I don't know how to clean an array from the elements which indexes stored in the other array. I need to complete the following C-program consisting of main(…) and function
void clear_MSBs( unsigned char dest_array[], unsigned char array_indices []).
Beginning of code:
#define N 8
#define M 5
int main()
{
unsigned char dest_array[N] = {248,249,250,251,252,253,254,255};
unsigned char array_indices[M] = {0,2,3,6,7}; // contains M=5 elements
clear_MSBs(dest_array, array_indices);
// print the modified dest_array[] here
return 0;
}
Note: It is guaranteed that all indices stored in the second array are within the
allowed range.
I would really appreciate your help.
If by cleaning, you mean marking the element as invalid (which is probably what you want), then you could just loop over the indices array, and use the i-th element of the indices array as the index for the destination array.
Example:
#include <stdio.h>
#define N 8
#define M 5
void clear_MSBs(unsigned char dest_array[], unsigned char array_indices [])
{
for(int i = 0; i < M; ++i)
dest_array[array_indices[i]] = 0;
}
int main()
{
unsigned char dest_array[N] = {248,249,250,251,252,253,254,255};
unsigned char array_indices[M] = {0,2,3,6,7}; // contains M=5 elements
clear_MSBs(dest_array, array_indices);
// print the modified dest_array[] here
for(int i = 0; i < N; ++i)
if(dest_array[i] != 0)
printf("%d ", dest_array[i]);
printf("\n");
return 0;
}
Output:
249 252 253
PS: The code assumes that an invalid element has the 0 value.
You just need to overwrite what you're deleting with the next value in the array, propagate that change, and then keep in mind where the new end is.
In C++, though, it would be better to use a std::vector:
std::vector<int> array; // initialize array... // delete element at index 2
array.erase(array.begin() + 2);
Related
Can we directly insert a 1D array to a 2D array?
For example I have this code:
void insert(int[]data , int**collection)
{
collection[1] = data
}
int main()
{
int data[2]= {1,3}
int collection[2][2];
insert(data,&collection);
}
Will this work?
You cannot insert a 1D array to 2D array the way you are doing. Use memcpy to copy the elements of 1D array to 2D array, like this:
memcpy(collection[1], data, 2 * sizeof(int));
this will copy the 2 integer elements of data array to collection[1].
Besides, a couple of problems in your code. Lets discuss them:
First:
insert(data,&collection);
^
You don't need to pass the address of collection. Note that, an array, when used in an expression, will convert to pointer to its first element (there are few exceptions to this rule). That means, when you pass collection, it will convert to type int (*)[2]. Just do:
insert(data, collection);
Second:
void insert(int[]data , int**collection)
int[]data is wrong. The first parameter of insert() should be int data[2] or int data[], both are equivalent to int * data. You can use either of them.
The second argument to insert() is collection array which is a 2D array of integers. When you pass it to insert(), it will decay to pointer whose type is int (*)[2]. The type of second parameter of insert() is int ** which is not compatible with the argument that you are passing to insert() function. The second parameter of insert() function should be int collection[2][2] or int collection[][2], both are equivalent to int (*collection)[2].
Putting these altogether, you can do:
#include <stdio.h>
#include <string.h>
#define ROW 2
#define COL 2
void insert(int data[ROW], int collection[ROW][COL]) {
//for demonstration purpose, copying elements of data array
//to all elements (1D array) of collection array.
for (int i = 0; i < ROW; ++i) {
memcpy(collection[i], data, COL * sizeof(int));
}
}
int main(void) {
int data[COL] = {1, 3};
int collection[ROW][COL];
insert(data, collection);
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < 2; ++j) {
printf("collection[%d][%d] : %d ", i, j, collection[i][j]);
}
printf("\n");
}
return 0;
}
Output:
# ./a.out
collection[0][0] : 1 collection[0][1] : 3
collection[1][0] : 1 collection[1][1] : 3
I declared a 2d array in main function and I want to get the number of elements in another function.
Let's say my main is like this.
#define col 100
#define row 10
int main(){
array[row][col] = {"abc1", "abc2", "abc3"};
int len = get2DLen(array);
}
and I copied this array in this form (*array)[column]
int get2DLen(char (*array)[col]){
int len;
//some operations..
return len;
}
Questions are..
how can I get col in func? In this case, 10.
how can I get the number of actual elements in the array? In this case, 3.
I'm aware of the fact that array means *array[0] in here.
I tried to iterate all the elements and check if it's '\0' like a normal char array string. But it's just absolutely wrong.. cause it was initialized anyway..
so any help would be appreciated
In C, it's not possible to determine the size of the array. Therefore, you can't get col in the function. You need to pass the dimension of the array to the function parameter.
#include <stdio.h>
#include <stddef.h>
#define col 100
#define row 10
int get2DLen(char (*array)[], size_t arr_row, size_t arr_col)
{
int len = arr_col;
return len;
}
int main()
{
char array[row][col] = {"abc1", "abc2", "abc3"};
int len = get2DLen(array, 3, 4);
printf("%d", len);
}
I am trying to convert a string into its equivalent matrix form in C. The matrix would have 3 rows and as many columns as required. The following code doesn't compile, and I haven't figured out what's going wrong.
The error that GCC throws is:
app.c:10:25: error: subscripted value is not an array, pointer, or vector
printf("%d\n", arr[i][k]);
~~~^~
1 error generated.
Main file (app.c):
#include <stdio.h>
#include "converter.h"
int main() {
char source[] = "This is the source. "; // placeholder text
int arr = convert(source);
for (int i = 0; i < 21; i++) {
for (int k = 0; k < 3; k++) {
printf("%d\n", arr[i][k]); // error occurs at this line.
}
}
return 0;
}
converter.c file:
// Converts an input string to its respective ASCII matrix.
#include <string.h>
#include <stdio.h>
#include "converter.h"
// Converts the entire string into an multi-dimensional array.
int convert(char text[]){
// copy the input text into a local store.
char store[strlen(text)];
strcpy(store, text);
// make sure the length of the input string is a multiple of 3 or make it so.
int excess = strlen(store)%3;
char excess_spaces[3] = " ";
if (excess != 0) {
strncat(store, excess_spaces, 3-excess);
}
// covert the source into an array
int arr[3][strlen(store)/3];
int steps = strlen(store)/3;
for (int i = 0; i < steps; i++) {
int t[3];
for (int k = 0; k < 3; k++) {
t[k] = (int) store[3*i+k];
arr[k][i] = t[k];
}
}
return arr;
}
converter.h file:
int convert(char text[]);
There are multiple issues in this code.
The allocating storage for string, one must include one byte for a null terminator. Replace:
char store[strlen(text)];
with:
char store[strlen(text) + 1];
Additionally store must be big enough to contain the excess which is up to 3 spaces.
char store[strlen(text) + 3 + 1];
In C you cannot use an array as a value. It is converted to a pointer to it's first element in pretty must every context. Therefore it is not possible to return an array directly. It could be workaround by wrapping an array with a struct but it a topic for another day.
As result return arr will be equivalent to return &arr[0] which is int (*)[XXX] a pointer to int array of size XXX.
Never ever return a pointer to an object with automatic storage. It's Undefined Behaviour. I know that the intention was returning an array not a pointer to it. Create an object with dynamic storage with malloc-like function to safely return a pointer.
Returning Variable Length Array (VLA) by value is not possible because Variably Modified (VM) types cannot be defined at file scope.
It looks that indices are swapped in:
printf("%d\n", arr[i][k]);
I guess it should be arr[k][i].
Now... let's solve it.
Returning VLA is tricky. One solution is to pass a pointer to VLA as an argument. See https://stackoverflow.com/a/14088851/4989451.
The issue with this solution is that the caller must be able to compute the dimensions.
The other way it to wrap the result of the convert() to a struct. Note that the function and the struct can share the name. The result with have the sizes of VLA as n and m members and the pointer to the data as arr. The caller need to cast it to proper VM type.
To cumbersome casts between the non-trivial pointer types, one can cast via void*.
When all work with the array is done, release it memory with free().
// Converts an input string to its respective ASCII matrix.
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
// Converts the entire string into an multi-dimensional array.
struct convert {
int n, m;
int *arr;
} convert(char text[]){
// copy the input text into a local store.
size_t textlen = strlen(text);
char store[textlen + 3 + 1];
strcpy(store, text);
// make sure the length of the input string is a multiple of 3 or make it so.
int excess = textlen % 3;
char excess_spaces[3] = " ";
if (excess != 0) {
strncat(store, excess_spaces, 3-excess);
}
size_t storelen = strlen(store);
// allocate VLA with dynamic storage
int (*arr)[storelen / 3] = malloc(3 * sizeof *arr);
// covert the source into an array
int steps = storelen / 3;
for (int i = 0; i < steps; i++) {
int t[3];
for (int k = 0; k < 3; k++) {
t[k] = (int) store[3*i+k];
arr[k][i] = t[k];
}
}
return (struct convert){ .n = 3, .m = steps, .arr = (int*)arr };
}
int main() {
char source[] = "This is the source. "; // placeholder text
struct convert res = convert(source);
int n = res.n, m = res.m;
int (*arr)[m] = (void*)res.arr;
for (int i = 0; i < n; i++, puts("")) {
for (int k = 0; k < m; k++) {
printf("%d ", arr[i][k]); // error occurs at this line.
}
}
free(arr);
return 0;
}
I'm tasked with writing a function that will identify all the even numbers in an sample array {10,2,9,3,1,98,8] and place them in an array called EvenNumbers. I have to allow the function so that it works with different combinations of numbers in the array not just the numbers in the sample array above.
I'm wondering is there any way to add numbers to an array that could be different every time? How would I extract the even numbers an place them into an array? Also
for the even array size its giving me an error that the expression must have a constant value but when I use const int it still gives me that error.
Here is the full question.
"Using the array of sample values {10,2,9,3,1,98,8}, write a function that will identify all the even numbers in an array and place it in an array called EvenNumbers. The function must work in all cases, not just in the case of the array shown. Assume that the array size is always available through a global constant called MAX"
Here is what I have so far. I've no idea how I will extract the even numbers from a for loop and place them in an array. I also dont know what the "expression must have a constant value" is about?
#include <stdio.h>
#include <stdlib.h>
void EvenNumber(int Array[], int size);
int main()
{
int array[7] = { 10,2,9,3,1,98,8 };
EvenNumber(array, 7);
}
void EvenNumber(int Array[], int size)
{
int i;
int EvenArraySize;
for (i = 0; i < size; i++)
{
if (Array[i] % 2 == 0)
{
EvenArraySize++;
}
}
int Even[EvenArraySize];
}
The right way to go is to use malloc to allocate just the right amount of memory.
Count the number of even numbers
Allocate the space needed to store them
Copy even numbers in this space
Do whatever you want with these numbers
Free the allocated space
Snippet:
#include <stdio.h>
#include <stdlib.h>
#define MAX 7
int
main()
{
int array[] = {10,2,9,3,1,98,8};
int *even_numbers;
int i, nb_even_numbers;
for (i = 0, nb_even_numbers = 0; i < MAX; i++)
{
if (array[i] % 2 == 0)
nb_even_numbers++;
}
even_numbers = malloc(sizeof(int) * nb_even_numbers);
if (!even_numbers)
{
perror("malloc");
return 1;
}
for (i = 0, nb_even_numbers = 0; i < MAX; i++)
{
if (array[i] % 2 == 0)
even_numbers[nb_even_numbers++] = array[i];
}
/* do your stuff here */
free(even_numbers);
return 0;
}
First, you can never return a statically declared array from a function (even though you don't explicitly try, your Even array is destroyed when EvenNumber returns) Why? The function stack frame for EvenNumber is released for reuse on return and any locally declared arrays are no longer valid.
You either need to pass a second array as a parameter to EvenNumber, or you can dynamically allocate storage for Even in EvenNumber (with, e.g. malloc or calloc or realloc) and return a pointer to the beginning of the array. (you must also have some way to return the size or use a constant for a max size).
There is no need to use % (modulo) to test whether a number is odd/even. All you need to do is look at bit-0 (little endian). If it is 0, then the number is odd, if it is 1, then its even. Much more efficient than calling modulo which incorporates division.
Finally, main is type int and therefore returns a value.
Putting those pieces together, you can do something simple like the following:
#include <stdio.h>
#include <stdlib.h>
void EvenNumber (int *array, int *even, int size, int *esize);
int main (void)
{
int array[] = { 10,2,9,3,1,98,8 },
i, n = sizeof array / sizeof *array,
even[n], /* a VLA of the same size as array is fine here */
esize = 0;
EvenNumber (array, even, n, &esize);
printf ("array: ");
for (i = 0; i < n; i++)
printf (" %2d", array[i]);
printf ("\neven : ");
for (i = 0; i < esize; i++)
printf (" %2d", even[i]);
putchar ('\n');
return 0;
}
void EvenNumber (int *array, int *even, int size, int *esize)
{
int i;
for (i = 0; i < size; i++)
if ((array[i] & 1) == 0) /* simply looking at bit-0 is all you need */
even[(*esize)++] = array[i];
}
Note: esize is passed as a pointer to EvenNumber and updated within the function so that the number of elements in even are available back in the calling function (main() here).
Example Use/Output
$ ./bin/arrayeven
array: 10 2 9 3 1 98 8
even : 10 2 98 8
Let me know if you have any further questions.
I am trying to copy values from one array to another in C. The original array is of type long int, and it is inputted to the function as *pixel_frequency. The array I want to copy to is temp, and it is a 2D array. However, when I try to copy the values, pixel_frequency is fine but temp gives strange results. Below is the relevant code and some sample output.
Code:
struct node *generate_nodes(long int *pixel_frequency) {
int i;
int temp[max_value + 1][2];
for (i = 0; i < (max_value + 1); i++) {
temp[i][0] = i;
temp[i][1] = pixel_frequency[i];
printf("Frequency for %d is %d\n", temp[i][0], temp[i][1]);
}
...
Output (each frequency is supposed to be 256):
Frequency for 0 is 150160
Frequency for 1 is 256
Frequency for 2 is 256
Frequency for 3 is 256
Frequency for 4 is 255
...
Frequency for 254 is 892677956
Frequency for 255 is 1868789101
Below is an example version of code using your code snippet, take care of the return type and other things.
Instead of using a global variable for array size, you can pass it as a function argument so , it will be easy to identify the size of array you passed.
void generate_nodes(long int *pixel_frequency, size_t size) {
size_t i;
long int temp[size][2];
for (i = 0; i < size; i++) {
temp[i][0] = i;
temp[i][1] = pixel_frequency[i];
printf("Frequency for %ld is %ld\n", temp[i][0], temp[i][1]);
}
}
If you have declared pixel_frequency as a local variable in a function and used the address of array after the variable went out of scope, will lead to undefined behaviour.
int main(void) {
size_t max_len = 5000;
size_t i;
long int* pixel_frequency = malloc(max_len*sizeof(long int));
for( i = 0; i < max_len; ++i) {
pixel_frequency[i] = (i%256);
}
generate_nodes(pixel_frequency, max_len);
return 0;
}