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I tried this code..As you can see the problem is the empty elements are zero. So, I tried to check with it but the thing is I can have 0 as an element.
int main()
{
int array[10] = {1, 2, 0, 3, 4};
printf("%d\n", sizeof(array)/ sizeof(*array)); // This is the size of array
int i = 0;
while(array[i] != 0 && i < 10) {
i++;
};
printf("%d\n", i);
return 0;
}```
You can't. int array[10] will always create an array of 10 elements and you can't ask the compiler which of them have been assigned.
What you could do is int array[] = {1, 2, 0, 3, 4} then the compiler will infer the number of elements for you and you'll have sizeof(array)/ sizeof(*array) == 5
First set the array to a number outside the range of your inputs. Like a negative number.
for(i = 0;i < 10;i++)
array[i] = -1;
or set it to INT_MAX or INT_MIN
int array[10] = {1 , 2, 0, 3} . How can I find out that there are 4 elements here?
How can you say there are 4 elements there as you declared that int array[10] with the size of 10 elements. This implies, you already know the no. of elements. Also, in this scenario, you can't use an if statement to determine the no. of elements as you probably know that in C, if you initialize an array of 10 elements with less than 10 values, rest of them will automatically be assigned to 0.
You have several options:
You know how many elements are in the initializer, so you create another variable that stores that number:int array[10] = {1, 2, 0, 3, 4};
int num_items = 5;
You'll need to update num_items as you "add" or "remove" items to the array. If you treat the array as a stack (only add or remove at the highest index), then this is easy-ish:array[num_items++] = 7; // adds 7 after 4
...
x = array[--num_items]; // x gets 7, 7 is "removed" from the array, need special
// case logic for element 0
You pick a value that isn't valid (say -1) and initialize the remaining elements explicitly:int array[10] = {1, 2, 0, 3, 4, -1, -1, -1, -1, -1 };
You size the array for the initializer, meaning it can only ever store that many elements:int array[] = {1, 2, 0, 3, 4};
Otherwise, you'll need to use a different data structure (such as a linked list) if you need a container that can grow or shrink as items are added or removed.
I built a simple function that, given two arrays aa[5] = {5, 4, 9, -1, 3} and bb[2] = {16, -11}, orders them in a third array cc[7].
#include<stdio.h>
void merge(int *, int *, int *, int, int);
int main(){
int aa[5] = {5, 4, 9, -1, 3};
int bb[2] = {16, -11};
int cc[7];
merge(aa, bb, cc, 5, 2);
return 0;
}
void merge(int *aa, int *bb, int *cc, int m, int n){
int i = 0, j = 0, k = 0;
while(i < m && j < n){
if(aa[i] < bb[j])
cc[k++] = aa[i++]; /*Smallest value should be assigned to cc*/
else
cc[k++] = bb[j++];
}
while(i < m) /*Transfer the remaining part of longest array*/
cc[k++] = aa[i++];
while(j < n)
cc[k++] = bb[j++];
}
The cc array is correctly filled, but the values are not ordered. Instead of the expected cc = {-11, -1, 3, 4, 5, 9, 16} it returns cc = {5, 4, 9, -1, 3, 16, 11}.
Like the assignments cc[k++] = aa[i++] and cc[k++] = bb[j++] do not work, somehow, or the logical test if aa[i] < bb[j] goes ignored.
I hypothesized operators priority problems, hence I tested with two different standard, with no differences:
gcc main.c -o main.x -Wall
gcc main.c -o main.x -Wall -std=c89
I checked the code many times, unable to find any relevant error. Any suggestion at this point would be appreciated.
You need to think your algorithm through properly. There's no obvious bug in it. The problem is your expectations. One way to make this clear is to think about what would happen if one array was empty. Would the function merge change the order of anything? It will not. In fact, if two elements a and b are from the same array - be it aa or bb - and a comes before b in that array, then a will also come before b in cc.
The function does what you expect on sorted arrays, so make sure they are sorted before. You can use qsort for this.
Other than that, when you use pointers to arrays you do not want to change, use the const qualifier.
void merge(const int *aa, const int *bb, int *cc, int m, int n)
There's no bug in your implementation (at least I don't see any, imho) The problem is that the merging you have done is not for two sorted arrays (it's for several bunches of sorted numbers). Case you had feed two already sorted arrays you'd have the result sorted correctly.
The merge sorting algorithm begins with splitting the input into two parts of sorted arrays. This is done by switching arrays when you detect the element is not in order (it is not greater to last number) You get the first ordered set to fill an array (the first a elements of initial list which happen to be in order, to put into array A, and the second bunch of elements to put them into array B. This produces two arrays that can be merged (because they are already in order) and this merging makes the result a larger array (this fact is what warrants the algorithm will make larger and larger arrays at each pass and warrants it will finish at some pass. You don't need to operate array by array, as at each pass the list has less and less packs of larger bunch of sorted elements. in your case:
1st pass input (the switching points are where the input
is not in order, you don't see them, but you switch arrays
when the next input number is less than the last input one):
{5}, {4, 9}, {-1, 3}, {16}, {-11} (See note 2)
after first split:
{5}, {-1, 3}, {-11}
{4, 9}, {16}
after first merge result:
{4, 5, 9}, {-1, 3, 16}, {-11}
after second pass split:
{4, 5, 9}, {-11}
{-1, 3, 16}
result:
{-1, 3, 4, 5, 9, 16}, {-11}
third pass split:
{-1, 3, 4, 5, 9, 16}
{-11}
third pass result:
{-11, -1, 3, 4, 5, 9, 16}
The algorithm finishes when you don't get two bunches of ordered streams (you don't switch arrays), and you cannot divide further the stream of data.
Your implementation only executes one pass of merge sorting algorithm you need to implement it completely to get sorted output. The algorithm was designed to make it possible to do several passes when input is not feasible to put in arrays (as you do, so it doesn't fully illustrate the thing with arrays). Case you have it read from files, you'll see the idea better.
NOTE
Sort programs for huge amounts of data use merging algorithm for bunchs of data that are quicksort'ed first, so we start with buckets of data that don't fit in an array all together.
NOTE 2
The number 16 after number 3 should have been in the same bunch as previous bunch, making it {-1, 3, 16}, but as they where in different arrays at first, and I have not found any way to put them in a list that splits into this arrangement, I have forced the buckets as if 16 < 3, switching artificially the arrays on splitting the input. This could affect the final result in making an extra pass through the data, but doesn't affect the final result, which is a sorted list of numbers. I have made this on purpose, and it is not a mistake (it has no relevance to explain how the algorithm works) Anyway, the algorithm switches lists (I don't like to use arrays when describing this algoritm, as normally merging algorithms don't operate on arrays, as arrays are random access, while lists must be accessed by some iterator means from beginning to end, which is the requirement of the merging sort algorithm) The same happens to {4, 9}, {16} after the first split, just imagine the result of the comparisons was the one shown, as after the first merge everything is correct.
If your program works fine, you can sort in O(N) by comparison. As it is not possible and mentioned in comments by #Karzes, your program works fine just for the sorted sub arrays. Hence, if you want to implement merge function for the merge sort, you should try your program for these two inputs:
int aa[5] = {-1, 3, 4, 5, 9};
int bb[2] = {-11, 16};
Not the most efficient cause it's bobble sort...
#include<stdio.h>
void merge(int *, int *, int *, int, int);
void sortarray(int array[], int arraySize)
{
int c,d,temp;
for (c = 0 ; c < arraySize-1; c++)
{
for (d = 0 ; d < arraySize - c - 1; d++)
{
if (array[d] > array[d+1]) /* For decreasing order use < */
{
temp = array[d];
array[d] = array[d+1];
array[d+1] = temp;
}
}
}
}
int main(){
int aa[5] = {5, 4, 9, -1, 3};
int bb[2] = {16, -11};
int cc[7];
int i;
sortarray(aa,sizeof(aa)/sizeof(aa[0]));
sortarray(bb,sizeof(bb)/sizeof(bb[0]));
merge(aa, bb, cc, 5, 2);
for(i=0;i<sizeof(cc)/sizeof(cc[0]);i++)
{
printf("%d,",cc[i]);
}
return 0;
}
void merge(int *aa, int *bb, int *cc, int m, int n){
int i = 0, j = 0, k = 0;
while(i < m && j < n)
{
if(aa[i] < bb[j])
cc[k++] = aa[i++]; /*Smallest value should be assigned to cc*/
else
cc[k++] = bb[j++];
}
while(i < m) /*Transfer the remaining part of longest array*/
cc[k++] = aa[i++];
while(j < n)
cc[k++] = bb[j++];
}
If I want to append a number to an array initialized to int, how can I do that?
int arr[10] = {0, 5, 3, 64};
arr[] += 5; //Is this it?, it's not working for me...
I want {0,5, 3, 64, 5} in the end.
I'm used to Python, and in Python there is a function called list.append that appends an element to the list automatically for you. Does such function exist in C?
int arr[10] = {0, 5, 3, 64};
arr[4] = 5;
EDIT:
So I was asked to explain what's happening when you do:
int arr[10] = {0, 5, 3, 64};
you create an array with 10 elements and you allocate values for the first 4 elements of the array.
Also keep in mind that arr starts at index arr[0] and ends at index arr[9] - 10 elements
arr[0] has value 0;
arr[1] has value 5;
arr[2] has value 3;
arr[3] has value 64;
after that the array contains garbage values / zeroes because you didn't allocated any other values
But you could still allocate 6 more values so when you do
arr[4] = 5;
you allocate the value 5 to the fifth element of the array.
You could do this until you allocate values for the last index of the arr that is arr[9];
Sorry if my explanation is choppy, but I have never been good at explaining things.
There are only two ways to put a value into an array, and one is just syntactic sugar for the other:
a[i] = v;
*(a+i) = v;
Thus, to put something as the element at index 4, you don't have any choice but arr[4] = 5.
For some people which might still see this question, there is another way on how to append another array element(s) in C. You can refer to this blog which shows a C code on how to append another element in your array.
But you can also use memcpy() function, to append element(s) of another array. You can use memcpy()like this:
#include <stdio.h>
#include <string.h>
int main(void)
{
int first_array[10] = {45, 2, 48, 3, 6};
int scnd_array[] = {8, 14, 69, 23, 5};
int i;
// 5 is the number of the elements which are going to be appended
memcpy(first_array + 5, scnd_array, 5 * sizeof(int));
// loop through and print all the array
for (i = 0; i < 10; i++) {
printf("%d\n", a[i]);
}
}
You can have a counter (freePosition), which will track the next free place in an array of size n.
If you have a code like
int arr[10] = {0, 5, 3, 64}; , and you want to append or add a value to next index, you can simply add it by typing a[5] = 5.
The main advantage of doing it like this is you can add or append a value to an any index not required to be continued one, like if I want to append the value 8 to index 9, I can do it by the above concept prior to filling up before indices.
But in python by using list.append() you can do it by continued indices.
Short answer is: You don't have any choice other than:
arr[4] = 5;
void Append(int arr[],int n,int ele){
int size = n+1; // increasing the size
int arrnew[size]; // Creating the new array:
for(int i = 0; i<size;i++){
arrnew[i] = arr[i]; // copy the element old array to new array:
}
arrnew[n] = ele; // Appending the element:
}
by above simple method you can append the value
I am new to thrust (cuda) and I want to do some array operations but I don´t find any similar example on the internet.
I have following two arrays (2d):
a = { {1, 2, 3}, {4} }
b = { {5}, {6, 7} }
I want that thrust compute this array:
c = { {1, 2, 3, 5}, {1, 2, 3, 6, 7}, {1, 2, 3, 5}, {1, 2, 3, 6, 7} }
I know how it works in c/c++ but not how to say thrust to do it.
Here is my idea how it wohl maybe could work:
Thread 1:
Take a[0] -> expand it with b.
Write it to c.
Thread 2:
Take a[1] -> expand it with b.
Write it to c.
But I have no idea how to do that. I could write the array a and b to an 1d array like:
thrust::device_vector<int> dev_a;
dev_a.push_back(3); // size of first array
dev_a.push_back(1);
dev_a.push_back(2);
dev_a.push_back(3);
dev_a.push_back(1); // size of secound array
dev_a.push_back(4);
thrust::device_vector<int> dev_b;
dev_b.push_back(1); // size of first array
dev_b.push_back(5);
dev_b.push_back(2); // size of secound array
dev_b.push_back(6);
dev_b.push_back(7);
And the pseudo-function:
struct expand
{
__host__ __device__
?? ?? (const array ai, const array *b) {
for bi in b: // each array in the 2d array
{
c.push_back(bi[0] + ai[0]); // write down the array count
for i in ai: // each element in the ai array
c.push_back(i);
for i in bi: // each element in the bi array
c.push_back(i);
}
}
};
Anyone any idea?
I guess you won't get any speed increase on the GPU in such kind of operation since it needs a lot oo memory accesses - a slow operation on GPU.
But if you anyway want to implement this:
I guess, for the reason I wrote previously, trust won't help you with ready-to-use algorithm. This means that you need to write your own kernel, however, you can leave memory management to thust.
It is always faster to create arrays in CPU memory and, when ready, copy the whole array to GPU. (CPU<->GPU copies are faster on long continiuos pieces of data)
Keep in mind that GPU runs hundreds of threads in parallel. Each thread need to know what to read and where to write.
Global memory operations are slow (300-400 clocks). Avoid thread reading the whole array from global memory to find out that it needed only the last few bytes.
So, as I can see you program.
Make your arrays 1D in a CPU memory look like this:
float array1[] = { 1, 2, 3, 4};
float array2[] = { 5, 6, 7};
int arr1offsets[] = {0, 2, 3, 1}; // position of the first element and length of subarray pairs
int arr2offsets[] = {0, 1, 1, 2};
Copy your arrays and offsets to GPU and allocate memory for result and it's offsets. I guess, you'll have to count max length of one joint subarray and allocate memory for the worst case.
Run the kernel.
Collect the results
The kernel may look like this (If I correctly understood your idea)
__global__ void kernel(float* arr1, int* arr1offset,
float* arr2, int* arr2offset,
float* result, int* resultoffset)
{
int idx = threadIdx.x+ blockDim.x*blockIdx.x;
int a1beg = arr1offset[Idx*2];
int a2beg = arr2offset[Idx*2];
int a1len = arr1offset[Idx*2+1];
int a2len = arr2offset[Idx*2+1];
resultoffset[idx*2] = idx*MAX_SUBARRAY_LEN;
resultoffset[idx*2+1] = a1len+a2len;
for (int k = 0; k < a1len; ++k) result[idx*MAX_SUBARRAY_LEN+k] = arr1[a1beg+k];
for (int k = 0; k < a2len; ++k) result[idx*MAX_SUBARRAY_LEN+a1len+k] = arr2[a2beg+k];
}
This code is not perfect, but should do the right thing.
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
nul terminating a int array
I'm trying to print out all elements in an array:
int numbers[100] = {10, 9, 0, 3, 4};
printArray(numbers);
using this function:
void printArray(int array[]) {
int i=0;
while(array[i]!='\0') {
printf("%d ", array[i]);
i++;
}
printf("\n");
}
the problem is that of course C doesn't differentiate between just another zero element in the array and the end of the array, after which it's all 0 (also notated \0).
I'm aware that there's no difference grammatically between 0 and \0 so I was looking for a way or hack to achieve this:
10 9 0 3 4
instead of this
10 9
The array could also look like this: {0, 0, 0, 0} so of course the output still needs to be 0 0 0 0.
Any ideas?
Don't terminate an array with a value that could also be in the array.
You need to find a UNIQUE terminator.
Since you didn't indicate any negative numbers in your array, I recommend terminating with -1:
int numbers[100] = {10, 9, 0, 3, 4, -1};
If that doesn't work, consider: INT_MAX, or INT_MIN.
As a last resort, code a sequence of values that are guaranteed not to be in your array, such as: -1, -2, -3 which indicates the termination.
There is nothing "special" about terminating with 0 or \0. Terminate with whatever works for your case.
If your array truly can hold ALL values in ANY order, then a terminator isn't possible, and you will have to keep track of the length of the array.
From your example, this would look like:
int numbers[100] = {10, 9, 0, 3, 4};
int Count = 5;
int i;
for(i=0; i<Count; ++i)
{
// do something with numbers[i]
}
The typical ways to implement this are to:
define a sentinel value (which others have suggested)
define a struct with an int (the actual value) and a bool (indicating if it's the sentinel) and make an array of those instead
pass the length of the array with the array
define a struct that contains both the array and the length and pass that instead
Note that the first and second items are nearly identical.
For an array declared as
int numbers[100] = {10, 9, 0, 3, 4};
there's absolutely no way to distinguish the explicit 0 in the initializer from the implicit zeros used to initialize the tail portion of the array. They are the same zeros. So, what you want to do cannot be done literally.
The only way you can do it is to select some int value as a reserved dedicated terminator value and always add it at the end of the array explicitly. I.e. if you choose -42 as a terminator value, you'd have to declare it as
int numbers[100] = {10, 9, 0, 3, 4, -42};
and iterate up to the first -42 in your cycles.