Save pointer to array in struct.
I would like to store the pointer to array of int into struct but I am having trouble.
Below my code with commentaries:
typedef struct _index {
int *data;
} Index;
void adder(struct _index *s){
int i;
int arr[maxint];
for(i=0; i<maxint; i++) {
arr[i] = i+42;
}
//i got a gibberish
//s->data = arr;
//works fine
s->data = (int *) malloc(maxint * sizeof(int));
s->data = memcpy(s->data, arr, maxint * sizeof(int));
)
int main() {
Index n;
int i;
//this method also works.
//int arr[maxint];
//for(i=0; i<maxint; i++) {
// arr[i] = i+42;
//
//}
//n.data = arr;
adder(&n);
for(i=0; i<maxint;i++) {
printf("%d-", n.data[i]);
}//testing
return 0;
}
when i make assignment, i got strange numbers:
117-118-119-120-12-0-22-2292964-0-2293008-127-0-129-130-131-0-0-0-0-0-0-138-0
but if i use malloc and memcpy all works fine
You got gibberish in your first case, because you tried to "return" the address of a local variable from a function through the pointer. Once the function finishes execution, the int arr[maxint]; won't be valid. In other words, after adder() finishes execution, int arr[maxint]; goes out of scope and it's lifetime is over. So, the (returned) pointer becomes invalid and using that further in the caller function will result in undefined behaviour.
Solution:
As you've done correctly, using dynamic memory.
use static variable (not a good approach, but possible).
In both the aforesaid approach, the lifetime of the variable ( static arr array/ malloc()ed memory) is not limited to function scope and thus, the pointer to the meory will be vaild in the caller function.
The array arr in the adder() function is on the stack and only exists as long as the code in that function is running. Once adder() returns that memory is re-used by the rest of the program and its content overwritten.
int arr[] is placed on the stack and gets removed from the stack when it goes out of scope. So you will point to garbage.
It works fine if you include it in your main because that way it hasn't gone out of scope yet.
Malloc works because you allocate memory and not just place it on the stack.
Related
My Doubt is regarding only memory allocation so don't think about program output
#include<stdio.h>
int main(){
for(int i=0;i<20;i++){
char *str=malloc(sizeof(char)*6); //assuming length of each string is 6
scanf("%s",str);
insertinlinkedlist(str);
}
}
whenever i allocate memory here as shown above only the base address of char array will pass to linked list,and that is the memory block allocated for char array is inside main only and i am storing the base address of that array in str which is local to main and is passed to insetinlinkedlist
I want to ask whenever memory is allocated inside loop than why the number of
memory blocks(no of char arrays declared ) are created equal to n (number of time loop runs) since variable name is same we should be directed to same memory location
Note I have checked in compiler by running the loop all the times when loop runs memory the value of str is different
is The above method is correct of allocating memory through loop and through same variable "Is the method ensures that every time we allocate memory in above manner their will be no conflicts while memory allocation and every time we will get the address of unique memory block"
Now above doubt also creates a doubt in my mind
That if we do something like that
int main(){
for(int i=0;i<n;i++){
array[50];
}
}
then it will also create 50 array inside stack frame
malloc returns a pointer to the first allocated byte. Internally it keeps track of how much memory was allocated so it knows how much to free (you do need to insert calls to free() or you'll leak memory, by the way). Usually, it does this by allocating a little bit of memory before the pointer it gives you and storing the length there, however it isn't required to do it that way.
The memory allocated by malloc is not tied to main in any way. Currently main is the only function whose local variables have a pointer to that memory, but you could pass the pointer to another function, and that function would also be able to access the memory. Additionally, when the function that called malloc returns, that memory will remain allocated unless manually freed.
The variable name doesn't matter. A pointer is (to first approximation) just a number. Much like how running int a = 42; a = 20; is permitted and replaces the previous value of a with a new one, int *p = malloc(n); p = malloc(n); will first assign the pointer returned by the first malloc call to p, then will replace it with the return value of the second call. You can also have multiple pointers that point to the same address:
int *a = malloc(42);
int *b = malloc(42);
int *c = a;
a = malloc(42);
At the end of that code, c will be set to the value returned by the first malloc call, and a will have the value returned by the last malloc call. Just like if you'd done:
//assume here that f() returns a different value each time
//it's called, like malloc does
int a = f();
int b = f();
int c = a;
a = f();
As for the second part of your question:
for(int i=0;i<n;i++){
int array[50];
}
The above code will create an array with enough space for 50 ints inside the current stack frame. It will be local to the block within the for loop, and won't persist between iterations, so it won't create n separate copies of the array. Since arrays declared this way are part of the local stack frame, you don't need to manually free them; they will cease to exist when you exit that block. But you could pass a pointer to that array to another function, and it would be valid as long as you haven't exited the block. So the following code...
int sum(int *arr, size_t n) {
int count = 0;
for (size_t i = 0; i < n; i++) {
count += arr[i];
}
return count;
}
for(int i=0;i<n;i++){
int array[50];
printf("%d\n", sum(array, 50));
}
...would be legal (from a memory-management perspective, anyway; you never initialize the array, so the result of the sum call is not defined).
As a minor side note, sizeof(char) is defined to be 1. You can just say malloc(6) in this case. sizeof is necessary when allocating an array of a larger type.
A "Deeltal" keeps track of how many dividers an integer has (count) and keeps them in an array (dividers).
Examples:
value = 8 -> count = 3 and dividers = {1,2,4}
value = 10, count = 3, dividers = {1,2,5}
Hope everything is clear, take a look at the following code:
typedef struct{
int value;
int count;
int* dividers;
} Deeltal;
void free_dividers(Deeltal *g){ /*Deletes the int* dividers of a given Deeltal*/
free (g - > dividers);
}
/* the following two functions have the same purpose: deleting the content of a
given amount of "Deeltal" AND deleting the pointer to it aswell*/
void free_amountOfdeeltal(Deeltal *d, int amount){
int i;
for (i = 0; i < amount; i++){
free_dividers(&d[i]);
}
free(d);
}
void free_amountOfdeeltalVersion2(Deeltal **g, int amount){
int i;
for(i = 0; i < amount; i++){
free_dividers(&(*g)[i]);
}
free(*g);
}
If my main looked something like this
int main(void){
/*EDIT 3/11/2017: forgot to allocate memory for *d and initializing g.
Thanks for pointing this out*/
Deeltal g = 0;
g.value = 6; g.count = 3; g.dividers = {1,2,3};
Deeltal *d = malloc(sizeof(Deeltal));
d->value = 6; d->count = 3; d->dividers = {1,2,3};
free_amountOfdeeltal(&g);
free_amountOfdeeltalVersion2(&d);
}
What is the difference between free_amountOfdeeltal and free_amountOfdeeltalVersion2?
Both should do the same thing: releasing the memory of a Deeltal and also deleting the pointer pointing to that memory.
On a sidenote:
How do you delete the memory as well as the pointer?
Not withstanding calling this function with invalid data as pointed out by others .. I'll attempt to answer the question I think you are asking.
On a sidenote: How do you delete the memory as well as the pointer?
You can't really "delete the pointer" in this context as a pointer is simply a variable that is assigned an address. You delete memory that was allocated to you by passing free a pointer to the memory. Note that free does not modify the value of the pointer at all. (It can't because the pointer is passed by value.) After the call to free the pointer still points to the same memory address.
If what you mean is "how can I assign a meaningful value to the pointer to identify that its memory has already been deleted," then you can use the second form of your function:
void free_amountOfdeeltalVersion2(Deeltal **g, int amount);
and set *g to NULL before returning. You can then use this information than the pointer is NULL to identify the memory has already been deleted.
You didn't allocate any memory for d so your pointer doesn't point to any structure. Therefor, you can't access its properties or free its memory because you didn't reserve it in the first place. There's no way this code could come remotely close to compiling.
First of all you should be allocating memory for a "Deeltal" structure like this:
Deeltal *d = malloc(sizeof(Deeltal));
I recommend you go back and relearn how pointers work, as you're doing some really weird stuff there.
I have a function which creates an array, of say, size 5.
Is it possible for the function to accept a pointer (or maybe it needs a pointer to a pointer?) and then point said pointer at an array, so that when the callee then looks at the pointer, it can see all values of the array.
Something along the lines of this (except this will not work):
#define LENGTH 5
void assignArray(int *pointer)
{
int arr[LENGTH] = {0,1,2,3,4};
// Point the pointer at the array, without manually copying each element
pointer = arr;
}
void main()
{
int *pointer;
pointer = malloc(sizeof(int) * LENGTH);
assignArray(pointer);
int i;
for (i = 0 ; i < LENGTH ; i++) printf("%d\n", pointer[i]);
}
C assign array without element by element copy
In C, arrays (compile-time allocated) cannot be assigned. You need to copy the elements from one array to another.
To avoid element-by-element copy, you can copy the whole array all at a time using library function.
I'm not very sure what you want to ask here, but it seems, you need to do memcpy() to achieve your goal.
If you have a secondary array arr to copy from, you can write
memcpy( pointer, arr, ( (sizeof arr[0]) * LENGTH ));
The code to do what you are describing might look like:
#define LENGTH 5
void assignArray(int **pp)
{
static int arr[LENGTH] = {0,1,2,3,4};
// Point the pointer at the array, without manually copying each element
*pp = arr;
}
int main()
{
int *pointer;
assignArray(&pointer);
for (int i = 0 ; i < LENGTH ; i++)
printf("%d\n", pointer[i]);
}
Note that one does not simply point *pp at a non-static local variable arr. That is because int arr[] = .... would go out of scope when assignArray returns.
If you want each call to assignArray to "return" a different array then of course you will have to allocate space and use memcpy each time you want to make a copy of the original array.
int arr[LENGTH] = {0,1,2,3,4}; will be stack allocated, so attempting to return the pointer to any of its elements will give you undefined behaviour as the whole thing will be out of scope when the function returns.
If you want to change what a pointer is pointing to then use 2 levels of indirection ** (i.e. pass a pointer to a pointer). You'll need to allocate the array arr on the heap using malloc or something similar.
As you are trying to do it, it is not possible due to the fact that your local arr is saved to the stack and is cleaned up after the function assignArry finished. As already mentioned you need to memcpy.
This answer will have two parts:
As mentioned in other answers, this is now how you're supposed to do it. A common construct in similar code is:
void assignArray(int *dest, size_t size)
{
int i;
// initialize with some data
for (i=0; i<size; i++)
dest[i] = i;
}
This way you're not wasting space and time with an intermediate buffer.
Second part of this answer is about wrapping arrays in a struct. It's a silly trick, that in a way achieves exactly what you asked, and also something that you probably don't want because of extra data copying.
Example code:
#include <stdio.h>
#include <stdlib.h>
#define LENGTH 5
struct foo { int arr[LENGTH]; };
struct foo assignArray()
{
struct foo bar = { .arr = {0,1,2,3,4} };
/* return the array wrapper in struct on stack */
return bar;
}
int main()
{
struct foo *pointer;
pointer = malloc(sizeof(*pointer));
*pointer = assignArray(); /* this will copy the data, not adjust pointer location */
int i;
for (i = 0 ; i < LENGTH ; i++) printf("%d\n", pointer->arr[i]);
return 0;
}
I want to use only studio.h library to convert from decimal number to binary number by using an array to store remainder but the result is not correct, maybe i have problem with memory allocation or return value is wrong, please help me to check it.
Thank you so much!
#include <stdio.h>
int n = 0;
int* DecimalToBinary(int number){
int a[10];
while(number!=0){
a[n++] = number%2;
number/=2;
}
return a;
}
void main(){
int *d1 = DecimalToBinary(5);
int *d2 = DecimalToBinary(10);
for(int i = n-1 ;i>=0;i--)
printf(" %d",d1[i]);
printf("\n");
for(int i = n-1 ;i>=0;i--)
printf(" %d",d2[i]);
}
You return a pointer to a local array. That local array is on the stack, and when the function returns the array goes out of scope and that stack memory will be reused when you call the next function. This means that the pointer will now point to some other data, and not the original array.
There are two solutions to this:
Declare the array in the function calling DecimalToBinary and pass it as an argument.
Create the array dynamically on the heap (e.g. with malloc) and return that pointer.
The problem with method 2 is that it might create a memory leak if you don't free the returned pointer.
As noted by Craig there is a third solution, to make the array static inside the function. However in this case it brings other and bigger problems than the two solutions I originally listed, and that's why I didn't list it.
There is also another serious problem with the code, as noted by Uchia Itachi, and that is that the array is indexed by a global variable. If the DecimalToBinary function is called with a too big number, or to many times, this global index variable will be to big for the array and will be out of bounds for the array.
Both the problem with dereferencing a pointer to an out-of-scope array and the indexing out of bounds leads to undefined behavior. Undefined behavior will, if you're lucky, just lead to the wrong result being printed. If you're unlucky it will cause the program to crash.
You are returning a pointer to a locally allocated array. It is allocated on the stack, and goes away when the function returns, leaving your pointer pointing to garbage.
You have a few options. You could pass an array in to fill:
void DecimalToBinary(int result[10],int number){
while(number!=0){
result[n++] = number%2;
number/=2;
}
return result;
}
// usage example:
int b[10];
DecimalToBinary(b, 42);
Or you could allocate an array on the heap:
int* DecimalToBinary(int number){
int *a = (int *)malloc(sizeof(int) * 10);
while(number!=0){
a[n++] = number%2;
number/=2;
}
return a;
}
// usage example
int *b = DecimalToBinary(42);
free(b); // when finished with it
Or you could wrap the array in a struct:
typedef struct {
int b[10];
} result;
result DecimalToBinary(int number){
result r;
while(number!=0){
r.b[n++] = number%2;
number/=2;
}
return r;
}
// usage example
result r = DecimalToBinary(42);
If you do the malloc() option, do not forget to free() the returned data when you're done with it, otherwise it will hang around. This is called a memory leak. In more complex programs, it can lead to serious issues.
Note: By the way, if your number is larger than 1023 (10 binary digits), you'll overrun the array. You may also wish to explicitly stop once you've stored 10 digits, or pass the size of the array in, or compute the required size first and allocate that much space. Also, you will get some odd results if your number is negative, you might want to use number&1 instead of number%2.
Note 2: As noted elsewhere, you should make n local, or at the very least reinitalize it to 0 each time the function is called, otherwise it will just accumulate and eventually you'll go past the end of the array.
int[10] is not the same as int *; not only is the former created on the stack, it is a different type alltogether. You need to create an actual int * like so:
int *a = malloc (10 * sizeof (int));
Of course, don't forget to free() it after use!
What you can also do and what is commonly done in C is creating the array where it is called and provide a pointer to that array to the function, this way when the array is on the stack of the function that calls it and not in the function self. We also have to specify the size of the array on to that function, since the function cannot know to how many elements the pointer points to
void DecimalToBinary( int number, int* output, unsigned size ) {
/*adapt this to your liking*/
int i;
for ( i = 0; i < size && number != 0; i++) {
output[i] = number%2;
number/2;
}
}
and in you main function you would call it like this:
int array[10];
DecimalToBinary( 5, array, sizeof(array)/sizeof(array[0]));
now array has the same result as a would have had in your example.
The problem in your code lies here..
int * DecimalToBinary(int number){
int a[10];
while(number!=0){
a[n++] = number%2;
number/=2;
}
return a;
}
The array a scope is only till this function. Once this function terminates, the memory allocated for this array will be released, either u need to use dynamic memory allocation or make array a global.
This is the correct program:
#include <stdio.h>
int n = 0;
int a[10] = {0};
int* DecimalToBinary(int number){
n = 0;
while(number!=0){
a[n++] = number%2;
number = number/2;
}
return a;
}
int main(){
int *d1;
int *d2;
int i;
d1 = DecimalToBinary(5);
for(i = n-1;i>=0;i--)
printf(" %d",d1[i]);
printf("\n");
d2 = DecimalToBinary(10);
for(i = n-1;i>=0;i--)
printf(" %d",d2[i]);
printf("\n");
}
I have a function that I pass an array into and an int into from my main function. I am doing operations to the array inside this new function, let's call it foo. In foo, I initialize another array with 52 cells all with 0. I do operations on the array that I passed from main, and transfer that data to the newly initialized array. I want to return the new array back to the main function. But of course, I can't return data structures like arrays. So I instead return an int pointer that points to this array. Inside the int main, I pass the pointer to have it point to various cells in the array. When I print the results of what the pointer is pointing to, it should either be pointing to 0 or an integer greater than 0. But instead, I get inconsistent results. For some reason, some of the values that SHOULD be 0, prints out garbage data. I've been trying to spot the bug for some time, but I just wanted a second hand look at it. Here is just the GENERAL idea for the code for this portion anyways...
int main(){
int *retPtr;
char input[] = "abaecedg";
retPtr = foo(input, size);
for(i=0; i<52; i++){
// error displayed here
printf("%d\n", *(retPr + i));
}
}
int foo(char input[], int size)
{
int arr[52] = {0}; // should initialize all 52 cells with 0.
int i=0, value; // looking for non-zero results in the end.
int *ptr = &arr[0];
for(i=0; i<size; i++){
if(arr[i] > 64 && arr[i] < 91){
value = input[i] - 65;
arr[value]++;
}
}
return ptr;
}
Hopefully this makes sense of what I'm trying to do. In the foo function, I am trying to find the frequency of certain alphabets. I know this might be a bit cryptic, but the code is quite long with comments and everything so I wanted to make it as succinct as possible. Is there any possible reason why I'm getting correct values for some (numbers > 0, 0) and garbage values in the other?
The reason you get garbage back is that the array created in foo is allocated in foos stack frame, and you then return a pointer into that frame. That frame is discarded when foo returns.
You should allocate the array on the heap (using malloc and friends) if you want it to remain after foo returns. Don't forget to free() it when you're done with the array.
int main(){
char input[] = "abaecedg";
int retPtr[] = foo(input, size); //An array and a pointer is the same thing
...
free(retPtr);
}
int *foo(char input[], int size)
{
int arr[] = calloc(52*sizeof(int); // should initialize all 52 cells with 0.
...
arr[value]++;
...
return arr;
}
Another way is to let foo take an array as a parameter and work with that, in this way:
int main(){
int ret[52] = {0};
...
foo(input, size, ret);
...
}
void foo(char input[], int size, int *arr)
{
...
arr[value]++;
...
return; //Don't return anything, you have changed the array in-place
}
The reason this works is because an array is the exact same thing as a pointer, so you are really passing the array by reference into foo. arr will be pointing to the same place as ret, into the stack frame of main.
In function foo the array arr is a local array, that is, allocated on the stack. You must not return any pointer of data allocated on the stack, since the stack is rewinded after you return from the function, and its content is no more guaratneed.
If you want to return an array you should allocate it on the heap using malloc, for example, and return the pointer malloc returned. But you will then have to free that memory somewhere in your program. If you fail to free it you will have what's called a "memory leak", which may or may not crash/disturb this program from running again, depending on your environment. A not clean situation, that's for sure.
That's why I consider C not so good for functional programing idioms, such as returning things from function (unless they are primitive types). I would achieve what you tried to do by passing another array to foo - an output array, companioned by a size variable, and fill that array.
Alternately, you could wrap the array within a struct and return that struct. Structs can be returned by value, in which case they are copied via the stack to the caller function's returned value.