until it reaches the null char at the end of the string
I am trying to build a list-like heap storage of C-strings.
This is a simplified part of the program.
However, with each iteration new brings up the same address.
#include <iostream>
class listStringContainer {
public:
listStringContainer(const char* c);//constructor
};
int main(){
listStringContainer lsc1 ("Lorem ipsum");// calling the constructor
}
listStringContainer::listStringContainer(const char* c) {//constructor
char * Memory_Address;
auto time{5};
while (--time>=0) {
Memory_Address = new char[16];
//the memory location is to be saved into a vector
std::cout << "Memory_Address: "<< &Memory_Address << std::endl;
}
}
Output:
Memory_Address: 0x62fed8
Memory_Address: 0x62fed8
Memory_Address: 0x62fed8
Memory_Address: 0x62fed8
Memory_Address: 0x62fed8
The same result on g++ and MSVS.
Why does new appoint the same location and how to make new appoint different addresses?
You need to cast the static_cast<void*>(Memory_Address) to get the value stored in Memory_Address.
Lets consider:
char * p;
p= new char[16];
strcpy(p, "Hello");
cout << *p << endl; // Prints 'H'
cout << &p << endl; // Prints address of p
cout << p << endl; // Prints "Hello"
cout << static_cast<void*>(p) << endl; // Prints address of p[0]
Consider the below same scenario but with Integer data type:
int * ptr;
ptr= new int[16];
ptr[0] = 10;
cout << *ptr << endl; // Prints 10
cout << &ptr << endl; // Prints ptr address
cout << ptr << endl; // Prints address of ptr[0]
Therefore, Integer doesn't require casting to void* for getting &ptr[0]
This question already has answers here:
cout << with char* argument prints string, not pointer value
(6 answers)
Closed 3 years ago.
If char* is a pointer (I find its size is always 4 bytes), how do I find its value (the address in hexa or decimal)? I tried &(*p) for char *p. It simply returned the initial string. If it is always 4 bytes, how is it that it can be initialized to long strings but point to the first character? In other words where is the string stored?
Is char* a weird pointer used for purposes other than what a pointer is intended to be?
// an initial pointer (usually its size is 32bit or 64bit, depending on CPU/OS).
// it's value is currently NULL (not pointing anywhere),
// so we can't do very much with this right now.
char* p = nullptr;
// for the sake of sanity, check the value (should be zero)
// we have to convert to intptr_t, otherwise we'd get the string
// value being printed out.
std::cout << "p address = " << intptr_t(p) << std::endl << std::endl;
// lets allocate a few chars to play with
p = new char[10];
// copy in some text value
std::strcpy(p, "Hello");
// and now if we print the address, the text string,
// and the char we are pointing at
std::cout << "p address = " << intptr_t(p) << std::endl;
std::cout << "p string = " << p << std::endl;
std::cout << "p dereferenced = " << *p << std::endl << std::endl;
// for fun, lets increment the pointer by 1
++p;
// this should have made a couple of changes here
std::cout << "p address = " << intptr_t(p) << std::endl;
std::cout << "p string = " << p << std::endl;
std::cout << "p dereferenced = " << *p << std::endl << std::endl;
// decrement again (so we can delete the correct memory allocation!)
--p;
// now free the original allocation
delete [] p;
// if we print again, notice it still has the memory location?
std::cout << "p address = " << intptr_t(p) << std::endl;
// This would be bad to access (we've just deleted the memory)
// So as a precaution, set the pointer back to null
p = nullptr;
// should be back where we started
std::cout << "p address = " << intptr_t(p) << std::endl;
So while studying for my exams I was trying to do a practice problem for pointers.
In the following code I'm trying to display the number of elements before the first occurrence of 0.
There is only one part that i didn't understand please see the 6th last line.
#include <iostream>
using namespace std;
int main()
{
int A[10];
for (int i = 0; i < 10; i++){
cout << "Please enter number " << i + 1 << " in the array: ";
cin >> A[i];
}
int *Aptr = A;
while(*Aptr !=0){
cout << *Aptr << "";
Aptr++;
}
cout << "\nThere are " << (Aptr - A) //Here is what i don't understand.
<< " numbers before the first occurrence of 0." << endl;
system("pause");
return 0;
}
So why exactly is (Aptr - A) giving me the number of elements instead of a memory location, and why is this even doable since Aptr is a pointer and A is an array?
Can someone explain to me in detail?
When used in an expression, like Aptr - A, the name of an array A will be implicitly converted to a pointer (equal to &A[0]).
Then the compiler is faced with subtracting two pointers of the same type (both of type int * in your case). That is specified as giving a value of type std::ptrdiff_t, which is, in turn "a signed integral type able to represent the result of subtracting two pointers".
Pointer arithmetic, when subtracting two pointers of type int (i.e. two int *s) gives the number of ints between the two pointers (assuming they are in the same object, which is true in this case, since Aptr points at an element of the array A).
Practically, if Aptr is equal to &A[i], the subtraction Aptr - &A[0] gives a std::ptrdiff_t equal to i.
Note: there is another problem in your code, as since the first (for) loop reads 10 values, while the second while loop keeps incrementing Aptr until it points at an int with value 0. If the user enters any zero values, the second loop will stop when it finds the first (even if the user enters non-zero elements after that). If the user enters no values equal to 0, then the while loop has undefined behaviour, since Aptr will keep walking through memory past the end of A until it happens to find memory that compares (as an int) equal to 0.
First of all, name of array A is associated to address of (pointer at) the first item in the array.
So why exactly is (Aptr - A) giving me the number of elements?
Because according to rules address arithmetic subtraction operation (also +, and similar) is performed based on the data type.
I mean, that compiler operating with int* makes ++, --, addition, subtraction an integer, etc. adds addresses needed for shifting to next/previous item.
If you really want to see how many bytes are located between addresses, just convert addresses to int before making subtraction:
cout << endl << "Address difference is " << int(Aptr) - int(A) << endl;
You can try that with different data types as follows:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int A[5];
short B[5];
unsigned char C[5];
cout << "Array (data type) | Syze of array | Size of item | Item distanse | Bytes distance" << endl;
cout << "A (int) :" << setw(10)
<< sizeof(A) << setw(15)
<< sizeof(A[0]) << setw(15)
<< &A[4] - A << setw(15)
<< int(&A[4]) - int(A) << endl;
cout << "B (short) :" << setw(10)
<< sizeof(B) << setw(15)
<< sizeof(B[0]) << setw(15)
<< &B[4] - B << setw(15)
<< int(&B[4]) - int(B) << endl;
cout << "C (un.char) :" << setw(10)
<< sizeof(C) << setw(15)
<< sizeof(C[0]) << setw(15)
<< &C[4] - C << setw(15)
<< int(&C[4]) - int(C) << endl;
system("pause");
return 0;
}
UPDATE
To be better prepared for your exam, consider the following example with pointers:
#include <iostream>
using namespace std;
int main()
{
int A[5] = {0}; // all items now are 0
int * P = A + 2; // the same as P = &A[2];
*P = 33; // writing to item A[2];
cout << A[2] << endl; // just to check in usual way
cout << *(A + 2) << endl; // using A as a pointer
cout << *(2 + A) << endl; // almost the same to previous
cout << 2[A] << endl; // quite strange, but it works
cout << 0[P] << endl; // and this is the same
return 0;
}
You must understand that 0[P] means for compiler *(0 + P), as well as 2[A] means - *(2 + A), but you should not write in your program in such style (exceptions are only cases when you want to confuse a reader).
And one more important thing - difference between array and pointer - are shown in the following example:
int A[] = {1, 2, 3, 4, 5};
int *P = A;
cout << "A = " << A << endl;
cout << "P = " << P << endl;
cout << "size of A = " << sizeof(A) << endl;
cout << "size of P = " << sizeof(P) << endl;
even if the addresses (vaules A and P) are equal, compiler works with array (A) in a different way than with pointer: sizeof(A) means memory allocated for whole array (5 items of sizeof(int) each), but sizeof(P) means memory allocated for data type int * (pointer to int). So, sizeof(P) depends only on compiler and OS platform (e.g. pointer can be 32-bit or 64-bit), but sizeof(A) depends on size of item (int may be not 32 bits) and NUMBER OF ITEMS in the array.
And you can "go to the next item" with pointer:
P++;
cout << *P << endl;
but you are not able to do:
A++;
because A is not variable of pointer type (it is just similar in sense of "address of the first item"), and compiler will say you something like:
error : '++' needs l-value
I've been learning C++ on my own for a while now, and I've come to a "roadblock" when it comes to pointers. I'm using this as my http://www.cplusplus.com/doc/tutorial/pointers/ learning material, but I'm still having an issue. So to test something I wanted to copy the contents of one array into another. I wrote the following.
char arrayA[15] = "abcdef";
char arrayB[15];
char *a = arrayA;
char *b = arrayB;
cout << "before loop:" << endl;
cout << a << endl;
cout << b << endl;
while (*a != '\0') {
// Copy the contents of a into b
*b = *a;
// Step
a++;
b++;
}
// Assign null to the end of arrayB
*b = '\0';
cout << "after loop:" << endl;
cout << a << endl;
cout << b << endl;
I get the following results.
before loop:
abcdef
after loop:
When I cout the contents before the loop I get the expected results. a contains "abcdef" and b is nothing, because there is no value yet. Now after the loop, both a and b show no results. This is where I am lost. I used * to dereference both a and b and assign the value of a into b. Where did I go wrong? Do I need to use the & with this?
Solution:
After the loop is complete, pointer *a is pointing to the end of arrayA and pointer *b is pointing to the end of arrayB. So to get the full results of arrayB simply cout << arrayB. Or create a pointer that never changes and always points to arrayB char *c = arrayB and cout << c at the end of the loop.
After the loop a and b have changed, they then point to the end of the string. You need to make a copy of the pointers to step through so that as you iterate you're not changing the location of a and b.
The problem is that you're outputting your temporary variables that were used to iterate through the array. They are now at the end of the copied data. You should output the value of arrayA and arrayB instead.
Remember begin of array. In this moment you are incrementing pointers and printing something which is pointed by them at the end of arrays, after loop ends.
char arrayA[15] = "abcdef";
char arrayB[15];
char *a_beg = arrayA;
char *b_beg = arrayB;
char *a;
char *b;
cout << "before loop:" << endl;
cout << a_beg << endl;
cout << b_beg << endl;
a = a_beg;
b = b_beg;
while (*a != '\0') {
// copy contents of a into b and increment
*b++ = *a++;
}
// assign null to the end of arrayB
*b = '\0';
cout << "after loop:" << endl;
cout << a_beg << endl;
cout << b_beg << endl;