Bit confused about usage of new to allocate memory dynamically.
e.g. If I need to allocate memory for 100 ints(assuming int is 4 bytes), should I say :
int *ptr = new int[100];
or
int *ptr = new int[100*4]; //assuming int is 4 bytes.
Basially new operator allocates memory in bytes or that many bytes of type T used while invoking the new operator?
If my class doesn't have a allocator member function defined, & i need to allocate an array of object of that class type, will the new oeprator find the sizeof(class type) and allocate accordingly or how would it work?
EDIT:
Sorry for clubbing multiple questions, but its related:
will this piece of code work fine if i want to allocate a 2D array of size [100][4] of ints
int *arr = new int [100][4];
thank you.
-AD
The size given to new is the number of items, not the memory size. However, consider using std::vector instead.
For example, new int[100] allocates at least 100 * sizeof(int) bytes (which is 400 when sizeof(int) is 4); any more it allocates will be due to implementation and runtime details which you (except for very rarely) cannot depend on.
If you don't have an operator new or operator new[] in your class (which you usually shouldn't), then the global versions will be used, and they will use sizeof(your_type) correctly.
Did you try the code in the update?
Multidimensional arrays are actually arrays of arrays; so new returns a pointer to the first item just as it does for single dimensional arrays, which is a pointer to an array:
typedef int int4[4]; // for clarity
int4 *arr = new int4[100];
// above two lines are identical to:
int (*arr)[4] = new int[100][4];
Again, you're almost always better off using a container than managing this yourself. That's vector, et. al., but also containers like boost::array and dedicated "matrix" types for "square" 2-dimensional arrays.
The confusion may come because of malloc which uses data size, not the number of items. To allocate an array of 100 integers, you can do:
int *ptr = (int *)malloc(100 * 4);
or (better, works for all platforms because an int is not always 4 bytes long):
int *ptr = (int *)malloc(100 * sizeof(int));
instead of:
int *ptr = new int[100];
which, as Fred says, automatically uses the data size to allocate the correct area. Of course malloc can only be used for built-in types such as int.
Related
Because segmentation fault related to malloc/free happens, I would like to convert malloc/free to new/delete.
Error occurred when malloc/free is converted to below.
Let me know how to solve it.
(original)
char *only_valid_data = static_cast<char*> (malloc (data_size));
(converted)
char *only_valid_data = new static_cast<char*> [data_size];
Just do
char* only_valid_data = new char[data_size];
to allocate.
To free you do
delete[] only_valid_data;
Important note: When you allocate memory with new it will allocate data_size elements, not data_size bytes (like malloc does). The size of an element is the size of the non-pointer base type, in your case e.g. sizeof(*only_valid_data). In this case the element size and the byte size is the same (as sizeof(char) is specified to always be 1), but if the base type is something else it will make a big difference.
For example, if you do
int* int_ptr = new int[10];
then ten integers will be allocated, not ten bytes. This is equivalent to
int* int_ptr = reinterpret_cast<int*>(malloc(10 * sizeof(*int_ptr)));
Also note that for complex types, like structures and classes, allocating with new (or new[]) does more than just allocating memory, it will also make sure that the allocated object(s) constructor is called. The malloc function only allocates memory, it doesn't call the constructor.
Final note: The problem you have with the segmentation fault is probably not caused by your allocation, no matter how you allocate the memory. The problem is more likely because of something else, something you do not show in your question, like writing out of bounds of the allocated memory or dereferencing a null-pointer.
You need run your program in a debugger to catch the crash in action, it will allow you to examine the function call stack, and if the crash doesn't happen in your code then you walk up the call stack until you reach your code. There you can examine the values of variables, to help you understand why the problem occurred.
The malloc family (malloc, realloc, calloc, free) is used almost always in C code, as C++ provides the new and delete operators which are a lot more reliable to use.
A problem with malloc for allocation is that you must specify the size of the type in bytes that you want to allocate. For example:
int* ptr = malloc(5);
Will not allocate space for 5 integers in memory; it will allocate 5 bytes of memory (the size of an integer is 4 bytes, so this would obviously cause problems when assigning).
To do it properly, it must be written as
int* ptr = malloc(5 * sizeof(int));
So that 20 bytes are allocated.
However, there are some exceptions to the case. char, for example only requires one byte of memory, so doing
char* ptr = malloc(5);
Will allocate enough memory to hold 5 characters, and in a way is more valid that writing:
char* ptr = malloc(5 * sizeof(char)); //5 * sizeof(char) == 5 * 1 == 5
However, the free function does not need to know the size of the pointer to be deallocated; a void* is only needed.
Note that in C++, the return of malloc must be cast properly to the type wanted; malloc returns a void* type, but C++ does not allow any pointer to assign a void* to any pointer type like C does:
int* ptr = malloc(5 * sizeof(int)); //valid C code, invalid C++
int* ptr2 = (int*)malloc(5 * sizeof(int)); //valid C code, valid C++
In C++, the new[] operator resolves the issue of remembering to add the sizeof operator.
int* ptr = new int[5];//allocates 5 integers
int* ptr2 = new int(5);//be careful: this allocates a single integer with value of 5
Note that if the new[] operator has been used, the delete[] operator must be used. Otherwise the delete operator must be used:
int* ptr = new int[5];//allocates 5 integers
delete[] ptr;//deallocate the 5 integers
int* ptr2 = new int(5);//be careful: this allocates a single integer with value of 5
delete ptr;//deallocate the integer
The problem with your code is that it does not fit the syntax of the new[] operator
The syntax could be described as:
T* p = new T[size];
Thus your code:
char *only_valid_data = new static_cast<char*> [data_size];
Should be corrected to:
char *only_valid_data = new char[data_size];
As static_cast<char*> is not a type.
Hope this helps :)
What is the difference between this two array definitions and which one is more correct and why?
#include <stdio.h>
#define SIZE 20
int main() {
// definition method 1:
int a[SIZE];
// end definition method 1.
// defintion method 2:
int n;
scanf("%d", &n);
int b[n];
// end definition method 2.
return 0;
}
I know if we read size, variable n, from stdin, it's more correct to define our (block of memory we'll be using) array as a pointer and use stdlib.h and array = malloc(n * sizeof(int)), rather than decalring it as int array[n], but again why?
It's not "more correct" or "less correct". It either is xor isn't correct. In particular, this works in C, but not in C++.
You are declaring dynamic arrays. Better way to declare Dynamic arrays as
int *arr; // int * type is just for simplicity
arr = malloc(n*sizeof(int*));
this is because variable length arrays are only allowed in C99 and you can't use this in c89/90.
In (pre-C99) C and C++, all types are statically sized. This means that arrays must be declared with a size that is both constant and known to the compiler.
Now, many C++ compilers offer dynamically sized arrays as a nonstandard extension, and C99 explicitly permits them. So int b[n] will most likely work if you try it. But in some cases, it will not, and the compiler is not wrong in those cases.
If you know SIZE at compile-time:
int ar[SIZE];
If you don't:
std::vector<int> ar;
I don't want to see malloc anywhere in your C++ code. However, you are fundamentally correct and for C that's just what you'd do:
int* ptr = malloc(sizeof(int) * SIZE);
/* ... */
free(ptr);
Variable-length arrays are a GCC extension that allow you to do:
int ar[n];
but I've had issues where VLAs were disabled but GCC didn't successfully detect that I was trying to use them. Chaos ensues. Just avoid it.
Q1 : First definition is the static array declaration. Perfectly correct.
It is when you have the size known, so no comparison with VLA or malloc().
Q2 : Which is better when taking size as an input from the user : VLA or malloc .
VLA : They are limited by the environment's bounds on the size of automatic
allocation. And automatic variables are usually allocated on the stack which is relatively
small.The limitation is platform specific.Also, this is in c99 and above only.Some ease of use while declaring multidimensional arrays is obtained by VLA.
Malloc : Allocates from the heap.So, for large size is definitely better.For, multidimensional arrays pointers are involved so a bit complex implementataion.
Check http://bytes.com/topic/c/answers/578354-vla-feature-c99-vs-malloc
I think that metod1 could be little bit faster, but both of them are correct in C.
In C++ first will work, but if you want to use a second you should use:
int size = 5;
int * array = new int[size];
and remember to delete it:
delete [] array;
I think it gives you more option to use while coding.
If you use malloc or other dynamic allocation to get a pointer. You will use like p+n..., but if you use array, you could use array[n]. Also, while define pointer, you need to free it; but array does not need to free.
And in C++, we could define user-defined class to do such things, and in STL, there is std::vector which do the array-things, and much more.
Both are correct. the declaration you use depends on your code.
The first declaration i.e. int a[size]; creates an array with a fixed size of 20 elements.
It is helpful when you know the exact size of the array that will be used in the code. for example, you are generating
table of a number n up till its 20th multiple.
The second declaration allows you to make an array of the size that you desire.
It is helpful when you will need an array of different sizes, each time the code is executed for example, you want to generate the fibonacci series till n. In that case, the size of the array must be n for each value of n. So say you have n = 5, in this case int a [20] will waste memory because only the first five slots will be used for the fibonacci series and the rest will be empty. Similarly if n = 25 then your array int a[20] will become too small.
The difference if you define array using malloc is that, you can pass the size of array dynamically i.e at run time. You input a value your program has during run time.
One more difference is that arrays created using malloc are allocated space on heap. So they are preserved across function calls unlike static arrays.
example-
#include<stdio.h>
#include<stdlib.h>
int main()
{
int n;
int *a;
scanf("%d",&n);
a=(int *)malloc(n*sizeof(int));
return 0;
}
Consider the following code
struct foo
{
const int txt_len;
const int num_len;
char * txt;
int * num;
foo(int tl, int nl): txt_len(tl), num_len(nl)
{
char * tmp = new char[txt_len * sizeof(char) + num_len * sizeof(int)];
txt = new (tmp) char [txt_len * sizeof(char)];
num = new (tmp + txt_len * sizeof(char)) int[num_len * sizeof(int)];
// is this the same as above?
// txt = tmp;
// num = (int *) (tmp + txt_len * sizeof(char));
}
~foo()
{
delete[] txt; // is this the right way to free the memory?
}
};
I want *txt and *num to be contiguous, is that the best way to do it?
also is there any difference between placement new and pointer arithmetic? which one should I use?
If you want a contiguous block of memory, you have to allocate it whole with a single call to operator new[] or malloc() or similar. Multiple calls to these functions do not guarantee any contiguity of allocated blocks whatsoever. You may allocate a big block and then carve parts from it as needed.
And you should delete and free() all blocks previously allocated with new and malloc(), otherwise you'll leak memory and probably make your program unstable (it will fail to allocate more memory at some point) and exert unnecessary pressure on memory in the OS, possibly slowing down other programs or making them unstable as well.
Placement new, however, does not actually allocate any memory. It simply constructs an object at the specified location and so you don't need to free that memory twice.
One problem that I see in your code is that it doesn't align ints. On some platforms reading or writing integers bigger than 1 byte from/to the memory must be aligned and if it's not, you can either read/write values from/to wrong locations or get CPU exceptions leading to termination of your program. The x86 is very permissive in this regard and won't mind, though may tax you with degraded performance.
You'll need to put the int data first, due to the alignment issues. But we can't then do delete num[] as the type is wrong - it must be cast to a char* before deleting.
char * tmp = new char[num_len * sizeof(int) + txt_len * sizeof(char)];
num = new (tmp) int[num_len];
txt = new (tmp + num_len * sizeof(int)) char [txt_len];
(This makes liberal use of the fact that sizeof(char)==1)
You might be tempted to do delete[] num, but num is of type int*, and it was new'ed as a char*. So you need to do;
delete[] (char*) num;
This is the same as long as you use POD types. And your delete is fine.
However, as David's comment states, you need to consider alignment problems.
Placement new is mostly use when you want to call constructor of class/struct on some preallocated memory blocks.
But for native types it makes no different to use placement new & pointer arithmetic.
Please correct me if I was wrong.
If txt and num always point to int and char, other built in types or other types not requiring construction, then no. You don't need placement new.
If on the other hand you were to change one of them to a class which requires construction, i.e. changes txt to type std::string, then using placement new is necessary.
Placement new allows you to call the constructor, building, if you like, the object at that address. Built in types have default constructors that do nothing if your not initializing.
In both cases you need to do pointer arithmetic, just one way you store the answer in a pointer, the other you pass the answer to placement new which gives it back to you for storage in the pointer, and then calls the constructor.
there is no problem here:
int *x;
x = (int *) malloc(0 * sizeof(int));
int value = 5;
x = (int *) realloc(x,(value)*sizeof(int));
But I cant do that for strings :\
I wanna do that for a y[] array, like this:
y[0]="hello"
y[1]="how are you"
how can I do that?
std::vector<std::string> y;
y.push_back("hello");
y.push_back("how are you");
Don't use malloc, or realloc, or free in C++. Don't use char* for purposes other than interop. Stay away from pointers until you actually need them (same for arrays).
You cannot use realloc on an array of std::string objects because realloc moves things around by bit copying and this is not allowed on general objects.
The standard class std::vector is a generic container for objects that moves and copies things around correctly (using copy constructors, assignments and similar methods) and that can change its size for example with the resize method. All the needed memory is allocated and freed automatically as needed.
With std::vector for example you can write code like...
std::vector<std::string> v; // An empty vector
v.resize(10); // Now size is 10 elements (all empty strings "")
v[0] = "Hello"; // First element is now the string "Hello"
v[1] = "world."; // Second element is now the string "world."
v.resize(2); // Now v.size() is 2
v.push_back("How's going"); // Now the size is 3 and third element is filled.
Do yourself a favor and pick up a good C++ book, reading it cover to cover. C++ is a powerful but complex language and if you try to learn it by experimenting with a compiler you're making a terrible mistake for many reasons.
What you're doing right now is not C++ ... you can do what you want with C-style strings, but you would need an array of pointers to type char that allow you to access the allocated memory for each string in the array. This can be done like so:
char* y[2];
y[0] = strdup("hello");
y[1] = strdup("how are you");
You also need to keep in mind that your y array now "owns" the pointers, so you must call free() on each pointer in the array in order to avoid any memory leaks should you decide to change the strings each pointer is pointing to.
If you want to go with an idiomatic C++ solution though, and not revert to C-style strings, then you should be using std::string and std::vector ... doing so avoids the issues with memory leaks as well as allocating and deallocating the memory associated with dynamically allocated C-strings.
You can actually do exactly what you need exactly like with integers:
typedef const char *c_string;
c_string *y;
y = (c_string *) malloc(0 * sizeof(c_string));
int value = 5;
y = (c_string *) realloc(y,(value)*sizeof(c_string));
y[0]="hello";
y[1]="how are you";
This won't work with non-const char * though, so this example is of limited usability.
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.