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What will be the output of the following C++ code?
#include<iostream>
using namespace std;
int &fun()
{
static int x = 10;
return x;
}
int main()
{
fun() = 30;
cout << fun();
return 0;
}
using namespace std;
int &fun()
{
static int x = 10;
return x;
}
int main()
{
fun() = 30;
cout << fun();
return 0;
}
- a)30
- b)10
- c)Error
- d)Segmentation fault
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
What will be the output of the following C++ code?#include<iostream...
A function returning value by reference can be used as lvalue i.e. it can be used on the left side of an expression. Here when we are doing fun() = 30 then we are changing the value of x(i.e. value returning) to and as x is static therefore it will not be initialized again so the value of x becomes 30 hence the output is 30.
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Community Answer
What will be the output of the following C++ code?#include<iostream...
Explanation:
Static Variable:
- The function `fun` returns a reference to a static variable `x`.
- Static variables are initialized only once and retain their value across function calls.
Main Function:
- In the `main` function, we assign the value `30` to the variable `x` by calling the function `fun()`.
- The statement `fun() = 30;` modifies the value of `x`.
Output:
- The output of the code will be `30` because we modified the value of the static variable `x` to `30` in the `main` function.
Therefore, the correct answer is option 'A) 30'.
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What will be the output of the following C++ code?#include<iostream>using namespace std;int &fun(){ static int x = 10; return x;}int main(){ fun() = 30; cout << fun(); return 0;}a)30b)10c)Errord)Segmentation faultCorrect answer is option 'A'. Can you explain this answer?
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