I have searched all over and have not found anything on this.
I have the following simple code:
#include<iostream>
using namespace std;
int main() {
int *newArray;
int size;
cout << "enter size: ";
cin >> size;
newArray = new int[size];
cin.get();
return 0;
For some reason, regardless of the number entered, the dynamic array always has only one element allocated. I am not getting any compiler errors or run time errors (other than the array issue). I have also tried
int *newArray;
newArray = new int[100];
and
const int SIZE = 100;
int *newArray;
newArray = new int[SIZE];
I get the same result.
I am using Visual Studio Community 2015 for my IDE if that makes a difference.
If anyone has any pointers (no pun intended) please let me know. Thank you.
newArray is not an array, it's just a pointer to some place in memory. It doesn't contain a number of elements, it's just an address in memory. When you do newArray = new int[n], sizeof(int)*n bytes of memory are allocated and the address of this allocated space is saved in newArray.
Visual Studio only knows that newArray is a pointer to an int. So it just shows the one number at that address. It cannot know how much memory has been allocated there.
If you want to see more than the first int, open the memory view, type newArray as the address, set up it to show ints. You will see plain memory interpreted as integers, so you will see numbers from your array. But there will be no end of these ints. After your numbers, there will be garbage.
Another option to try (I'm not sure if it works though) is to add (int[100])newArray to the Watch window.
So memory is allocated, feel free to write and read it and don't forget to delete[] it.
Related
I used the malloc function for allocating 10^9 memory locations as part of an array.
The Code 1 gets executed successfully.
Code 1:
int main(){
int size = (int)(1e9);
int* arr = (int*) malloc( size * sizeof(int) );
for(int i=0;i<size;i++){
arr[i] = i;
}
return 0;
}
But when I tried to access the particular memory location or index value = 12345678 (which is < 1e9), I got segmentation fault
Code 2:
int main(){
int size = (int)(1e9);
int* arr = (int*) malloc( size * sizeof(int) );
for(int i=0;i<size;i++){
arr[i] = i;
}
cout<<arr[12345678]<<endl; //added this line of code, which gives segmentation fault
return 0;
}
My guess is, this occurs due to memory fragmentation, but I am not sure about this. Can anyone kindly explain the correct reason.
It's definitely
Compiler optimisation
Limit on memory allocation as per standard sandbox
Modern operating systems are doing something called "lazy allocation", this means that when you ask to allocate memory, they give it to you without actually allocating the memory.
This is exactly the reason for the Virtual memory vs Physical memory taken you see while using top.
The memory is only really allocated when you try to use the buffer itself.
In your example, you've allocated a lot of memory (4GB), the OS let you allocate it but did not actually free up the memory needed for it, once you've tried using it, it tried to actually provide the memory you asked for, couldn't do it, and therefore crashed.
I am doing an assignment for my university course and having the following trouble in my code:
cout << "Enter number of values you want to enter" << endl;
int Arraysize;
cin >> Arraysize;
int input_Arr[Arraysize];
The compiler gives an array saying that the array size has to be a constant. I have tried dynamically giving an array size, but that gives the same error. The only way it allows a variable as array size is , when the variable is made a const.
Can you tell me what is the way to get array size from the user and then declare array of that size?
Also , if you do so by dynamic memory allocation, please explain your code at each step.
Variable sized arrays are not standard C++, thus the compiler is not happy.
You have two options:
1) Use an std::vector, which dynamically grows in size automatically. However, since this is HW, you might not be allowed to use it just yet.
2) Use dynamic allocation of memory for your array, by using new and delete, like this:
#include <iostream>
using namespace std;
int main() {
int Arraysize;
cin >> Arraysize;
int *input_Arr = new int[Arraysize];
// Here you have an array called `input_Arr`, of size `Arraysize`, ready to be used (eg filled).
// when you are done with using the array, you must free the dynamic memory
delete [] input_Arr;
return 0;
}
The two points that require your attention are:
int *input_Arr = new int[Arraysize];
What it does? It dynamically allocates memory for an array called input_Arr, that will store ints. Its size is Arraysize.
Every time memory is allocated dynamically, it must gets free'd by the program. To do that, we use this code:
delete [] input_Arr;
which deletes an array called input_Arr.
If you want to use Dynamic allocation in C++, use keywords new and delete
code for this case will be
{
int size;
cin>>size;
int *my_array = new int[size]; //dynamic allocation
...
delete [] my_array; //don't forget to use delete at the end of your code
}
If I take a block of memory by the following line .
int* a = new int[10];
Then for freeing the memory , the code would be
delete [] a;
But if I take a pointer of single memory segment like the following
int* a = new int;
And then insert a array of data like following .
for(int i=0;i<10;i++)
{
a[i]= i ;
}
So to free the first memory segment that pointer "a" is pointing, the code would be like following
delete a;
But Here I inserted 9 more data from the memory that pointer "a" is pointing .So I am using actually 10 memory segment here . how can I free all this 10 memory ? Please help me to get the answer .
how can I free all this 10 memory ?
You can't and you shouldn't because the moment you tried to "insert a array of data like following" you have entered Undefined Behavior land for writing to a location that you didn't allocate with new in the first place. You asked for a single int, you got a single int. Don't write past it.
You allocated only one int. Memory from a[1] to a[9] may be assigned to other objects and you may corrupt them.
P.S. Btw you can not free memory that you did not allocate in any case.
new int allocates space for one int value. You cannot legally pretend that it's an array of ten int values, and the compiler won't generate code to expand the allocated memory if you go out of bounds. The code you wrote produces undefined behavior. The fact that it compiled and ran doesn't change that; sooner or later it will cause problems.
To allocate an array that can hold 10 int values, use your first expression: new int[10]. That will allocate space for ten int values.
To allocate an array that can be expanded at will, use std::vector<int>.
I'm trying to allocate a new array if integers (See HwGrades allocation below)
When I put the HwNum=2, the new function creates an array of size 1 only!
and when the for loop iterates 2 times it doesnt give me access violation
Help would be appreciated..
Here's the constructor
EE_Course::EE_Course(int Course_ID, char * Course_Name, int Hw_Num, double Hw_Weigh,int Factor_)
{
CourseID = Course_ID;
CourseName = new char[strlen(Course_Name) + 1];
strcpy(CourseName, Course_Name);
HwNum = Hw_Num;
HwWeigh = Hw_Weigh;
HwGrades = new int [HwNum]; // STARTING FROM HERE
for (int i = 0; i < Hw_Num; i++) { //UNTIL HERE
HwGrades[i] = 0;
}
Factor_ = 0;
ExamGrade = 0;
}
And those are the Course class private variables :
protected:
int CourseID;
int HwNum;
char* CourseName;
double HwWeigh;
int ExamGrade;
int* HwGrades;
};
The debugger does not show the whole array if it is a pointer. It shows the address of the array and the first element the array is pointing. So there is nothing wrong with your code.
You could see it if it was defined as an array:
int HwGrades[100];
If you really want to use a pointer and see it's content, you have two choices:
Define it as an array, debug it, fix/verify your code and turn back to pointer.
I don't know what is you environment, but usually there is a memory view option. You can check what's in the array any time you want. Just open the memory view of your IDE and watch the address of your pointer.
EDIT:
Apparently there is a third(and the best) option. See Rabbi Shuki's answer.
The debugger just shows one element. Here's why:
The type of HwGrades is int*. So when showing the contents of HwGrades what should the debugger do? The debugger does not know, that the pointer is actually pointing to the first element of an array. It assumes it just points to an int. Therefore, the debugger shows just the first element of the array that is actually of size 2.
If you're using the Visual Studio debugger, you can write HwGrades,2 in the watch window to see the first two elements of the array. Replace 2 by whatever your tickles your fancy. ;)
However, generally I would strongly advice to use the STL container std::vector for dynamic arrays. It will be easier to program and the debugger will be your friend without the hassle.
If you want to see the next cells of the array in the watch screen you can put the name and add a comma and the number of cells you want to see.
I.E.
HwGrades, 2
I've been reading through some books, and when it comes to Class/Functions using Pointers/Dynamic Memory (or heap or w/e they call it) I start to get confused.
Does anyone have a simple....like easy example they can show, because the books im using are using overly complex examples (large classes or multiple functions) and it makes it hard to follow. Pointers have always been my weak point anyways but I understand BASIC pointers, just classes/functions using them is a little bit confusing.
Also.....when would you use them is another question.
Stack allocation:
char buffer[1000];
Here the 1000 must be a constant. Memory is automatically freed when buffer goes out of scope.
Heap Allocation:
int bufsz = 1000;
char* buffer = new char[bufsz];
//...
delete [] buffer;
Here bufsz can be a variable. Memory must be freed explicitly.
When to use heap:
You don't know how much space you will need at compile time.
You want the memory/object to persist beyond the current scope.
You need a large chunk of memory (stack space is more limited than heap space)
Your computer's RAM is a big pile of bytes ordered one after another, and each one of those bytes can be accesed independently by it's address: an integer number startig from zero, upwards. A pointer is just a variable to hold that address of a single place in memory.
Since the RAM is a big chunk of bytes, the CPU ussually divides that big pile of bytes on several chunks. The most important ones are:
Code
Heap
Stack
The Code chunk is where the Assembly code lies. The Heap is a big pool of bytes used to allocate:
Global variables
Dynamic data, via the new operation on C++, or malloc() on C.
The stack is the chunk of memory that gets used to store:
Local variables
Function parameters
Return values (return statement on C/C++).
The main difference between the Stack and Heap is the way it is used. While the Heap is a big pool of bytes, the Stack "grows" like a stack of dishes: you can't remove the dish on the bottom unless there are no more dishes on it's top.
That's how recursion is implemented: every time you call a function recursively, memory grows on the stack, allocating parameters, local variables and storing return values of the returning functions, one on top of the others just like the stack of dishes.
Data living on the Stack have different "Life Span" than the data living on the Heap. Once a function exits, the data on the local variables get lost.
But if you allocate data on the Heap, that data won't get lost util you explicitly free that data with the delete or free() operations.
A pointer is basically a variable that contains the memory address of another variable (or in other cases to a function, but lets focus on the first).
That means that if I declare int[] x = {5,32,82,45,-7,0,123,8}; that variable will be allocated to memory at a certain address, lets say it got allocated on address 0x00000100 through 0x0000011F however we could have a variable which indicates a certain memory address and we can use that to access it.
So, our array looks like this
Address Contents
0x00000100 1
0x00000104 32
0x00000108 82
0x0000010B 45
0x00000110 -7
0x00000114 0
0x00000118 123
0x0000011B 8
If, for example, we were to create a pointer to the start of the array we could do this: int* p = &x; imagine this pointer variable got created a memory address 0x00000120 that way the memory at that address would contain the memory location for the start of array x.
Address Contents
0x00000120 0x00000100
You could then access the contents at that address through your pointer by dereferencing the pointer so that int y = *p would result in y = 1. We can also move the pointer, if we were to do p += 3; the pointer would be moved 3 addresses forward (note, however, that it moves 3 times the size of the type of object it is pointing to, here I am making examples with a 32 bit system in which an int is 32 bits or 4 bytes long, therefore the address would move by 4 bytes for each increment or 12 bytes in total so the pointer would end up pointing to 0x0000010B), if we were to dereference p again by doing y = *p; then we'd end up having y = 45. This is just the beginning, you can do a lot of things with pointers.
One of the other major uses is to pass a pointer as a parameter to a function so that it can do operations on certain values in memory without having to copy all of them over or make changes that will persist outside of the function's scope.
Warning: Don't do this. This is why we have vectors.
If you wanted to create an array of data, and return if from a function, how would you do it?
Obviously, this does not work:
int [10] makeArray(int val)
{
int arr[10];
for(int i=0; i<10; ++i)
arr[i] = val;
return arr;
}
You cannot return an array from a function. We can use pointers to refer to the first element of an array, like this:
int * makeArray(int val)
{
int arr[10];
for(int i=0; i<10; ++i)
arr[i] = val;
return &(arr[0]); // Return the address of the first element.
// Not strictly necessary, but I don't want to confuse.
}
This, however, also fails. arr is a local variable, it goes on the stack. When the function returns, the data is no longer valid, and now you have a pointer pointing to invalid data.
What we need to do is declare an array that will survive even after the function exits. For that, we use keyword new which creates that array, and returns the address to us, which needs to be stored in a pointer.
int * makeArray(int val)
{
int * arr = new int[10];
for(int i=0; i<10; ++i)
arr[i] = val;
return arr;
}
Then you can call that function and use that array like this:
int * a = makeArray(7);
for(int i=0; i<10; ++i)
std::cout << a[i] << std::endl;
delete [] a; // never forget this. Obviously you wouldn't do it right
// away like this, but you need to do it sometime.
Using pointers with new also gives you the advantage that you can determine the size of the array at runtime, something you can't do with local static arrays(though you can in C):
int * makeArray(int size, int val)
{
int * arr = new int[size];
for(int i=0; i<size; ++i)
arr[i] = val;
return arr;
}
That used to be one of the primary purposes for pointers. But like I said at the top, we don't do that anymore. We use vector.
One of the last vestiges of pointers is not for dynamic arrays. The only time I ever use them, is in classes where I want one object to have access to another object, without giving it ownership of that object. So, Object A needs to know about Object B, but even when Object A is gone, that doesn't affect Object B. You can also use references for this, but not if you need to give Object A the option to change which object it has access to.
(not tested, just writing down. and keeping things intentionally primitive, as requested.)
int* oneInt = new int; // allocate
*oneInt = 10; // use: assign a value
cout << *oneInt << endl; // use: retrieve (and print) the value
delete oneInt; // free the memory
now an array of ints:
int* tenInts = new int[10]; // allocate (consecutive) memory for 10 ints
tenInts[0] = 4353; // use: assign a value to the first entry in the array.
tenInts[1] = 5756; // ditto for second entry
//... do more stuff with the ints
delete [] tenInts; // free the memory
now with classes/objects:
MyClass* object = new MyClass(); // allocate memory and call class constructor
object->memberFunction("test"); // call a member function of the object
delete object; // free the object, calling the destructor
Is that what you wanted? I hope it helps.
I think this is what you're asking about:
Basically C++ doesn't allow variable-sized arrays. Any array in C++ has to be given a very specific size. But you can use pointers to work around that. Consider the following code:
int *arry = new int[10];
That just created an array of ints with 10 elements, and is pretty much the same exact thing as this:
int arry[] = int[10];
The only difference is that each one will use a different set of syntax. However imagine trying to do this:
Class class:
{
public:
void initArry(int size);
private:
int arry[];
};
void class::initArry(int size)
{
arry = int[size]; // bad code
}
For whatever reason C++ was designed to not allow regular arrays to be assigned sizes that are determined at runtime. Instead they have to be assigned sizes upon being coded. However the other way to make an array in C++ - using pointers - does not have this problem:
Class class:
{
public:
~class();
void initArry(int size);
private:
int *arry;
};
class::~class()
{
delete []arry;
}
void class::initArry(int size)
{
arry = new int[size]; // good code
}
You have to do some memory cleanup in the second example, hence why I included the destructor, but by using pointers that way you can size the array at runtime (with a variable size). This is called a dynamic array, and it is said that memory here is allocated dynamically. The other kind is a static array.
As far as 2-dimensional arrays go, you can handle it kind of like this:
Class class:
{
public:
~class();
void initArrays(int size1, int size2);
private:
int **arry;
};
class::~class()
{
delete [] arry[0];
delete [] arry[1];
delete [] arry;
}
void class::initArrays(int size1, int size2)
{
arry = new int*[2];
arry[0] = new int[size1];
arry[1] = new int[size2];
}
Disclaimer though: I haven't done much with this language in a while, so I may be slightly incorrect on some of the syntax.