#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void main() {
int i,j;
int *u = malloc(10000 * 10000 * sizeof(int));
for (i=0; i<10000; i++)
{
for(j=0;j<10000;j++)
{
u[i][j]=i+j;
}
}
free(u);
}
I edited my program. when compiling this program, I get an error "subscripted value is neither array nor pointer nor vector".
how can i allocate memory?
You have allocated memory for a single dimensional array and you are trying to use it as a two-dimensional array. There is a slight alteration which you need to do to your code:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main() {
int i,j;
int *u = malloc(10 * 10 * sizeof(int));
for (i=0; i<10; i++)
{
for(j=0;j<10;j++)
{
u[10*i +j]=i+j; // this is how you can use it
}
}
for (i=0; i<10; i++)
{
for(j=0;j<10;j++)
{
printf("%d ",u[10 *i +j]);
}
printf("\n");
}
free(u);
return 0;
}
Note that I have used size 10*10, you can do the same for whatever size you need.
Check-here
You can't allocate large arrays on heap directly in the declaration.
You can allocate large arrays using malloc as follows
#include <stdlib.h>
int *matrix = malloc(ROW * COLUMNs * sizeof(int));
Always use column major order to search for elements.
Explanation for column major order can be found here Accessing elements in a matrix
Here size = 10000
Always after you complete your task,free the memory
free(matrix);
Related
Is there a way to initialize all int array elements to zero except a for loop where we loop through to set values to zero. Here the size of array is decided by input of user.
#include <stdio.h>
int main() {
int num_cases = 0;
scanf("%d", & num_cases);
int arr_counter[num_cases];
for (int x = 0; x < num_cases; x++) {
arr_counter[x] = 0;
}
}
Yes, you can do that in multiple ways under C standard. For example:
memset() [Stack and Heap]
calloc() [Heap Only]
loops, e.g., do-while, while and for [Stack and Heap]
{ } [Stack Only]
1
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
int arr[10];
size_t len_arr = sizeof(arr) / sizeof(*arr);
memset(arr, 0, sizeof(arr));
for(size_t i = 0; i < len_arr; i++)
printf("%d\n", arr[i]);
return EXIT_SUCCESS;
}
Note: We can use memset() to set all values as 0 or -1 for integral data types also. It will not work if we use it to set as other values. The reason is simple, memset() works byte by byte.
2
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
int *arr = calloc(10, sizeof(int));
if(!arr)
{
fprintf(stderr, "bad ptr");
return EXIT_FAILURE;
}
for(size_t i = 0; i < 10; i++)
printf("%d\n", arr[i]);
free(arr);
return EXIT_SUCCESS;
}
Note: You need to keep track of arr maximum length.
Note: malloc() leaves garbage value in your pointer, whereas calloc() uses memset() to initialize them to 0.
Note: You need to free the heap allocated resource.
3
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
int arr[10];
size_t len_arr = sizeof(arr) / sizeof(*arr);
for(size_t i = 0; i < len_arr; i++)
arr[i] = 0;
for(size_t i = 0; i < len_arr; i++)
printf("%d\n", arr[i]);
return EXIT_SUCCESS;
}
Note: You can use any of your favorite loop.
4
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int arr[10] = {};
size_t len_arr = sizeof(arr) / sizeof(*arr);
for(size_t i = 0; i < len_arr; i++)
printf("%d\n", arr[i]);
return EXIT_SUCCESS;
}
Use calloc function available in stdlib.h
#include<stdio.h>
#include<stdlib.h>
int main(){
int num_cases;
scanf("%d", &num_cases);
int* arr = (int*)calloc(num_cases,sizeof(int));
return 0;
}
To initialize each element of Array you have two approaches:
If you are going for static memory allocation, you can initialize it like this:
int arr[10] = {};
If you are going for dynamic memory allocation, you can use calloc function.
int *arr = (int) calloc(numberOfElementsInArray,sizeOfEachElement);
In your case, it would be like:
int *arr_counter = (int*) calloc(num_cases,sizeof(int));
NOTE: You need to include malloc.h header file to use calloc function.
because I am new in C, I am not sure how to ask it, but here is my Code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ARRAY_SIZE 500
int main(int argc, char *argv[]) {
for (int j=0; j<ARRAY_SIZE; ++j) {
printf("Memory Size: %d\n", j);
int bytes = (1024*1024);
char *data;
data = (char *) malloc(bytes);
for(int i=0;i<bytes;i++){
data[i] = (char) rand();
}
}
//Free all Char*data that I have declared inside the for loop here
return 0;
}
So I need to free my data variables that I have allocated inside the for loop. How is it possible? I am testing some portion of my memory blocks. So I am running it because I wanna see how far it goes. So the above code gets me to that point. Now I am trying to run a loop below threshold point so that I can assure, the memory that I am working with is good and can sustain. To do so, I need to clear the memory that I have created inside the loop.
Thanks in advance
I think you'll want an array of pointers and then free those pointers after the loop.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ARRAY_SIZE 500
int main(int argc, char *argv[]) {
char * data[ARRAY_SIZE] = {0};
for (int j=0; j<ARRAY_SIZE; ++j) {
printf("Memory Size: %d\n", j);
int bytes = (1024*1024);
data[j] = (char *) malloc(bytes);
for(int i=0;i<bytes;i++){
data[j][i] = (char) rand();
}
}
for (int j=0; j<ARRAY_SIZE; ++j) {
free(data[j]);
}
return 0;
}
Since you assigned the return value of malloc to data which is defined inside of the loop, that memory is lost at the end of each iteration of the loop.
You need to add a call to free at the bottom of the loop.
for (int j=0; j<ARRAY_SIZE; ++j) {
printf("Memory Size: %d\n", j);
int bytes = (1024*1024);
char *data;
data = (char *) malloc(bytes);
for(int i=0;i<bytes;i++){
data[i] = (char) rand();
}
free(data);
}
I'm trying to count the number of elements in an array as a pointer as the code followed:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int *ptr = (int *) malloc(sizeof(int*));
for(int i=0; i<8; i++)
{
printf("The number: " );
scanf("%d", &ptr[i]);
}
int size = sizeof(ptr)/sizeof(int);
printf("%d\n", size);
return 0;
}
I have tried the syntax for an array size = sizeof(ptr)/sizeof(int);but I got the wrong answer which is 1 for all cases. I don't know how to get the correct answer, which is 8 for this case
Unfortunately, you cannot get the size of an array allocated with malloc (because you are actually getting the size of a pointer). You must always store that somwhere else. Since you always allocate 8 elements, why not make it a static array?
#include <stdio.h>
int main()
{
int arr[8];
for(int i=0; i<8; i++)
{
printf("The number: " );
scanf("%d", &arr[i]);
}
int size = sizeof(arr)/sizeof(int);
printf("%d\n", size);
return 0;
}
My following code sucessfully runs in sample input but gives segmentation faults in 13 test cases.
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
int n;
int k;
int q;
int index[q];
scanf("%d %d %d",&n,&k,&q);
int *a = (malloc(sizeof(int) * n));
for(int a_i = 0; a_i < n; a_i++){
scanf("%d",&a[a_i]);
}
for(int a0 = 0; a0 < q; a0++){
int m;
scanf("%d",&m);
index[a0] = m;
}
for(int i=0; i<k; i++){
int ap = a[n-2];
for(int p=1; p<n-1; p++){
a[p] = a[p-1];
}
a[0] = a[n-1];
a[n-1] = ap;
}
for(int j=0; j<q;j++){
printf("%d\n", a[index[j]]);
}
return 0;
}
I am unable to find where the segmentation fault is. Also check out this:where I asked about declaring a as a pointer using malloc
There might have been chances that using malloc() to declare a would have lead to segmentation faults since it does not check for allocation error, but even when i defined a as an array the problem still remained.
q is not initialized, but is used as the argument to the declaration of an array.
You should use malloc to allocate the index array, after reading the value of q.
I found a segmentation fault in my C code and could not find a good explanation or solution for it after searching.
This first code gives me segmentation fault after printing 0.
#include <stdlib.h>
#include <stdio.h>
int main() {
int **defs = malloc(16 * sizeof *defs);
int i;
for (i = 0; i < 16; i++) {
printf("%d\n", i);
*defs[i] = i;
}
free(defs);
return 0;
}
This second code works fine.
#include <stdlib.h>
#include <stdio.h>
int main() {
int *defs = malloc(16 * sizeof defs);
int i;
for (i = 0; i < 16; i++) {
printf("%d\n", i);
defs[i] = i;
}
free(defs);
return 0;
}
These are just examples, not my actual code. I also tried doing pointer arithmetic but same result.
Could someone please explain this? Thank you.
In first code you have not allocated each of the int* memory blocks.
So, before assigning values to defs[i], you have to populate it with memory of type int*.
defs[i] = (int*)malloc(sizeof(int) * number_of_elements);
And then defs[i][some_index] = value.