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.
Related
I have been reading that malloc is used for dynamic memory allocation. But if the following code works...
int main(void) {
int i, n;
printf("Enter the number of integers: ");
scanf("%d", &n);
// Dynamic allocation of memory?
int int_arr[n];
// Testing
for (int i = 0; i < n; i++) {
int_arr[i] = i * 10;
}
for (int i = 0; i < n; i++) {
printf("%d ", int_arr[i]);
}
printf("\n");
}
... what is the point of malloc? Isn't the code above just a simpler-to-read way to allocate memory dynamically?
I read on another Stack Overflow answer that if some sort of flag is set to "pedantic", then the code above would produce a compile error. But that doesn't really explain why malloc might be a better solution for dynamic memory allocation.
Look up the concepts for stack and heap; there's a lot of subtleties around the different types of memory. Local variables inside a function live in the stack and only exist within the function.
In your example, int_array only exists while execution of the function it is defined in has not ended, you couldn't pass it around between functions. You couldn't return int_array and expect it to work.
malloc() is used when you want to create a chunk of memory which exists on the heap. malloc returns a pointer to this memory. This pointer can be passed around as a variable (eg returned) from functions and can be used anywhere in your program to access your allocated chunk of memory until you free() it.
Example:
'''C
int main(int argc, char **argv){
int length = 10;
int *built_array = make_array(length); //malloc memory and pass heap pointer
int *array = make_array_wrong(length); //will not work. Array in function was in stack and no longer exists when function has returned.
built_array[3] = 5; //ok
array[3] = 5; //bad
free(built_array)
return 0;
}
int *make_array(int length){
int *my_pointer = malloc( length * sizeof int);
//do some error checking for real implementation
return my_pointer;
}
int *make_array_wrong(int length){
int array[length];
return array;
}
'''
Note:
There are plenty of ways to avoid having to use malloc at all, by pre-allocating sufficient memory in the callers, etc. This is recommended for embedded and safety critical programs where you want to be sure you'll never run out of memory.
Just because something looks prettier does not make it a better choice.
VLAs have a long list of problems, not the least of which they are not a sufficient replacement for heap-allocated memory.
The primary -- and most significant -- reason is that VLAs are not persistent dynamic data. That is, once your function terminates, the data is reclaimed (it exists on the stack, of all places!), meaning any other code still hanging on to it are SOL.
Your example code doesn't run into this problem because you aren't using it outside of the local context. Go ahead and try to use a VLA to build a binary tree, then add a node, then create a new tree and try to print them both.
The next issue is that the stack is not an appropriate place to allocate large amounts of dynamic data -- it is for function frames, which have a limited space to begin with. The global memory pool, OTOH, is specifically designed and optimized for this kind of usage.
It is good to ask questions and try to understand things. Just be careful that you don't believe yourself smarter than the many, many people who took what now is nearly 80 years of experience to design and implement systems that quite literally run the known universe. Such an obvious flaw would have been immediately recognized long, long ago and removed before either of us were born.
VLAs have their place, but it is, alas, small.
Declaring local variables takes the memory from the stack. This has two ramifications.
That memory is destroyed once the function returns.
Stack memory is limited, and is used for all local variables, as well as function return addresses. If you allocate large amounts of memory, you'll run into problems. Only use it for small amounts of memory.
When you have the following in your function code:
int int_arr[n];
It means you allocated space on the function stack, once the function will return this stack will cease to exist.
Image a use case where you need to return a data structure to a caller, for example:
Car* create_car(string model, string make)
{
Car* new_car = malloc(sizeof(*car));
...
return new_car;
}
Now, once the function will finish you will still have your car object, because it was allocated on the heap.
The memory allocated by int int_arr[n] is reserved only until execution of the routine ends (when it returns or is otherwise terminated, as by setjmp). That means you cannot allocate things in one order and free them in another. You cannot allocate a temporary work buffer, use it while computing some data, then allocate another buffer for the results, and free the temporary work buffer. To free the work buffer, you have to return from the function, and then the result buffer will be freed to.
With automatic allocations, you cannot read from a file, allocate records for each of the things read from the file, and then delete some of the records out of order. You simply have no dynamic control over the memory allocated; automatic allocations are forced into a strictly last-in first-out (LIFO) order.
You cannot write subroutines that allocate memory, initialize it and/or do other computations, and return the allocated memory to their callers.
(Some people may also point out that the stack memory commonly used for automatic objects is commonly limited to 1-8 mebibytes while the memory used for dynamic allocation is generally much larger. However, this is an artifact of settings selected for common use and can be changed; it is not inherent to the nature of automatic versus dynamic allocation.)
If the allocated memory is small and used only inside the function, malloc is indeed unnecessary.
If the memory amount is extremely large (usually MB or more), the above example may cause stack overflow.
If the memory is still used after the function returned, you need malloc or global variable (static allocation).
Note that the dynamic allocation through local variables as above may not be supported in some compiler.
Most examples using structs in C use malloc to assign the required size block of memory to a pointer to that struct. However, variables with basic types (int, char etc.) are allocated to the stack and it is assumed that enough memory will be available.
I understand the idea behind this is that memory may not be available for larger structs so we use malloc to ensure we do indeed have enough memory but in the case of our struct being small is this really necessary? For example if a struct only consists of three ints, surely I am always fine to assume there is enough memory?
So really my question boils down to what are the best practises in C regarding when it is necessary to malloc variables and what is the justification?
The only time you don't have to allocate memory is when you statically allocate memory, which is what happens when you have a statement like:
int number = 5;
You can always write it as:
int *pNumber = malloc(sizeof(int));
but you have to make sure to free it or you will be leaking memory.
You can do the same thing with a struct (instead of dynamically allocating memory for it, statically allocate):
struct some_struct_t myStruct;
and access members by:
myStruct.member1 = 0;
etc...
The big difference between dynamic allocation and static is whether that data is available outside of your current scope. With static allocation, it's not. With dynamic it is, but you have to make sure to free it.
Where you run into trouble is when you have to return a structure (or a pointer to it) from a function. You either have to dynamically allocate inside the function which is returning it or you have to pass in a pointer to an externally (dynamically or statically) allocated structure which the function can then work with.
Good code gets re-used. Good code have few size limitations. Write good code.
Use malloc() whenever there is anything more than trivial buffer sizes.
Buffer size to write an int: The needed buffer size is at most sizeof(int)*CHAR_BIT/3 + 3. Use a fixed buffer.
Buffer size to write a double as in sprintf(buf, "%f",...: The needed buffer size could be thousands of bytes: use malloc(). Or use sprintf(buf, "%e",... and use a fixed buffer.
Forming a file path name could involve thousands of char. Use malloc().
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.
May be similar question found on SO. But, I didn't found that, here is the scenario
Case 1
void main()
{
char g[10];
char a[10];
scanf("%[^\n] %[^\n]",a,g);
swap(a,g);
printf("%s %s",a,g);
}
Case 2
void main()
{
char *g=malloc(sizeof(char)*10);
char *a=malloc(sizeof(char)*10);
scanf("%[^\n] %[^\n]",a,g);
swap(a,g);
printf("%s %s",a,g);
}
I'm getting same output in both case. So, my question is when should I prefer malloc() instead of array or vice-verse and why ?? I found common definition, malloc() provides dynamic allocation. So, it is the only difference between them ?? Please any one explain with example, what is the meaning of dynamic although we are specifying the size in malloc().
The principle difference relates to when and how you decide the array length. Using fixed length arrays forces you to decide your array length at compile time. In contrast using malloc allows you to decide the array length at runtime.
In particular, deciding at runtime allows you to base the decision on user input, on information not known at the time you compile. For example, you may allocate the array to be a size big enough to fit the actual data input by the user. If you use fixed length arrays, you have to decide at compile time an upper bound, and then force that limitation onto the user.
Another more subtle issue is that allocating very large fixed length arrays as local variables can lead to stack overflow runtime errors. And for that reason, you sometimes prefer to allocate such arrays dynamically using malloc.
Please any one explain with example, what is the meaning of dynamic although we are specifying the size.
I suspect this was significant before C99. Before C99, you couldn't have dynamically-sized auto arrays:
void somefunc(size_t sz)
{
char buf[sz];
}
is valid C99 but invalid C89. However, using malloc(), you can specify any value, you don't have to call malloc() with a constant as its argument.
Also, to clear up what other purpose malloc() has: you can't return stack-allocated memory from a function, so if your function needs to return allocated memory, you typically use malloc() (or some other member of the malloc familiy, including realloc() and calloc()) to obtain a block of memory. To understand this, consider the following code:
char *foo()
{
char buf[13] = "Hello world!";
return buf;
}
Since buf is a local variable, it's invalidated at the end of its enclosing function - returning it results in undefined behavior. The function above is erroneous. However, a pointer obtained using malloc() remains valid through function calls (until you don't call free() on it):
char *bar()
{
char *buf = malloc(13);
strcpy(buf, "Hello World!");
return buf;
}
This is absolutely valid.
I would add that in this particular example, malloc() is very wasteful, as there is more memory allocated for the array than what would appear [due to overhead in malloc] as well as the time it takes to call malloc() and later free() - and there's overhead for the programmer to remember to free it - memory leaks can be quite hard to debug.
Edit: Case in point, your code is missing the free() at the end of main() - may not matter here, but it shows my point quite well.
So small structures (less than 100 bytes) should typically be allocated on the stack. If you have large data structures, it's better to allocate them with malloc (or, if it's the right thing to do, use globals - but this is a sensitive subject).
Clearly, if you don't know the size of something beforehand, and it MAY be very large (kilobytes in size), it is definitely a case of "consider using malloc".
On the other hand, stacks are pretty big these days (for "real computers" at least), so allocating a couple of kilobytes of stack is not a big deal.
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.