People seem to say how malloc is so great when using arrays and you can use it in cases when you don't know how many elements an array has at compile time(?). Well, can't you do that without malloc? For example, if we knew we had a string that had max length 10 doesn't the following do close enough to the same thing?... Besides being able to free the memory that is.
char name[sizeof(char)*10];
and
char *name = malloc(sizeof(char)*10);
The first creates an array of chars on the stack. The length of the array will be sizeof(char)*10, but seeing as char is defined by the standard of being 1 in size, you could just write char name[10];
If you want an array, big enough to store 10 ints (defined per standard to be at least 2 bytes in size, but most commonly implemented as 4 bytes big), int my_array[10] works, too. The compiler can work out how much memory will be required anyways, no need to write something like int foo[10*sizeof(int)]. In fact, the latter will be unpredictable: depending on sizeof(int), the array will store at least 20 ints, but is likely to be big enough to store 40.
Anyway, the latter snippet calls a function, malloc wich will attempt to allocate enough memory to store 10 chars on the heap. The memory is not initialized, so it'll contain junk.
Memory on the heap is slightly slower, and requires more attention from you, who is writing the code: you have to free it explicitly.
Again: char is guaranteed to be size 1, so char *name = malloc(10); will do here, too. However, when working with heap memory, I -and I'm not alone in this- prefer to allocate the memory like so some_ptr = malloc(10*sizeof *some_ptr); using *some_ptr, is like saying 10 times the size of whatever type this pointer will point to. If you happen to change the type later on, you don't have to refactor all malloc calls.
General rule of thumb, to answer your question "can you do without malloc", is that you don't use malloc, unless you have to.
Stack memory is faster, and easier to use, but it is less abundant. This site was named after a well-known issue you can run into when you've pushed too much onto the stack: it overflows.
When you run your program, the system will allocate a chunk of memory that you can use freely. This isn't much, but plenty for simple computations and calling functions. Once you run out, you'll have to resort to allocating memory from the heap.
But in this case, an array of 10 chars: use the stack.
Other things to consider:
An array is a contguous block of memory
A pointer doesn't know/can't tell you how big a block of memory was allocated (sizeof(an_array)/sizeof(type) vs sizeof(a_pointer))
An array's declaration does not require the use of sizeof. The compiler works out the size for you: <type> my_var[10] will reserve enough memory to hold 10 elements of the given type.
An array decays into a pointer, most of the time, but that doesn't make them the same thing
pointers are fun, if you know what you're doing, but once you start adding functions, and start passing pointers to pointers to pointers, or a pointer to a pointer to a struct, that has members that are pointers... your code won't be as jolly to maintain. Starting off with an array, I find, makes it easier to come to grips with the code, as it gives you a starting point.
this answer only really applies to the snippets you gave, if you're dealing with an array that grows over time, than realloc is to be preferred. If you're declaring this array in a recursive function, that runs deep, then again, malloc might be the safer option, too
Check this link on differences between array and pointers
Also take a look at this question + answer. It explains why a pointer can't give you the exact size of the block of memory you're working on, and why an array can.
Consider that an argument in favour of arrays wherever possible
char name[sizeof(char)*10]; // better to use: char name[10];
Statically allocates a vector of sizeof(char)*10 char elements, at compile time. The sizeof operator is useless because if you allocate an array of N elements of type T, the size allocated will already be sizeof(T)*N, you don't need to do the math. Stack allocated and no free needed. In general, you use char name[10] when you already know the size of the object you need (the length of the string in this case).
char *name = malloc(sizeof(char)*10);
Allocates 10 bytes of memory in the heap. Allocation is done at run time, you need to free the result.
char name[sizeof(char)*10];
The first one is allocated on the stack, once it goes out of scope memory gets automatically freed. You can't change the size of the first one.
char *name = malloc(sizeof(char)*10);
The second one is allocated on the heap and should be freed with free. It will stick around otherwise for the lifetime of your application. You can reallocate memory for the second one if you need.
The storage duration is different:
An array created with char name[size] exists for the entire duration of program execution (if it is defined at file scope or with static) or for the execution of the block it is defined in (otherwise). These are called static storage duration and automatic storage duration.
An array created with malloc(size) exists for just as long as you specify, from the time you call malloc until the time you call free. Thus, it can be made to use space only while you need it, unlike static storage duration (which may be too long) or automatic storage duration (which may be too short).
The amount of space available is different:
An array created with char name[size] inside a function uses the stack in typical C implementations, and the stack size is usually limited to a few megabytes (more if you make special provisions when building the program, typically less in kernel software and embedded systems).
An array created with malloc may use gigabytes of space in typical modern systems.
Support for dynamic sizes is different:
An array created with char name[size] with static storage duration must have a size specified at compile time. An array created with char name[size] with automatic storage duration may have a variable length if the C implementation supports it (this was mandatory in C 1999 but is optional in C 2011).
An array created with malloc may have a size computed at run-time.
malloc offers more flexibility:
Using char name[size] always creates an array with the given name, either when the program starts (static storage duration) or when execution reaches the block or definition (automatic).
malloc can be used at run-time to create any number of arrays (or other objects), by using arrays of pointers or linked lists or trees or other data structures to create a multitude of pointers to objects created with malloc. Thus, if your program needs a thousand separate objects, you can create an array of a thousand pointers and use a loop to allocate space for each of them. In contrast, it would be cumbersome to write a thousand char name[size] definitions.
First things first: do not write
char name[sizeof(char)*10];
You do not need the sizeof as part of the array declaration. Just write
char name[10];
This declares an array of 10 elements of type char. Just as
int values[10];
declares an array of 10 elements of type int. The compiler knows how much space to allocate based on the type and number of elements.
If you know you'll never need more than N elements, then yes, you can declare an array of that size and be done with it, but:
You run the risk of internal fragmentation; your maximum number of bytes may be N, but the average number of bytes you need may be much smaller than that. For example, let's say you want to store 1000 strings of max length 255, so you declare an array like
char strs[1000][256];
but it turns out that 900 of those strings are only 20 bytes long; you're wasting a couple of hundred kilobytes of space1. If you split the difference and stored 1000 pointers, then allocated only as much space as was necessary to store each string, then you'd wind up wasting a lot less memory:
char *strs[1000];
...
strs[i] = strdup("some string"); // strdup calls malloc under the hood
...
Stack space is also limited relative to heap space; you may not be able to declare arbitrarily large arrays (as auto variables, anway). A request like
long double huge[10000][10000][10000][10000];
will probably cause your code to crash at runtime, because the default stack size isn't large enough to accomodate it2.
And finally, most situations fall into one of three categories: you have 0 elements, you have exactly 1 element, or you have an unlimited number of elements. Allocating large enough arrays to cover "all possible scenarios" just doesn't work. Been there, done that, got the T-shirt in multiple sizes and colors.
1. Yes, we live in the future where we have gigabytes of address space available, so wasting a couple of hundred KB doesn't seem like a big deal. The point is still valid, you're wasting space that you don't have to.
2. You could declare very large arrays at file scope or with the static keyword; this will allocate the array in a different memory segment (neither stack nor heap). The problem is that you only have that single instance of the array; if your function is meant to be re-entrant, this won't work.
Related
I want to create an array of 100 words and each word has 10 max characters but I also want to save in a variable the number of words in this array.
char array[1000];
Is this the most efficient way or can I use malloc() like this:
char *k;
k=(char *)malloc(1000 * sizeof(char));
How can I accomplish my task?
If you have an array of 100 words, perhaps you better use 2 dimesional array, like char array[100][11]. I'm specifying the second dimension as 11 because I'm taking into the consideration the null character (a word with max 10 chars + 1 null), the majority of string handling functions in C expect the strings to be null terminated.
it depends on how you want to use the stored data. What is its lifetime?
your first one is ambiguous:
char array[1000] outside a function is static data; inside a function its on the stack
the second one puts it on the heap.
You should read up on the differences between the characteristics of these 3 different allocation types (they all have pluses and minuses)
Stack, Static, and Heap in C++ is one place
In general, if you're looking for performance the declaration of char array[1000] is going to win because it's allocated on the stack.
When you allocate memory using malloc, it's coming from the heap. Also, it's good practice to always do static allocation if you can get away with it because it drastically reduces your chances of a memory leak, the memory will automatically be freed when you leave the scope.
When shall i use malloc instead of normal array definition in C?
I can't understand the difference between:
int a[3]={1,2,3}
int array[sizeof(a)/sizeof(int)]
and:
array=(int *)malloc(sizeof(int)*sizeof(a));
In general, use malloc() when:
the array is too large to be placed on the stack
the lifetime of the array must outlive the scope where it is created
Otherwise, use a stack allocated array.
int a[3]={1,2,3}
int array[sizeof(a)/sizeof(int)]
If used as local variables, both a and array would be allocated on the stack. Stack allocation has its pros and cons:
pro: it is very fast - it only takes one register subtraction operation to create stack space and one register addition operation to reclaim it back
con: stack size is usually limited (and also fixed at link time on Windows)
In both cases the number of elements in each arrays is a compile-time constant: 3 is obviously a constant while sizeof(a)/sizeof(int) can be computed at compile time since both the size of a and the size of int are known at the time when array is declared.
When the number of elements is known only at run-time or when the size of the array is too large to safely fit into the stack space, then heap allocation is used:
array=(int *)malloc(sizeof(int)*sizeof(a));
As already pointed out, this should be malloc(sizeof(a)) since the size of a is already the number of bytes it takes and not the number of elements and thus additional multiplication by sizeof(int) is not necessary.
Heap allocaiton and deallocation is relatively expensive operation (compared to stack allocation) and this should be carefully weighted against the benefits it provides, e.g. in code that gets called multitude of times in tight loops.
Modern C compilers support the C99 version of the C standard that introduces the so-called variable-length arrays (or VLAs) which resemble similar features available in other languages. VLA's size is specified at run-time, like in this case:
void func(int n)
{
int array[n];
...
}
array is still allocated on the stack as if memory for the array has been allocated by a call to alloca(3).
You definately have to use malloc() if you don't want your array to have a fixed size. Depending on what you are trying to do, you might not know in advance how much memory you are going to need for a given task or you might need to dynamically resize your array at runtime, for example you might enlarge it if there is more data coming in. The latter can be done using realloc() without data loss.
Instead of initializing an array as in your original post you should just initialize a pointer to integer like.
int* array; // this variable will just contain the addresse of an integer sized block in memory
int length = 5; // how long do you want your array to be;
array = malloc(sizeof(int) * length); // this allocates the memory needed for your array and sets the pointer created above to first block of that region;
int newLength = 10;
array = realloc(array, sizeof(int) * newLength); // increase the size of the array while leaving its contents intact;
Your code is very strange.
The answer to the question in the title is probably something like "use automatically allocated arrays when you need quite small amounts of data that is short-lived, heap allocations using malloc() for anything else". But it's hard to pin down an exact answer, it depends a lot on the situation.
Not sure why you are showing first an array, then another array that tries to compute its length from the first one, and finally a malloc() call which tries do to the same.
Normally you have an idea of the number of desired elements, rather than an existing array whose size you want to mimic.
The second line is better as:
int array[sizeof a / sizeof *a];
No need to repeat a dependency on the type of a, the above will define array as an array of int with the same number of elements as the array a. Note that this only works if a is indeed an array.
Also, the third line should probably be:
array = malloc(sizeof a);
No need to get too clever (especially since you got it wrong) about the sizeof argument, and no need to cast malloc()'s return value.
Well, I can't understand when and why it is needed to allocate memory using malloc.
Here is my code:
#include <stdlib.h>
int main(int argc, const char *argv[]) {
typedef struct {
char *name;
char *sex;
int age;
} student;
// Now I can do two things
student p;
// Or
student *ptr = (student *)malloc(sizeof(student));
return 0;
}
Why is it needed to allocate memory when I can just use student p;?
malloc is used for dynamic memory allocation. As said, it is dynamic allocation which means you allocate the memory at run time. For example, when you don't know the amount of memory during compile time.
One example should clear this. Say you know there will be maximum 20 students. So you can create an array with static 20 elements. Your array will be able to hold maximum 20 students. But what if you don't know the number of students? Say the first input is the number of students. It could be 10, 20, 50 or whatever else. Now you will take input n = the number of students at run time and allocate that much memory dynamically using malloc.
This is just one example. There are many situations like this where dynamic allocation is needed.
Have a look at the man page malloc(3).
You use malloc when you need to allocate objects that must exist beyond the lifetime of execution of the current block (where a copy-on-return would be expensive as well), or if you need to allocate memory greater than the size of that stack (i.e., a 3 MB local stack array is a bad idea).
Before C99 introduced VLAs, you also needed it to perform allocation of a dynamically-sized array. However, it is needed for creation of dynamic data structures like trees, lists, and queues, which are used by many systems. There are probably many more reasons; these are just a few.
Expanding the structure of the example a little, consider this:
#include <stdio.h>
int main(int argc, const char *argv[]) {
typedef struct {
char *name;
char *sex;
char *insurance;
int age;
int yearInSchool;
float tuitionDue;
} student;
// Now I can do two things
student p;
// Or
student *p = malloc(sizeof *p);
}
C is a language that implicitly passes by value, rather than by reference. In this example, if we passed 'p' to a function to do some work on it, we would be creating a copy of the entire structure. This uses additional memory (the total of how much space that particular structure would require), is slower, and potentially does not scale well (more on this in a minute). However, by passing *p, we don't pass the entire structure. We only are passing an address in memory that refers to this structure. The amount of data passed is smaller (size of a pointer), and therefore the operation is faster.
Now, knowing this, imagine a program (like a student information system) which will have to create and manage a set of records in the thousands, or even tens of thousands. If you pass the whole structure by value, it will take longer to operate on a set of data, than it would just passing a pointer to each record.
Let's try and tackle this question considering different aspects.
Size
malloc allows you to allocate much larger memory spaces than the one allocated simply using student p; or int x[n];. The reason being malloc allocates the space on heap while the other allocates it on the stack.
The C programming language manages memory statically, automatically, or dynamically. Static-duration variables are allocated in main memory, usually along with the executable code of the program, and persist for the lifetime of the program; automatic-duration variables are allocated on the stack and come and go as functions are called and return. For static-duration and automatic-duration variables, the size of the allocation must be compile-time constant (except for the case of variable-length automatic arrays[5]). If the required size is not known until run-time (for example, if data of arbitrary size is being read from the user or from a disk file), then using fixed-size data objects is inadequate. (from Wikipedia)
Scope
Normally, the declared variables would get deleted/freed-up after the block in which it is declared (they are declared on the stack). On the other hand, variables with memory allocated using malloc remain till the time they are manually freed up.
This also means that it is not possible for you to create a variable/array/structure in a function and return its address (as the memory that it is pointing to, might get freed up). The compiler also tries to warn you about this by giving the warning:
Warning - address of stack memory associated with local variable 'matches' returned
For more details, read this.
Changing the Size (realloc)
As you may have guessed, it is not possible by the normal way.
Error detection
In case memory cannot be allocated: the normal way might cause your program to terminate while malloc will return a NULL which can easily be caught and handled within your program.
Making a change to string content in future
If you create store a string like char *some_memory = "Hello World"; you cannot do some_memory[0] = 'h'; as it is stored as string constant and the memory it is stored in, is read-only. If you use malloc instead, you can change the contents later on.
For more information, check this answer.
For more details related to variable-sized arrays, have a look at this.
malloc = Memory ALLOCation.
If you been through other programming languages, you might have used the new keyword.
Malloc does exactly the same thing in C. It takes a parameter, what size of memory needs to be allocated and it returns a pointer variable that points to the first memory block of the entire memory block, that you have created in the memory. Example -
int *p = malloc(sizeof(*p)*10);
Now, *p will point to the first block of the consecutive 10 integer blocks reserved in memory.
You can traverse through each block using the ++ and -- operator.
In this example, it seems quite useless indeed.
But now imagine that you are using sockets or file I/O and must read packets from variable length which you can only determine while running. Or when using sockets and each client connection need some storage on the server. You could make a static array, but this gives you a client limit which will be determined while compiling.
int numbers*;
numbers = malloc ( sizeof(int) * 10 );
I want to know how is this dynamic memory allocation, if I can store just 10 int items to the memory block ? I could just use the array and store elemets dynamically using index. Why is the above approach better ?
I am new to C, and this is my 2nd day and I may sound stupid, so please bear with me.
In this case you could replace 10 with a variable that is assigned at run time. That way you can decide how much memory space you need. But with arrays, you have to specify an integer constant during declaration. So you cannot decide whether the user would actually need as many locations as was declared, or even worse , it might not be enough.
With a dynamic allocation like this, you could assign a larger memory location and copy the contents of the first location to the new one to give the impression that the array has grown as needed.
This helps to ensure optimum memory utilization.
The main reason why malloc() is useful is not because the size of the array can be determined at runtime - modern versions of C allow that with normal arrays too. There are two reasons:
Objects allocated with malloc() have flexible lifetimes;
That is, you get runtime control over when to create the object, and when to destroy it. The array allocated with malloc() exists from the time of the malloc() call until the corresponding free() call; in contrast, declared arrays either exist until the function they're declared in exits, or until the program finishes.
malloc() reports failure, allowing the program to handle it in a graceful way.
On a failure to allocate the requested memory, malloc() can return NULL, which allows your program to detect and handle the condition. There is no such mechanism for declared arrays - on a failure to allocate sufficient space, either the program crashes at runtime, or fails to load altogether.
There is a difference with where the memory is allocated. Using the array syntax, the memory is allocated on the stack (assuming you are in a function), while malloc'ed arrays/bytes are allocated on the heap.
/* Allocates 4*1000 bytes on the stack (which might be a bit much depending on your system) */
int a[1000];
/* Allocates 4*1000 bytes on the heap */
int *b = malloc(1000 * sizeof(int))
Stack allocations are fast - and often preferred when:
"Small" amount of memory is required
Pointer to the array is not to be returned from the function
Heap allocations are slower, but has the advantages:
Available heap memory is (normally) >> than available stack memory
You can freely pass the pointer to the allocated bytes around, e.g. returning it from a function -- just remember to free it at some point.
A third option is to use statically initialized arrays if you have some common task, that always requires an array of some max size. Given you can spare the memory statically consumed by the array, you avoid the hit for heap memory allocation, gain the flexibility to pass the pointer around, and avoid having to keep track of ownership of the pointer to ensure the memory is freed.
Edit: If you are using C99 (default with the gnu c compiler i think?), you can do variable-length stack arrays like
int a = 4;
int b[a*a];
In the example you gave
int *numbers;
numbers = malloc ( sizeof(int) * 10 );
there are no explicit benefits. Though, imagine 10 is a value that changes at runtime (e.g. user input), and that you need to return this array from a function. E.g.
int *aFunction(size_t howMany, ...)
{
int *r = malloc(sizeof(int)*howMany);
// do something, fill the array...
return r;
}
The malloc takes room from the heap, while something like
int *aFunction(size_t howMany, ...)
{
int r[howMany];
// do something, fill the array...
// you can't return r unless you make it static, but this is in general
// not good
return somethingElse;
}
would consume the stack that is not so big as the whole heap available.
More complex example exists. E.g. if you have to build a binary tree that grows according to some computation done at runtime, you basically have no other choices but to use dynamic memory allocation.
Array size is defined at compilation time whereas dynamic allocation is done at run time.
Thus, in your case, you can use your pointer as an array : numbers[5] is valid.
If you don't know the size of your array when writing the program, using runtime allocation is not a choice. Otherwise, you're free to use an array, it might be simpler (less risk to forget to free memory for example)
Example:
to store a 3-D position, you might want to use an array as it's alwaays 3 coordinates
to create a sieve to calculate prime numbers, you might want to use a parameter to give the max value and thus use dynamic allocation to create the memory area
Array is used to allocate memory statically and in one go.
To allocate memory dynamically malloc is required.
e.g. int numbers[10];
This will allocate memory statically and it will be contiguous memory.
If you are not aware of the count of the numbers then use variable like count.
int count;
int *numbers;
scanf("%d", count);
numbers = malloc ( sizeof(int) * count );
This is not possible in case of arrays.
Dynamic does not refer to the access. Dynamic is the size of malloc. If you just use a constant number, e.g. like 10 in your example, it is nothing better than an array. The advantage is when you dont know in advance how big it must be, e.g. because the user can enter at runtime the size. Then you can allocate with a variable, e.g. like malloc(sizeof(int) * userEnteredNumber). This is not possible with array, as you have to know there at compile time the (maximum) size.
Am I correct in thinking that:
char *buff[500];
... creates a stack variable, and:
char *buff = (char *)malloc(500);
... creates a heap variable?
If that's correct, when and why would you use heap variables over stack variables and vice versa. I understand the stack is faster is there anything else.
One last question, is the main function a stack frame on the stack?
Yes, first one creates an array of char pointers in the stack, about 500*4 bytes and second one allocates 500 chars in the heap and points a stack char ptr to them.
Allocating in the stack is easy and fast, but stack is limited, heap is slower but much bigger. Apart from that, stack allocated values are "deleted" once you leave the scope, so it is very good for small local values like primitive variables.
If you allocate too much in the stack you might run out of stack and die, main as all the functions you execute has a stack frame in the stack and all the local variables to the function are stored there, so going too deep into function calling might get you into a stackoverflow as well.
In general is a good rule of thumb to allocate anything that you use often and is bigger than a hundred bytes in the heap, and small variables and pointers in the stack.
Seeing that you wrote
char *buff = (char *)malloc(500);
you probably meant
char buff[500]; instead of
char* buff[500];
in your first example (so you have a char-array, not an array of pointers to chars)
Yes, "allocation" on the stack is faster because you just increase a pointer stored in the ESP register.
You need heap-variables if you want:
1) more memory than fits in the stack (generally much earlier)
2) pass memory that was allocated by a called function to the calling function.
Your buffs are not equivalent.
The first one (char *buff[500]) is an array of 500 pointers; the 2nd one (char *buff = (char *)malloc(500)) is a pointer.
The pointer (on the stack) points to 500 bytes of memory (if the malloc call succeeded) on the heap.
The array of pointers is on the stack. Its pointers are not initialized.
Unless using C99, using the stack the size of your array must be known at compile-time. That means you cannot do:
int size = 3; // somewhere, possibly from user input
char *buff[size];
But using "the heap" (dynamic allocation), you can provide any dimensions you like. That's because the memory allocation is performed at run-time, rather than hardcoded into the executable.
The C standard contains neither the words heap nor stack. What we have here instead are two storage durations (of 4 in total): automatic and allocated:
char buff[500]; // note the missing * to match the malloc example
within a function declares the object buff as an array of char and having automatic storage duration. The object will cease to be when the block where the object was declared, is exited.
char *buff = malloc(500); // no cast necessary; this is C
will declare buff as a pointer to char. malloc will reserve 500 continuous bytes and return a pointer to it. The returned 500-byte object will exist until it is explicitly freed with a call to free. The object is said to have allocated storage duration.
That's all the C standard says. It doesn't specify that the char buff[500] needs to be allocated from a "stack", or that there needs to be a stack. It doesn't specify that the malloc needs to use some "heap". On the contrary, a compiler might internally implement the char buff[500] like
{
char *buff = malloc(500);
free(buff);
}
Or it can deduce that the allocated memory is not used, or that it is only used once, and use a stack-based allocation instead of actually calling malloc.
In practice, most current compilers and environments will use a memory layout called stack for automatic variables, and the objects with allocated storage duration are said to come from "heap" which is a metaphor for the unorganized mess that it is compared to the orderly stack, but it is not something that has to be so.
Heap variables can be created dynamically, ie you can ask a size to your user and malloc a new variable with this size.
The size of a stack variable must be known at compile time.
Like you said, stack variable are faster allocated and accessed. So i'll recommend using them every time you know the size at compile time. Otherwise you don't have the choice, you must use malloc()
This is indeed a variable allocated on the stack:
char buff[500]; // You had a typo here?
and this is on the heap:
char *buff = (char *)malloc(500);
Why would you use one vs the other?
In char *buff[500], the 500 needs to be a compile-time constant. You can't use it if 500 is computed at runtime.
On the other hand, stack allocations are instantaneous while heap allocations take time (thus they incur a runtime performance cost).
Space on the stack is limited by the thread's stack size (typically 1MB before you get a stack overflow), while there's much more available on the heap.
If you allocate an array on the stack big enough to take up more than 2 pages of virtual memory as managed by the OS, and access the end of the array before doing anything else, there's the possibility of getting a protection fault (this depends on the OS)
Finally: every function called has a frame on the stack. The main function is no different. It isn't even any more special than the other functions in your program, since when your program starts running the first code that runs is inside the C runtime environment. After the runtime is ready to begin execution of your own code, it calls main just as you would call any other function.
Those two aren't equivalent. The first is an array of size 500 (on the stack) with pointers to characters. The second is a pointer to a memory chunk of 500 which can be used with the indexing operator.
char buff[500];
char *buff = (char *)malloc(sizeof(char)*500);
Stack variables should be preferred because they require no deallocation. Heap variables allow passing of data between scopes as well as dynamic allocation.