Hackerrank Solution - Variadic functions in C:
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Welcome back Guys!!
In this post we will solve Variadic Function in C Hackerrank problem.
It is a medium level problem using concept of Variadic Function ...
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What is variadic Function:
We typically use variadic function when we don’t know the total number of arguments that will be used for a function. Basically one single function could potentially have n number of arguments.
Problem Statement to Variadic functions in C hackerrank problem is as follows:
Variadic functions are functions which take a variable number of arguments. In C programming, a variadic function will contribute to the flexibility of the program that you are developing.
The declaration of a variadic function starts with the declaration of at least one named variable, and uses an ellipsis as the last parameter, e.g.
int printf(const char* format, ...);
In this problem, you will implement three variadic functions named , and to calculate sums, minima, maxima of a variable number of arguments. The first argument passed to the variadic function is the count of the number of arguments, which is followed by the arguments themselves.
Input Format
- The first line of the input consists of an integer .
- Each test case tests the logic of your code by sending a test implementation of 3, 5 and 10 elements respectively.
- You can test your code against sample/custom input.
- The error log prints the parameters which are passed to the test implementation. It also prints the sum, minimum element and maximum element corresponding to your code.
Constraints
.
Output Format
"Correct Answer" is printed corresponding to each correct execution of a test implementation."Wrong Answer" is printed otherwise.
Sample Input 0
1
Sample Output 0
Correct Answer
Correct Answer
Correct Answer
Solution code for Variadic functions in C hackerrank problem is as follows:
There is no built in way for a variadic function to know how many parameters have been passed. You don't have to use them all, but if you request too many, you will get garbage. The variadic functions in this problem use a standard method of passing the number of parameters to follow as the first parameter of the function. At least one fixed parameter must precede the elipsis so two problems are solved with variable.
So, how does one use the variable list of arguments? First declare a variable of type
va_list which is called ap in the Tester's Code below. Next initialize the list with it's length (count) as va_start(ap, count). Call the generator va_arg(ap, int) telling it source and data type to retrieve the next argument in the list. Finally, when the list has been processed, it is good form to use va_end(ap).
#include <stdarg.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MIN_ELEMENT 1
#define MAX_ELEMENT 1000000
int sum (int count,...) {
va_list ap;
int i, sum;
va_start (ap, count); /* Initialize the argument list. */
sum = 0;
for (i = 0; i < count; i++)
sum += va_arg (ap, int); /* Get the next argument value. */
va_end (ap); /* Clean up. */
return sum;
}
int min(int count,...) {
va_list ap;
int i,m,a[count];
va_start(ap,count);
for(i=0;i<count;i++)
{
a[i]=va_arg(ap,int);
}
m=a[0];
for(i=0;i<count;i++)
{
if(a[i]<m)
{
m=a[i];
}
}
return m;
}
int max(int count,...) {
va_list ap;
int i,m,a[count];
va_start(ap,count);
for(i=0;i<count;i++)
{
a[i]=va_arg(ap,int);
}
m=a[0];
for(i=0;i<count;i++)
{
if(a[i]>m)
{
m=a[i];
}
}
return m;
}
int test_implementations_by_sending_three_elements() {
srand(time(NULL));
int elements[3];
elements[0] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[1] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[2] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
fprintf(stderr, "Sending following three elements:\n");
for (int i = 0; i < 3; i++) {
fprintf(stderr, "%d\n", elements[i]);
}
int elements_sum = sum(3, elements[0], elements[1], elements[2]);
int minimum_element = min(3, elements[0], elements[1], elements[2]);
int maximum_element = max(3, elements[0], elements[1], elements[2]);
fprintf(stderr, "Your output is:\n");
fprintf(stderr, "Elements sum is %d\n", elements_sum);
fprintf(stderr, "Minimum element is %d\n", minimum_element);
fprintf(stderr, "Maximum element is %d\n\n", maximum_element);
int expected_elements_sum = 0;
for (int i = 0; i < 3; i++) {
if (elements[i] < minimum_element) {
return 0;
}
if (elements[i] > maximum_element) {
return 0;
}
expected_elements_sum += elements[i];
}
return elements_sum == expected_elements_sum;
}
int test_implementations_by_sending_five_elements() {
srand(time(NULL));
int elements[5];
elements[0] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[1] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[2] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[3] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[4] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
fprintf(stderr, "Sending following five elements:\n");
for (int i = 0; i < 5; i++) {
fprintf(stderr, "%d\n", elements[i]);
}
int elements_sum = sum(5, elements[0], elements[1], elements[2], elements[3], elements[4]);
int minimum_element = min(5, elements[0], elements[1], elements[2], elements[3], elements[4]);
int maximum_element = max(5, elements[0], elements[1], elements[2], elements[3], elements[4]);
fprintf(stderr, "Your output is:\n");
fprintf(stderr, "Elements sum is %d\n", elements_sum);
fprintf(stderr, "Minimum element is %d\n", minimum_element);
fprintf(stderr, "Maximum element is %d\n\n", maximum_element);
int expected_elements_sum = 0;
for (int i = 0; i < 5; i++) {
if (elements[i] < minimum_element) {
return 0;
}
if (elements[i] > maximum_element) {
return 0;
}
expected_elements_sum += elements[i];
}
return elements_sum == expected_elements_sum;
}
int test_implementations_by_sending_ten_elements() {
srand(time(NULL));
int elements[10];
elements[0] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[1] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[2] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[3] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[4] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[5] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[6] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[7] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[8] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[9] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
fprintf(stderr, "Sending following ten elements:\n");
for (int i = 0; i < 10; i++) {
fprintf(stderr, "%d\n", elements[i]);
}
int elements_sum = sum(10, elements[0], elements[1], elements[2], elements[3], elements[4],
elements[5], elements[6], elements[7], elements[8], elements[9]);
int minimum_element = min(10, elements[0], elements[1], elements[2], elements[3], elements[4],
elements[5], elements[6], elements[7], elements[8], elements[9]);
int maximum_element = max(10, elements[0], elements[1], elements[2], elements[3], elements[4],
elements[5], elements[6], elements[7], elements[8], elements[9]);
fprintf(stderr, "Your output is:\n");
fprintf(stderr, "Elements sum is %d\n", elements_sum);
fprintf(stderr, "Minimum element is %d\n", minimum_element);
fprintf(stderr, "Maximum element is %d\n\n", maximum_element);
int expected_elements_sum = 0;
for (int i = 0; i < 10; i++) {
if (elements[i] < minimum_element) {
return 0;
}
if (elements[i] > maximum_element) {
return 0;
}
expected_elements_sum += elements[i];
}
return elements_sum == expected_elements_sum;
}
int main ()
{
int number_of_test_cases;
scanf("%d", &number_of_test_cases);
while (number_of_test_cases--) {
if (test_implementations_by_sending_three_elements()) {
printf("Correct Answer\n");
} else {
printf("Wrong Answer\n");
}
if (test_implementations_by_sending_five_elements()) {
printf("Correct Answer\n");
} else {
printf("Wrong Answer\n");
}
if (test_implementations_by_sending_ten_elements()) {
printf("Correct Answer\n");
printf("Wrong Answer\n");
}
}
return 0;}
Please read the code and analysis it . It's a big one I know it will take time to grasp.
Feel free to leave your comments and doubts in the comment section below.
See you next time.
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Variadic functions in C – Hacker Rank Solution
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