How can I let the user choose a number say n and then create an array with the size of n?
Can I just say int a[]=malloc (n*sizeof(int))?
There are two ways to do that. If the array size is small then you can use variable length array
/* Valid in C99 and later */
int n;
scanf("%d", &n);
int a[n];
This will allocate memory on stack. Other way is you can use dynamic memory allocation which will allocate memory on the heap
int *a = malloc(n*sizeof(int));
Your idea is nearly correct:
int a[] = malloc(n*sizeof(int));
Using malloc is the correct way.
But you cannot assign the returned address to an array.
You must use a pointer variable instead:
int *a = malloc(n*sizeof(int));
Yes if u want to set the size of the array at run-time.
Then u should go for dynamic memory allocation(malloc/calloc).
int a[]=malloc (n*sizeof(int));//this not possible.
int *a =malloc (n*sizeof(int)); // this is possible.
There are two basic ways for allocating the memory to create an array where the size to the array is determined as input:
The first one is,
allocating the memory for array in the 'stack' segment of memory where the size of array is taken as input ant then the array of that particular size is defined and granted memory accordingly.
int n;
scanf("%d",&n); //scanning the size
int arr[n]; //declaring the array of that particular size here
The second one is,
allocating the required memory in the 'heap' segment of memory.It is the memory allocated during runtime (execution of the program)
So,another way of declaring an array where size is defined by user is
int n,*arr;
scanf("%d",&n);
arr=malloc(n*sizeof(int)); //malloc function provides a contiguous space
or
arr=calloc(n,sizeof(int)); //calloc function is similar,initializes as 0
to use both these functions make sure to include stdlib.h.
Variable length arrays (VLAs) were added to C with C99, but made optional with C11. They are still widely supported, though. This is the simplest way to define an array with user-selected size at runtime.
Other than that VLAs may not be available on all platforms, they also may fail silently when there is an allocation failure. This is a disadvantage that malloc() avoids when used correctly.
You can't assign to an array in C, and instead you need to store the value returned by malloc() in a pointer. Note that malloc() returns NULL when there is an allocation failure, allowing code to check for failure and proceed accordingly. The actual allocation might look like this:
int *a_dyn = malloc(sizeof *a_dyn * arr_sz);
This is an idiomatic way of calling malloc(). Note that there is no need to cast the result of malloc(), and note that the operand to sizeof is not an explicit type, but rather an expression involving a_dyn. The sizeof operator uses the type of the expression *a_dyn, which is in fact int (there is no dereference made). This is less error-prone and easier to maintain when types change during the life of a program than coding with explicit types. Also note that the sizeof expression comes before arr_sz. This is a good practice to follow: sometimes you might have a call like:
int *arr = malloc(sizeof *arr * nrows * ncols);
Placing sizeof first forces the multiplication to be done using size_t values, helping to avoid overflow issues in the multiplication.
Don't forget to free any memory allocated with malloc() when it is no longer needed, avoiding memory leaks.
Whether you use a VLA or malloc(), you must validate user input before using it to avoid undefined behavior. Attempting to allocate an array of non-positive size leads to undefined behavior, and attempting to allocate too much memory will lead to an allocation failure.
Here is an example program that illustrates all of this:
#include <stdio.h>
#include <stdlib.h>
#define ARR_MAX 1024 // some sensible maximum array size
int main(void)
{
int arr_sz;
int ret_val;
/* validate user input */
do {
printf("Enter array size: ");
ret_val = scanf("%d", &arr_sz);
} while (ret_val != 1 || arr_sz < 1 || arr_sz > ARR_MAX);
/* use a VLA */
int a_vla[arr_sz];
for (int i = 0; i < arr_sz; i++) {
a_vla[i] = i;
printf("%d ", a_vla[i]);
}
putchar('\n');
/* use malloc() */
int *a_dyn = malloc(sizeof *a_dyn * arr_sz);
if (a_dyn == NULL) { // malloc failure?
fprintf(stderr, "Unable to allocate memory\n");
} else { // malloc success
for (int i = 0; i < arr_sz; i++) {
a_dyn[i] = i;
printf("%d ", a_dyn[i]);
}
putchar('\n');
}
/* avoid memory leaks */
free(a_dyn);
return 0;
}
Related
I try something like below but all the time I have a segmentation fault.
I don't really want to use (e.g.) #define N 1000 and then declare int buffer[N].
Just in case..I'm not allowed to use any headers except stdio.h as well as dynamic memory.
void input (int *buffer, int *length);
int main()
{
int length, *buffer = NULL, *numbers = NULL;
input(buffer, &length);
}
void input(int *buffer, int *length) {
scanf("%d", length);
if (*length < 0) {
error = 1;
return;
}
for (int i = 0; i < *length; i++) {
scanf("%d", *buffer[i]);
}
}
How to pass an array with unknown 1-d dimension into function
In C, arrays cannot exist until their size is known.
There are other approaches though.
In C, code cannot pass an array to a function. some_function(some_array) converts the array some_array to the address of the first element of the array: &some_array[0]. That is what the function receives, a pointer, not an array. The original size information of the array is not passed, thus also pass the length to the function.
Sample:
Read desired length.
{
int length = 0;
scanf("%d", &length);
Form a variable length array, length >= 1.
if (length <= 0) {
return NULL;
}
int buffer[length];
Now call a function, passing the length and the address of the first element of the array.
// Do stuff with length and buf, like read data
foo1(length, buffer);
// foo1() receives the length & address of the first element of the array as an int *
// Do more stuff with length and buf, like write data
foo2(length, buffer);
}
At the end of the block }, buffer no longer available.
In C, you can't create an array if you can't know its size at compile time (or at least not in certain implementations and standards), so doing something like buffer[length] won't work (again at least not in certain implementations/standards).
What you need to do to make sure this works everywhere is to use a pointer (as I see you're trying to use here). However, what you're doing wrong here that causes your segfault with the pointers is you assign them the value of NULL. This also won't work due to how when you assign a pointer an arbitrary value, there is no memory allocated for the pointer (This applies for everything other than addresses of "regular" variables using the & operator and assigning other pointers that are checked to be OK). Your pointers are just pointing to address 0 and can't be used for anything.
What you need to do here to fix the pointers is to use dynamic memory allocation, so you can have a truly variable-sized array. Specifically, you need to use a function like malloc or calloc to allocate memory for the pointers so they are usable. In your case, using calloc and reading its documentation, we see that it takes 2 parameters: The number of elements it should allocate memory for and the size of each element. We also know that it returns a pointer to the starting address of the allocated memory and that in case of failure (which can only happen if you're out of memory), it returns NULL. Using this, we understand that in your case the call to calloc would be like this:
int *buffer = (int *) calloc(length, sizeof(int));
The sizeof() function returns the size of a data type in bytes. Here you allocated enough memory for the pointer to hold length integers (since you'll use it as an array you need enough memory for all the integers, you're not just pointing to 1 integer but storing all of them), and calloc is also noted to initialize every allocated element to 0, so you have an array of integers that are all initialized to 0 (Also note that type casting has been used to make sure the allocated memory block is appropriate for use with an integer array, you can read more about type casting in this small article from Tutorialspoint if you'd like). Then, after this has been allocated, you can start reading your integers into the array. The complete code looks like this:
void input (int *buffer, int *length);
int main() {
// NOTE: I don't see the numbers pointer used here, maybe remove it?
int length, *buffer, *numbers;
input(buffer, &length);
}
void input(int *buffer, int *length) {
scanf("%d", length);
if (*length < 0) {
// Consider printing the exact error here
error = 1;
return;
}
buffer = (int *) calloc(length, sizeof(int));
if (buffer == NULL) {
printf("Couldn't allocate memory for buffer\n");
error = 1;
return;
}
// Accessing the elements of an array doesn't need * and in fact * here can (and probably will) cause terrible things
for (int i = 0; i < *length; i++) {
scanf("%d", buffer[i]);
}
}
Also don't forget to call free() on the pointer after you're done using it to avoid memory leaks (in your case that'd be after the call to input()).
Hope this helped, good luck!
You cannot use arrays because their memory size must be known to the compiler at compile time. Also you can't use Variable Length Arrays because they are allocated at the point of declaration and deallocated when the block scope containing the declaration exits.
The solution to your problem might be to use malloc
Here is the code I'm using:
#include <stdio.h>
#include <stdlib.h>
int main() {
int *arr;
int sz = 100000;
arr = (int *)malloc(sz * sizeof(int));
int i;
for (i = 0; i < sz; ++i) {
if (arr[i] != 0) {
printf("OK\n");
break;
}
}
free(arr);
return 0;
}
The program doesn't print OK. malloc isn't supposed to initialize the allocated memory to zero. Why is this happening?
malloc isn't supposed to initialize the allocated memory to zero. Why is this happening?
This is how it was designed more than 40 years ago.
But, at the same time, the calloc() function was created that initializes the allocated memory to zero and it's the recommended way to allocate memory for arrays.
The line:
arr = (int *)malloc(sz * sizeof(int));
Should read:
arr = calloc(sz, sizeof(int));
If you are learning C from an old book it teaches you to always cast the value returned by malloc() or calloc() (a void *) to the type of the variable you assign the value to (int * in your case). This is obsolete, if the value returned by malloc() or calloc() is directly assigned to a variable, the modern versions of C do not need that cast any more.
The man page of malloc says:
The malloc() function allocates size bytes and returns a pointer to
the allocated memory. The memory is not initialized. If size is 0,
then malloc() returns either NULL, or a unique pointer value that can
later be successfully passed to free().
So malloc() returns uninitialized memory, the contents of which is indeterminate.
if (arr[i] != 0)
In your program, You have tried to access the content of a memory block, which is invoked undefined behavior.
malloc isn't supposed to initialize the allocated memory to zero.
Memory allocated by malloc is uninitialised. Value at these locations are indeterminate. In this case accessing that memory can result in an undefined behavior if the value at that location is to be trap representation for the type.
n1570-ยง6.2.6.1 (p5):
Certain object representations need not represent a value of the object type. If the stored value of an object has such a representation and is read by an lvalue expression that does not have character type, the behavior is undefined. [...]
and footnote says:
Thus, an automatic variable can be initialized to a trap representation without causing undefined behavior, but the value of the variable cannot be used until a proper value is stored in it.
Nothing good can be expected if the behavior is undefined. You may or may not get expected result.
From the C Standard 7.22.3.4:
Synopsis
#include <stdlib.h>
void *malloc(size_t size);
Description
The malloc function allocates space for an object whose size is
specified by size and whose value is indeterminate.
The value is indeterminate. So, every compiler is free to behave how it wants. For example, in Microsoft Visual C++, in Debug mode, the area of allocated memory by malloc() is all set to 0xCDCDCDCD and when in Release mode it is random. In modern versions of GCC, it is set to 0x000000 if you don't enable code optimizations, and random otherwise. I don't know about other compilers, but you get the idea.
void *malloc(size_t size) is just supposed to keep aside the specified amount of space. That's all. There is no guarantee as to what will be present in that space.
Quoted from the man pages:
The malloc() function allocates size bytes and returns a pointer to
the allocated memory. The memory is not initialized. If size is 0,
then malloc() returns either NULL, or a unique pointer value that can
later be successfully passed to free().
Apart from calloc() you can use the memset() function to zero out a block of memory.
The first time you call malloc(3), it asks to the operating system to get memory for the heap space.
For security reasons, the unix/linux kernel (and many other operating systems) in general zeroes the page contents that is to be given to a process, so no process can access that memory's previous contents and do nasty things with it (like searching for old passwords, or similar things).
If you do several allocations and deallocations of memory, when the malloc module reuses the previous memory, you'll see garbage coming from malloc(3).
Zero's are assigned to page contents at first time in linux kernel.
Below program explains the memory initialisation difference in malloc and calloc:
#include<stdio.h>
#include<stdlib.h>
#define SIZE 5
int main(void) {
int *mal = (int*)malloc(SIZE*sizeof(int));
int *cal = (int*)calloc(SIZE, sizeof(int));
mal[4] = cal[4] = 100;
free(mal); free(cal);
mal = (int*)malloc(SIZE*sizeof(int));
cal = (int*)calloc(SIZE, sizeof(int));
for(int i=0; i<SIZE; i++) {
printf("mall[%d] = %d\n", i, mal[i]);
}
for(int i=0; i<SIZE; i++) {
printf("call[%d] = %d\n", i, cal[i]);
}
}
I use malloc to allocate everything from the heap(dynamic memory) while i should use calloc instead nowaday , and memset is great for filling you memory segment with any chosen character.
Compile and work great with GCC:
#include <stdio.h>
#include <stdlib.h>
#include <mem.h>
int main()
{
int *arr;
int sz = 100000;
arr = (int *)malloc(sz * sizeof(int));
memset(arr, 0, sz*sizeof(int) );
int i;
for (i = 0; i < sz; ++i) {
if (arr[i] != 0) {
printf("OK\n");
break;
}
}
free(arr);
return 0;
}
ref: http://www.cplusplus.com/reference/cstring/memset/
well, the value is not initialized in malloc.
And it does print "OK" in VS Code.
so in VS Code, the output is : "OK" followed by a garbage value.
in a web based compiler (here's the link : https://www.programiz.com/c-programming/online-compiler/ ),
the output was
"LOL" followed by '0'
so some compilers do initialize the value..but actually the value in malloc is not intialized. so it will return a garbage value when printed as in the above example in VS Code.
int main()
{
int *arr;
int sz = 100000;
arr = (int *)malloc(sz * sizeof(int));
int i;
for (i = 0; i < sz; i++)
{
if (arr[i] != 0)
{
printf("OK\n");
break;
}
else
{
printf("LOL \n");
break;
}
}
printf("%d", arr[0]);
free(arr);
I am unsure if I am using the realloc function correctly.
In my program, I first ask the user for the size of the array and allocate memory for it using malloc, then initialise it with some values.
Then I want to make the same array twice it's size, using realloc. Here is my code. Am I using realloc to resize int *A correctly?
#include <stdio.h>
#include <stdlib.h>
int main(){
int n;
printf("Enter size of array\n");
scanf("%d", &n);
int *A = (int*)malloc(n*sizeof(int)); //dynamically allocated array
for (int i = 0; i < n; i++) //assign values to allocated memory
{
A[i] = i + 1;
}
A = (int*)realloc(A, 2*sizeof(int)); //make the array twice the size
free(A);
}
When using malloc() , don't cast the return value as said here
You are not using the right size. In hereint *A = (int*)malloc(n*sizeof(int)); the size given to malloc is n*sizeof(int). If you want twice that size, you should call realloc() with n*sizeof(int)*2 instead of 2*sizeof(int)
Handle realloc() failure. In case realloc(A, new_size) fails, A == NULL and you will have memory leak. So, use a different pointer B, check if (B != NULL) and then assign A = B (old_size = new_size). If B == NULL deal with the allocation fail
In this case it's easier to double the n before malloc so you don't have the use realloc, because you know, that you gonna double the arraysize. Using realloc can slow the working of the program, because if you make it longer, and the adresses after the currently allocated memories aren't free, then the whole array will be moved. Also you have the change the last line as it was suggested before me.
My problem is that i am not sure how to allocate memory properly. I have researched it but to no avail. Some help regarding malloc would be great.
int main(int argc, char *argv[]) {
int i, a[5000], c = 1, m, n, input;
scanf("%d", &input);
for (i = 0; i <= 9999; i += 2) {
a[c] = i;
c++;
}
for (n = 2; n < 1118; n++) {
for (m = a[n]; m < a[5000]; m++) {
a[m] = a[m+1];
}
}
printf("%d", a[input]);
free (*a);
return 0;
}
'a' is allocated on stack therefore no need to free it.
You only need to free variables allocated by *alloc family of functions.
First of all, C arrays have 0-based indexing. By setting the intial value of c to 1 and then using as index inside the loop, you're going off-by-one. This invokes undefined behavior.
After that, you don't need to do free (*a);, a is an array, not a pointer returned by a memory allocator functions, malloc() or family.
That said, in this code, *a does not give you a pointer, at all, it is same as a[0] which is of type int.
Finally, without any bound checking from user supplied value of input, using a[input] may very well be accessing out of bound memory, causing UB.
FWIW, passing a pointer to free() which is not returned previously by malloc() and family also invokes undefined behavior.
malloc takes one argument - the number of bytes to allocate. It returns a void pointer (which is a pointer to a section of memory that can hold any data type).
Here's an example.
int *array = malloc(sizeof(int) * 10);
This allocates a 10-element array of integers. Note it leaves your data uninitialized, so the contents of the array are undefined. There's a function called calloc that does initialize it to zeros.
Also, a style tip. You might try to cast the result of a malloc call to a pointer for the type of data you will store in it (for example, int *array = (int *)malloc(sizeof(int) * 10);. This is frowned upon by C programmers for reasons explained in this post.
I just installed visual studio 2013 and gonna try a simple code. So I write this one:
int main(void) {
int length = 0;
int *array ;
printf("Enter the number of input: ");
scanf_s("%d", &length);
printf("%d", length);
for (int i = 0; i < length; i++){
scanf_s("%d", array + i);
}
printf("shoting ........ ");
getc;
return 0;
}
When I compile this code, it says local variable "array" is not initialized .
Am I missing something ?
The statement int *array ; declares a pointer. But you don't know where it points to. You can allocate some space using memory allocation functions like malloc() etc. or make it point to some other pre-allocated space. As you are asking user for the input length (i.e. don't know your memory requirement at compile time), dynamic memory allocation like malloc() would be the way to go.
Note: This answer is applicable if you are compiling in C,not C++
int *array;
Declares a pointer of type int*. Since it is not initialized, it points to some random location. This is what the compiler is trying to tell you. Just allocate memory dynamically using malloc. Add
array=malloc(length*sizeof(int));
Just after the first scanf_s to allocate enough memory. Note that you need to include stdlib.h to use malloc.
Also change
getc;
To
getchar();
It is also a good idea to check the return values of malloc,scanf_s etc to see if they are successful.
You need to Initialize the array
such as
int *array = maloc(sizeof(int) * 100);
or
int *array = new int[];