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Consider the following code snippet:
#include <iostream>
void incrementValue(int& num) {
    num++;
}
int main() {
    int x = 5;
    incrementValue(x);
    std::cout << x;
    return 0;
}
What will be the output of the above code?
  • a)
    5
  • b)
    6
  • c)
    Error
  • d)
    Undefined behavior
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Consider the following code snippet:#include <iostream>void incr...
The 'incrementValue' function takes an integer reference as a parameter and increments its value by 1. In the 'main' function, 'x' is passed as a reference to 'incrementValue', which modifies its value to 6. Therefore, the output will be 6.
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Community Answer
Consider the following code snippet:#include <iostream>void incr...
Explanation:

Passing arguments by reference:
- In the given code snippet, the function `incrementValue` takes an integer reference as an argument.
- This means that any changes made to the reference inside the function will affect the actual variable passed to the function.

Incrementing the value:
- Inside the `incrementValue` function, the value of `num` (which is a reference to `x` in the `main` function) is incremented by 1.

Output:
- After calling the `incrementValue` function with `x` as the argument, the value of `x` in the `main` function will be updated to 6.
- Therefore, the output of the code will be 6.
Therefore, the correct answer is option 'B) 6'.
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Consider the following code snippet:#include <iostream>void incrementValue(int& num) { num++;}int main() { int x = 5; incrementValue(x); std::cout << x; return 0;}What will be the output of the above code?a)5b)6c)Errord)Undefined behaviorCorrect answer is option 'B'. Can you explain this answer?
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