I am new to threads and I have a program that uses threads to find the minimum number out of a 2d array and later on, it finds the distance that the other elements of the array have from the minimum number and stores them in another array.
The user should enter the size of the array and the number of threads he wants to use.
I tried the program below for 1d array and it worked just fine. When I converted it to work for a 2d array it started crashing and throwing a segmentation fault. I, however, cannot find which part of the 2d declaration is wrong.
Any help is really appreciated.
Here is my code:
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/wait.h>
#include <string.h>
#include <pthread.h>
struct Parameters
{
// input
int s,p; //n is size of array, p is number of threads
int** array; //array with elements
int start;
int end;
// output
int smallest;
int pos; //position if minimum
int** B; //array that holds the distances
};
void* min(void* args)
{
struct Parameters* p = (struct Parameters*)args;
int **array = p->array;
int **B1 = p->B;
int start = p->start;
int end = p->end;
int smallest = array[start][start];
int pos = p->pos;
int distance;
//find the smallest
for (int i = start; i < end; i++)
{
for(int j = start; j < end; j++)
{
if (array[i][j] < smallest)
{
smallest = array[i][j];
pos = i;
}
}
}
//find the distances
for(int i = 0; i < ((struct Parameters*)args) -> s; i++)
{
for(int j = 0; j < ((struct Parameters*)args) -> s; j++)
{
distance = abs(pos - i);
B1[i][j] = distance;
}
}
params->smallest = smallest;
params->B = B1;
return NULL;
}
int main()
{
int smallest,pos;
int s,p;
struct Parameters *ptr = (struct Parameters *)malloc(sizeof(struct Parameters));
if(ptr == NULL)
{
printf("Not enough. Try again \n");
exit(0);
}
printf("Type s\n");
scanf("%d",&(ptr->s));
printf("Type p\n");
scanf("%d", &(ptr->p));
// declare an array of threads and associated parameter instances
pthread_t threads[(ptr->p)];
struct Parameters thread_parameters[(ptr->p)] ;
int arr[ptr->s][ptr->s];
int B2[ptr->s][ptr->s];
// intialize the array
for(int i=0; i< ptr->s; i++)
{
for(int j=0; j< ptr->s; j++)
{
printf("Type a \n");
scanf("%d",&arr[i][j]);
}
}
// smallest needs to be set to something
smallest = arr[0][0];
// start all the threads
for (int i = 0; i < ptr->p; i++)
{
memcpy(arr, thread_parameters[i].array, sizeof(arr));
thread_parameters[i].s = ptr->s;
memcpy(Bb, thread_parameters[i].B, sizeof(B2));
thread_parameters[i].start = i * (ptr->s / ptr->p);
thread_parameters[i].end = (i+1) * (ptr->s / ptr->p);
pthread_create(&threads[i], NULL, min, &thread_parameters[i]);
}
// wait for all the threads to complete
for (int i = 0; i < ptr->p; i++)
{
pthread_join(threads[i], NULL);
}
// Now aggregate the "smallest" and "largest" results from all thread runs
for (int i = 0; i < ptr->p; i++)
{
if (thread_parameters[i].smallest < smallest)
{
smallest = thread_parameters[i].smallest;
}
}
printf("Smallest is %d\n", smallest);
thread_parameters[ptr->p].B[ptr->s][ptr->s];
for (int i = 0; i < 1; i++)
{
for(int j = 0; j < ptr->s;j++)
{
for(int k = 0; k < ptr->s; k++)
{
printf("Element %d is %d away from min\n",j,thread_parameters[i].B[j][k]);
}
}
}
return 0;
}
Thank you!!
The issue with your code might also come from :
memcpy(arr, thread_parameters[i].array, sizeof(arr));
...
memcpy(Bb, thread_parameters[i].B, sizeof(B2));
as thread_parameters[i].array and thread_parameters[i].B are not allocated, if you are only reading the array it might b fine to only pass them by address
thread_parameters[i].array = arr
but for thread_parameters[i].B you would need to allocate the arrays and perform a deep copy (memcpy would not work)
The below text does not answer the question but does provide some insight on VLA usage
One reason for causing the segmentation with a declaration of a Variable Length Array is that the value is to large to allocate the array on the stack (some compiler choose this option, this choice might have performance reason).
The is not much option to recover cleanly from failure to allocate memory on the stack as there is little way to clean up stack memory during runtime within the same stack context.
You can mitigate the issue by allocating your 2D arrays on the heap instead, some of the strategies are available here(thanks #Lundin) and here.
int** alloc_2d_int_array(size_t rows, size_t cols) {
int **result = malloc(rows * sizeof(int *));
if(result == NULL) {
// could not allocate more memory
return NULL;
}
size_t row_size = cols * sizeof(int);
for(int i=0; i < rows; ++i) {
result[i] = malloc(row_size);
if(result[i] == NULL) {
// could not allocate more memory
// cleanup
return NULL;
}
}
return result;
}
the above implementation have not been tested, but does compile, there are still risk of integer overflow.
Then use the above define function as following:
int **arr = alloc_2d_int_array(ptr->s, ptr->s);
int **B2 = alloc_2d_int_array(ptr->s, ptr->s);
easier implementation (see here(thanks #Lundin))
int **arr = malloc(sizeof(int[ptr->s][ptr->s]);
int **B2 = malloc(sizeof(int[ptr->s][ptr->s]);
I am trying to practice with C by making a bubble sort program. The problem until now seems to be that the for loop that is giving values to the cells of the array is stuck after the condition is no longer fulfilled but it doesn't seem to be executing the commands in the loop. I don't know what is happening exactly and I have added some extra lines to see what is happening an these were my conclusions. Here is the code:
#include <stdio.h>
#include <stdlib.h>
void swap(int *x, int *y)
{
int temp = *x;
*x = *y;
*y = temp;
}
int *sort(int *array)
{
int finish = 1;
while (finish = 1)
{
finish = 0;
for (int i = 0; i <= sizeof(array); i++)
{
if ((array + i) > (array + i + 1))
{
swap(array + i, array + i + 1);
finish = 1;
}
}
}
return array;
}
int main()
{
int s, res;
printf("Give me the size of the array being sorted(larger than 1) : ");
do
{
res = scanf("%d", &s);
if (res != 1)
{
printf("Wrong Input!\n");
exit(1);
}
if (s < 2)
printf("Only numbers equal or larger than 2\n");
} while (s < 2);
int array[s];
for (int i = 0; i < s; i += 1)
{
scanf("%d", array + i);
printf("%d %d %d\n\n", *(array + i), i, i < s); // I used this to check if my values were ok
}
printf("end of reading the array"); //I added this line to see if I would exit the for loop. I am not seeing this message
sort(array);
printf("\n");
for (int i = 0; i < sizeof(array); i++)
printf("%d\n\n", array + i);
printf("Array has been sorted! Have a nice day!\n\n************************************************************");
return 0;
}
See the annotations in the code:
#include <stddef.h> // size_t 1)
#include <stdio.h>
#include <stdlib.h>
void swap(int *x, int *y)
{
int temp = *x;
*x = *y;
*y = temp;
}
int *sort(int *array, size_t size) // needs an extra parameter to know the size of the array
{
int finish = 1;
while (finish /* = 1 * you don't want assignment, you want comparison: */ == 1)
{
finish = 0;
for (int i = 0; i /* <= sizeof(array) */ < size - 1; i++) // i should be of type size_t
{
// if ((array + i) > (array + i + 1)) you are not dereferencing:
if(array[i] > array[i + 1])
{
// swap(array + i, array + i + 1); // easier to read imho:
swap(&array[i], &array[i + 1]);
finish = 1;
}
}
}
return array; // why does this function return anything? it is never used.
}
int main()
{
int s; /* , res; no need for an extra variable res */
printf("Give me the size of the array being sorted(larger than 1) : ");
do
{
// res = scanf("%d", &s);
// if (res != 1)
if (scanf("%d", &s) != 1)
{
printf("Wrong Input!\n");
// exit(1); // should be EXIT_FAILURE. Use return instead of exit() when in main().
return EXIT_FAILURE;
}
if (s < 2)
printf("Only numbers equal or larger than 2\n");
} while (s < 2);
int array[s];
for (int i = 0; i < s; /* i += 1* idiomatic: */ ++i) // size_t would be the correct type for s and i.
{
scanf("%d", /* array + i use indexes: */ &array[i]);
printf("%d %d %d\n\n", array[i], i, i < s); // again: indexes. i < s is allready ensured by the condition of the for-loop
}
printf("end of reading the array");
// sort(array); // sort will have no idea about the size of array use
sort(array, s); // instead.
printf("\n");
for (int i = 0; i < /* sizeof(array) 2) */ s; i++)
printf("%d\n\n", /* array + i * again you don't dereference */ array[i]);
printf("Array has been sorted! Have a nice day!\n\n************************************************************");
return 0;
}
1) size_t is the type that is guaranteed to be big enough to hold all sizes of objects in memory and indexes into them. The conversion specifier for scanf() is "%zu".
2) sizeof(array) in main() will yield the number of bytes in array, but you want the number of elements so you'd have to use sizeof(array) / sizeof(*array). But thats not needed since you already know its size. It is s.
This line
printf("end of reading the array");
has no line feed at the end of the string. This is a problem because printf is part of the family of functions called "buffered IO". The C library maintains a buffer of the things you want to print and only sends them to the terminal if the buffer gets full or it encounters \n in the stream of characters. You will not see, end of reading the array on your screen until after you have printed a line feed. You only do this after calling sort(). So all you know is your program is getting into an infinite loop at some point before the end of sort.
So there are actually three loops that could be infinite: the for loop you identified, the while loop in sort and the for loop inside the while loop. As the other answers point out, you have made the classic mistake of using assignment in the while conditional
while (finish = 1)
// ^ not enough equals signs
Unless your C compiler is really old, it is probably outputting a warning on that line. You should heed warnings.
Also, you should learn to use a debugger sooner rather than later. Believe me, it will save you a lot of time finding bugs.
In the sort function sizeof(array) returns the size of the pointer. (you can check it by yourself using printf("%d", sizeof(array).
The solution is to change your function to:
int sort(int* array, size_t size) { ... }
and call it with the correct array size:
sort(array, s);
i'm trying to figure out how to return an array from a function in the main().
I'm using C language.
Here is my code.
#include <stdio.h>
int *initArray(int n){
int i;
int *array[n];
for(i = 0; i < n; i++){
array[i] = i*2;
}
return array;
}
main(){
int i, n = 5;
int *array[n];
array[n] = initArray(n);
printf("Here is the array: ");
for(i = 0; i < n; i++){
printf("%d ", array[i]);
}
printf("\n\n");
}
And this is the errors the console gives me:
2.c: In function ‘initArray’:
2.c:8:13: warning: assignment makes pointer from integer without a cast [enabled by default]
array[i] = i*2;
^
2.c:11:3: warning: return from incompatible pointer type [enabled by default]
return array;
^
2.c:11:3: warning: function returns address of local variable [-Wreturn-local-addr]
2.c: In function ‘main’:
2.c:23:4: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat=]
printf("%d ", array[i]);
^
It's impossible!
I hate being a noob :(
If you could help, with explanations, I would appreciate! :D
Edit: iharob's answer is better than mine. Check his answer first.
Edit #2: I'm going to try to explain why your code is wrong
Consider the 2nd line of main() in your question:
int *array[n];
Let's try to read it backwards.
[n]
says we have an array that contains n elements. We don't know what type those elements are and what the name of the array is, but we know we have an array of size n.
array[n]
says your array is called array.
* array[n]
says you have a pointer to an array. The array that is being pointed to is called 'array' and has a size of n.
int * array[n];
says you have a pointer to an integer array called 'array' of size n.
At this point, you're 3/4 way to making a 2d array, since 2d arrays consist of a list of pointers to arrays. You don't want that.
Instead, what you need is:
int * array;
At this point, we need to examine your function, initArray:
int *initArray(int n){
int i;
int *array[n];
for(i = 0; i < n; i++){
array[i] = i*2;
}
return array;
}
The second line of initArray has the same mistake as the second line of main. Make it
int * array;
Now, here comes the part that's harder to explain.
int * array;
doesn't allocate space for an array. At this point, it's a humble pointer. So, how do we allocate space for an array? We use malloc()
int * array = malloc(sizeof(int));
allocates space for only one integer value. At this point, it's more a variable than an array:
[0]
int * array = malloc(sizeof(int) * n);
allocates space for n integer variables, making it an array:
e.g. n = 5:
[0][0][0][0][0]
Note:The values in the real array are probably garbage values, because malloc doesn't zero out the memory, unlike calloc. The 0s are there for simplicity.
However, malloc doesnt always work, which is why you need to check it's return value:
(malloc will make array = NULL if it isn't successful)
if (array == NULL)
return NULL;
You then need to check the value of initArray.
#include <stdio.h>
#include <stdlib.h>
int *initArray(int n){
int i;
int *array = malloc(sizeof(int) * n);
if (array == NULL)
return NULL;
for(i = 0; i < n; i++){
array[i] = i*2;
}
return array;
}
int main(){
int i, n = 5;
int *array = initArray(n);
if (array == NULL)
return 1;
printf("Here is the array: ");
for(i = 0; i < n; i++){
printf("%d ", array[i]);
}
free(array);
printf("\n\n");
return 0;
}
You can't just return an array like that. You need to make a dynamically allocated array in order to do that. Also, why did you use a 2d array anyway?
int array[5];
is basically (not completely) the same as:
int * array = malloc(sizeof(int) * 5);
The latter is a bit more flexible in that you can resize the memory that was allocated with malloc and you can return pointers from functions, like what the code I posted does.
Beware, though, because dynamic memory allocation is something you don't wanna get into if you're not ready for tons of pain and debugging :)
Also, free() anything that has been malloc'd after you're done using it and you should always check the return value for malloc() before using a pointer that has been allocated with it.
Thanks to iharob for reminding me to include this in the answer
Do you want to initialize the array? You can try it like this.
#include <stdio.h>
void initArray(int *p,int n)
{
int i;
for(i = 0; i < n; i++)
{
*(p+i) = i*2;
}
}
void main(void)
{
int i, n = 5;
int array[n];
initArray(array,n);
printf("Here is the array: ");
for(i = 0; i < n; i++)
{
printf("%d ", array[i]);
}
printf("\n\n");
}
If you don't want to get in trouble learning malloc and dynamic memory allocation you can try this
#include <stdio.h>
void initArray(int n, int array[n]) {
int i;
for (i = 0 ; i < n ; i++) {
array[i] = i * 2;
}
}
int main() { /* main should return int */
int i, n = 5;
int array[n];
initArray(n, array);
printf("Here is the array: ");
for(i = 0 ; i < n ; i++) {
printf("%d ", array[i]);
}
printf("\n\n");
return 0;
}
as you see, you don't need to return the array, if you declare it in main(), and pass it to the function you can just modify the values directly in the function.
If you want to use pointers, then
#include <stdio.h>
int *initArray(int n) {
int i;
int *array;
array = malloc(n * sizeof(*array));
if (array == NULL) /* you should always check malloc success */
return NULL;
for (i = 0 ; i < n ; i++) {
array[i] = i * 2;
}
return array;
}
int main() { /* main should return int */
int i, n = 5;
int *array;
array = initArray(n);
if (array == NULL) /* if null is returned, you can't dereference the pointer */
return -1;
printf("Here is the array: ");
for(i = 0 ; i < n ; i++) {
printf("%d ", array[i]);
}
free(array); /* you sould free the malloced pointer or you will have a memory leak */
printf("\n\n");
return 0;
}
This question already has answers here:
Algorithm: efficient way to remove duplicate integers from an array
(34 answers)
Closed 8 years ago.
I want small clarification in array concept in C.
I have array:
int a[11]={1,2,3,4,5,11,11,11,11,16,16};
I want result like this:
{1,2,3,4,5,11,16}
Means I want remove duplicates.
How is it possible?
You can't readily resize arrays in C - at least, not arrays as you've declared that one. Clearly, if the data is in sorted order, it is straight-forward to copy the data to the front of the allocated array and treat it as if it was of the correct smaller size (and it is a linear O(n) algorithm). If the data is not sorted, it gets messier; the trivial algorithm is quadratic, so maybe a sort (O(N lg N)) followed by the linear algorithm is best for that.
You can use dynamically allocated memory to manage arrays. That may be beyond where you've reached in your studies, though.
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
static int intcmp(const void *pa, const void *pb)
{
int a = *(int *)pa;
int b = *(int *)pb;
if (a > b)
return +1;
else if (a < b)
return -1;
else
return 0;
}
static int compact(int *array, int size)
{
int i;
int last = 0;
assert(size >= 0);
if (size <= 0)
return size;
for (i = 1; i < size; i++)
{
if (array[i] != array[last])
array[++last] = array[i];
}
return(last + 1);
}
static void print(int *array, int size, const char *tag, const char *name)
{
int i;
printf("%s\n", tag);
for (i = 0; i < size; i++)
printf("%s[%d] = %d\n", name, i, array[i]);
}
int main(void)
{
int a[11] = {1,2,3,4,5,11,11,11,11,16,16};
int a_size = sizeof(a) / sizeof(a[0]);
print(a, a_size, "Before", "a");
a_size = compact(a, a_size);
print(a, a_size, "After", "a");
int b[11] = {11,1,11,3,16,2,5,11,4,11,16};
int b_size = sizeof(b) / sizeof(b[0]);
print(b, b_size, "Before", "b");
qsort(b, b_size, sizeof(b[0]), intcmp);
print(b, b_size, "Sorted", "b");
b_size = compact(b, b_size);
print(b, b_size, "After", "b");
return 0;
}
#define arraysize(x) (sizeof(x) / sizeof(x[0])) // put this before main
int main() {
bool duplicate = false;
int a[11] = {1,2,3,4,5,11,11,11,11,16,16}; // doesnt have to be sorted
int b[11];
int index = 0;
for(int i = 0; i < arraysize(a); i++) { // looping through the main array
for(int j = 0; j < index; j++) { // looping through the target array where we know we have data. if we haven't found anything yet, this wont loop
if(a[i] == b[j]) { // if the target array contains the object, no need to continue further.
duplicate = true;
break; // break from this loop
}
}
if(!duplicate) { // if our value wasn't found in 'b' we will add this non-dublicate at index
b[index] = a[i];
index++;
}
duplicate = false; // restart
}
// optional
int c[index]; // index will be the number of objects we have in b
for(int k = 0; k < index; k++) {
c[k] = b[k];
}
}
If you really have to you can create a new array where that is the correct size and copy this into it.
As you can see, C is a very basic (but powerful) language and if you can, use a vector to but your objects in instead (c++'s std::vector perhaps) which can easily increase with your needs.
But as long as you only use small numbers of integers you shouldn't loose to much. If you have big numbers of data, you can always allocate the array on the heap with "malloc()" and pick a smaller size (maybe half the size of the original source array) that you then can increase (using realloc()) as you add more objects to it. There is some downsides reallocating the memory all the time as well but it is a decision you have to make - fast but allocation more data then you need? or slower and having the exact number of elements you need allocated (which you really cant control since malloc() might allocate more data then you need in some cases).
//gcc -Wall q2.cc -o q2 && q2
//Write a program to remove duplicates from a sorted array.
/*
The basic idea of our algorithm is to compare 2 adjacent values and determine if they
are the same. If they are not the same and we weren't already looking previusly at adjacent pairs
that were the same, then we output the value at the current index. The algorithm does everything
in-place and doesn't allocate any new memory. It outputs the unique values into the input array.
*/
#include <stdio.h>
#include <assert.h>
int remove_dups(int *arr, int n)
{
int idx = 0, odx = -1;
bool dup = false;
while (idx < n)
{
if (arr[idx] != arr[idx+1])
{
if (dup)
dup = false;
else
{
arr[++odx] = arr[idx];
}
} else
dup = true;
idx++;
}
return (odx == -1) ? -1 : ++odx;
}
int main(int argc, char *argv[])
{
int a[] = {31,44,44,67,67,99,99,100,101};
int k = remove_dups(a,9);
assert(k == 3);
for (int i = 0;i<k;i++)
printf("%d ",a[i]);
printf("\n\n");
int b[] = {-5,-3,-2,-2,-2,-2,1,3,5,5,18,18};
k = remove_dups(b,12);
assert(k == 4);
for (int i = 0;i<k;i++)
printf("%d ",b[i]);
printf("\n\n");
int c[] = {1,2,3,4,5,6,7,8,9};
k = remove_dups(c,9);
assert(k == 9);
for (int i = 0;i<k;i++)
printf("%d ",c[i]);
return 0;
}
you should create a new array and you should check the array if contains the element you want to insert before insert new element to it.
The question is not clear. Though, if you are trying to remove duplicates, you can use nested 'for' loops and remove all those values which occur more than once.
C does not have a built in data type that supports what you want -- you would need to create your own.
int a[11]={1,2,3,4,5,11,11,11,11,16,16};
As this array is sorted array, you can achieve very easily by following code.
int LengthofArray = 11;
//First elemnt can not be a duplicate so exclude the same and start from i = 1 than 0.
for(int i = 1; i < LengthofArray; i++);
{
if(a[i] == a[i-1])
RemoveArrayElementatIndex(i);
}
//function is used to remove the elements in the same as index passed to remove.
RemoveArrayElementatIndex(int i)
{
int k = 0;
if(i <=0)
return;
k = i;
int j =1; // variable is used to next item(offset) in the array from k.
//Move the next items to the array
//if its last item then the length of the array is updated directly, eg. incase i = 10.
while((k+j) < LengthofArray)
{
if(a[k] == a[k+j])
{
//increment only j , as another duplicate in this array
j = j +1 ;
}
else
{
a[k] = a[k+j];
//increment only k , as offset remains same
k = k + 1;
}
}
//set the new length of the array .
LengthofArray = k;
}
You could utilise qsort from stdlib.h to ensure your array is sorted into ascending order to remove the need for a nested loop.
Note that qsort requires a pointer to a function (int_cmp in this instance), i've included it below.
This function, int_array_unique returns the duplicate free array 'in-place' i.e. it overwrites the original and returns the length of the duplicate free array via the pn pointer
/**
* Return unique version of int array (duplicates removed)
*/
int int_array_unique(int *array, size_t *pn)
{
size_t n = *pn;
/* return err code 1 if a zero length array is passed in */
if (n == 0) return 1;
int i;
/* count the no. of unique array values */
int c=0;
/* sort input array so any duplicate values will be positioned next to each
* other */
qsort(array, n, sizeof(int), int_cmp);
/* size of the unique array is unknown at this point, but the output array
* can be no larger than the input array. Note, the correct length of the
* data is returned via pn */
int *tmp_array = calloc(n, sizeof(int));
tmp_array[c] = array[0];
c++;
for (i=1; i<n; i++) {
/* true if consecutive values are not equal */
if ( array[i] != array[i-1]) {
tmp_array[c] = array[i];
c++;
}
}
memmove(array, tmp_array, n*sizeof(int));
free(tmp_array);
/* set return parameter to length of data (e.g. no. of valid integers not
* actual allocated array length) of the uniqe array */
*pn = c;
return 0;
}
/* qsort int comparison function */
int int_cmp(const void *a, const void *b)
{
const int *ia = (const int *)a; // casting pointer types
const int *ib = (const int *)b;
/* integer comparison: returns negative if b > a
and positive if a > b */
return *ia - *ib;
}
Store the array element with small condition into new array
**just run once 100% will work
!)store the first value into array
II)store the another element check with before stored value..
III)if it exists leave the element--and check next one and store
here the below code run this u will understand better
int main()
{
int a[10],b[10],i,n,j=0,pos=0;
printf("\n enter a n value ");
scanf("%d",&n);
printf("\n enter a array value");
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);//gets the arry value
}
for(i=0;i<n;i++)
{
if(check(a[i],pos,b)==0)//checks array each value its exits or not
{
b[j]=a[i];
j++;
pos++;//count the size of new storing element
}
}
printf("\n after updating array");
for(j=0;j<pos;j++)
{
printf("\n %d",b[j]);
} return 0;
}
int check(int x,int pos,int b[])
{ int m=0,i;
for(i=0;i<pos;i++)//checking the already only stored element
{
if(b[i]==x)
{
m++; //already exists increment the m value
}
}
return m;
}
I am using realloc to allocated memory at runtime in dynamic array. Firstly, I allocated a memory with calloc with sizeof a random integer a. In my program, I have taken a=2. After that I want to store some 14 random values generated, so I have to resize the memory using realloc. I am doing the same in a for loop. FOr 1 iteration, realloc works but after that size doesnt increase and a error occurs "corruption in heap". I am not able to understand the problem. Pls help me if you can, in understanding where the problem is occuring and how to solve it.
Thanks a lot.
Below is my code:
j=j*a; //a=3
numbers = (int*) calloc(b, j); //b=14, no of elements I want to store
printf("Address:%p\n",numbers);
if (numbers == NULL)
{
printf("No Memory Allocated\n");
}
else
{
printf("Initial array size: %d elements\n", a);
printf("Adding %d elements\n", b);
}
srand( (unsigned) time( NULL ) );
for(count = 1; count <= b ; count++)
{
if(i <= j)
{
numbers[count] = rand() % 100 + 1;
printf( "Adding Value:%3d Address%p\n", numbers[count],numbers[count] );
i++;
}
if (i > j)
{
printf("Increasing array size from %d bytes to %d bytes\n",j,j*a);
j=j*a;
numbers = (int*) realloc(numbers,j);
printf("Address:%p\n",numbers);
if(numbers == NULL)
{
printf("No Memory allocated\n");
}
}
}
free(numbers);
return 0;
}
The initial array length (length and size are not the same) is b, not a.
Adding b elements? I don't think you are.
Arrays are zero-based in C. You loop should be for(count=0; count<b ; count++).
count is a terrible name for a loop variable. count should hold the number of elements and not be a loop variable.
It's hard to imagine what j could be. Since you use it as the element size in your call to calloc it ought be at least be a multiple of 4, the size of in int. What is it?!
The realloc doesn't seem to bear any relation to the calloc.
I'm sure there are lots of other problems. If you want more help then a clear statement of what your goal is would be required.
EDIT
It sounds like you want something like this:
int capacity = 10;
int count = 40;
int i;
int* array = (int*)malloc(capacity*sizeof(int));
for (i=0; i<count; i++)
{
if (i==capacity)
{
capacity *= 2;
array = (int*)realloc(array, capacity*sizeof(int));
}
array[i] = RandomIntInRange(1, 100);
}
free(array);
Notes:
No error checking. In production code you would check that the allocations succeeded, and the realloc done this way would leak if it failed. But there's no point confusing the message with error checking when you are still at this level of understanding.
No reading input - you can do that.
No writing output - you can do that.
The integer "j" is not initialized in your code, resulting in a = 0 * 3, meaning a will be zero and no memory will be allocated. The segfault is due to you not handling that numbers is NULL. Change to and set j to something meaningful
#include <stdlib.h>
#include <stdio.h>
void
main (int argc, char *argv[])
{
int a = 3;
int j = 1 * a; //a=3
int b = 14;
int *numbers = calloc (b, j); //b=14, no of elements I want to store
int count = 0, i = 0;
printf ("Address:%p\n", numbers);
if (numbers == NULL)
{
printf ("No Memory Allocated\n");
return;
}
else
{
printf ("Initial array size: %d elements\n", a);
printf ("Adding %d elements\n", b);
}
srand ((unsigned) time (NULL));
for (count = 1; count <= b; count++)
{
if (i <= j)
{
numbers[count] = rand () % 100 + 1;
printf ("Adding Value:%3d Address%p\n", numbers[count],
&(numbers[count]));
i++;
}
if (i > j)
{
printf ("Increasing array size from %d bytes to %d bytes\n", j,
j * a);
j = j * a;
numbers = (int *) realloc (numbers, j);
printf ("Address:%p\n", numbers);
if (numbers == NULL)
{
printf ("No Memory allocated\n");
}
}
}
free (numbers);
}