[英]Measuring time in C
我正在為編程課程編寫實驗作業。 我需要在C中執行:)。 我選擇的任務如下所示:我在C中創建了三種數據類型:鏈表,二進制樹和avl樹。 所有數據處理5,000,000個元素。 我需要測量在特定步驟中查找和未找到元素所花費的時間。 對於陽性和陰性結果,我都有固定的值。
這是我的一些代碼,其中包含用於創建5,000,000個數字的數組的功能和用於二叉樹添加和查找節點的兩個功能。 我希望代碼很清楚。
#include <stdio.h>
#include <stdlib.h>
#define SIZE 5000000
typedef struct Node {
int data;
struct Node *left, *right;
} Node;
void evenArrayGen(int num, int even_nums[]);
void shuffle(int numbers[]);
void addNode(Node **root, int value);
Node* findNode(Node *root, int value);
/** Recursively adds given value to the tree
*
* @param root
* @param value
*
*/
void addNode(Node **root, int value) {
Node *temp;
///check statement to avoid infinite loop in adding
if(*root == NULL) {
temp = (Node*)malloc(sizeof(Node));
temp->data = value;
temp->left = temp->right = NULL;
*root = temp;
return;
}
///recursively check where to add node
///if smaller than the root add to the left
///if bigger than the root add to the right
else {
temp = *root;
if(value < temp->data) {
addNode(&(temp->left), value);
} else {
addNode(&(temp->right), value);
}
return;
}
}
/** Recursively searches for Node with given value
*
* @param root
* @param value
* @return Node or NULL (if Node was not found)
*
*/
Node* findNode(Node *root, int value) {
Node *temp;
///check statement to avoid infinite loop in searching
///if it reachese that point given value is not in the tree
if(root == NULL) {
// printf("Given value is not in the tree\n");
return NULL;
}
temp = root;
if(temp->data == value) {
// printf("Searched value is: %i\n", temp->data);
return temp;
} else if(value < temp->data) {
findNode(temp->left, value);
} else {
findNode(temp->right, value);
}
}
/** Generates array of ordered even numbers from 2
*
* @param num number of uniqe even digits
* @param array of size equal to the number
*
*/
void evenArrayGen(int num, int even_nums[]) {
int i, current;
i = current = 0;
for(; i<num; i++) {
even_nums[i] = current += 2;
}
return;
}
/** Shuffle the array in random order. Radomly gets the index between 0 and
* current last index. Swaps number at random index with last one and
* decreses the current_last index by 1.
*
* @param array of numbers to be shuffled
*
*/
void shuffle(int nums[]) {
int i, len, current_last, index, temp;
///set the current_last to length of the array to track index for
///swaping nums
current_last = len = SIZE;
for (i=0; i<len; i++) {
srand(time(NULL));
index = rand()%(current_last);
temp = nums[index];
nums[index] = nums[current_last];
nums[current_last] = temp;
current_last--;
}
return;
}
int main() {
//initialize root for the tree
Node *root;
//intilialize array of 5,000,000 elements, and scores
static int nums[SIZE];
int i; //iterator
//initialize positive and negative numbers for find method
int positive_num, negative_num;
//initilaize timer
clock_t start_for_adding, start_for_finding;
//add elements to the array
evenArrayGen(SIZE, nums);
shuffle(nums);
//set positive number to one of the first 5000 elements
positive_num = nums[3222];
negative_num = 345; //all numbers in num are even so negative need to be odd
root = NULL; //set the root Node to NULL
start_for_adding = clock(); //zero the timer
//now iterate trough all elements in nums array and add each to
//the binary tree
for(i=0; i<SIZE; i++) {
//check if i reached proper value
if (i == 5000 || i == 50000 || i == 500000 || i == (5000000-1)) {
//add the adding time
printf("\nIndex: %d\n", i);
printf("Adding: %.5f\n", (double) clock() - start_for_adding);
start_for_finding = clock(); //zero the timer
//search for positive num
findNode(root, positive_num);
printf("Finding positive: %.5f\n",
(double) clock() - start_for_finding);
start_for_finding = clock(); //zero the timer
//search for negative num
findNode(root, negative_num);
printf("Finding negative: %.5f\n",
(double) clock() - start_for_finding);
start_for_adding = clock(); //reset the timer for adding
}
addNode(&root, nums[i]);
}
return;
}
查找元素的時間都指向零(在兩種情況下)
我很迷茫,不知道現在應該去哪里(我的任務最簡單的部分顯示這是最困難的部分...)
clock
的分辨率很可能會粗略地衡量findNode
調用findNode
函數所花費的時間。 通常,對時間進行測量以多次執行此類呼叫,然后將該時間除以執行呼叫的次數。
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