Select all the correct answersPredict the output and give your views for the given code snippet in c#include<stdio.h>void Function(int arr[], int n) { int i; for (i = 0; i < n; i++) printf("%d ", arr[i]);}int main() { int arr[5] = {0}; Function(arr, 5); return 0;}OptionsIt will print 0 five timesIt will print 0 infinite timesThe output of the code is compiler dependentint arr[5] = {0}; statement throws and errorIt will print 0 only once and rest 4 times it will print some garbage value
Question
Select all the correct answersPredict the output and give your views for the given code snippet in c#include<stdio.h>void Function(int arr[], int n) { int i; for (i = 0; i < n; i++) printf("%d ", arr[i]);}int main() { int arr[5] = {0}; Function(arr, 5); return 0;}OptionsIt will print 0 five timesIt will print 0 infinite timesThe output of the code is compiler dependentint arr[5] = {0}; statement throws and errorIt will print 0 only once and rest 4 times it will print some garbage value
Solution
The correct answer is "It will print 0 five times".
Here's the step by step explanation:
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The code defines a function named "Function" which takes an array and its size as arguments. It then loops over the array and prints each element.
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In the main function, an array of size 5 is declared and initialized with all elements as 0.
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The "Function" is then called with this array and its size as arguments.
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As the array contains only 0s, the function will print 0 five times, once for each element in the array.
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The statement "int arr[5] = {0};" does not throw an error. It is a valid way to declare and initialize an array in C.
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The output of the code is not compiler dependent. It will always print 0 five times regardless of the compiler used.
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The code will not print 0 infinite times. The loop in the "Function" runs exactly n times, where n is the size of the array passed as an argument.
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The code will not print any garbage values as all elements in the array are initialized to 0.
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