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What will be the output of the following code?
#include <iostream>
int myFunction(int x) {
    return x * x;
}
int myFunction(int& x) {
    return ++x;
}
int main() {
    int value = 5;
    int result = myFunction(value);
    std::cout << result << std::endl;
    return 0;
}
  • a)
    25
  • b)
    6
  • c)
    Compilation error
  • d)
    Undefined behavior
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
What will be the output of the following code?#include <iostream>...
The code defines two overloaded functions named myFunction. The first myFunction takes an integer x and returns its square. The second myFunction takes an integer reference x and increments its value. In main, an integer variable value is declared and assigned a value of 5. The first myFunction is called with value as an argument, resulting in a return value of 25. The value 25 is then printed.
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Community Answer
What will be the output of the following code?#include <iostream>...
Explanation:

Function Overloading:
- In the given code, two functions with the same name `myFunction` are defined but with different parameters (one takes an integer and the other takes an integer reference).
- This is an example of function overloading in C++, where multiple functions can have the same name but different parameters.

myFunction(int x):
- When `myFunction(int x)` is called with the value `5`, it returns the square of `5`, which is `25`.

Output:
- Therefore, the output of the code will be 25.
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What will be the output of the following code?#include <iostream>int myFunction(int x) { return x * x;}int myFunction(int& x) { return ++x;}int main() { int value = 5; int result = myFunction(value); std::cout << result << std::endl; return 0;}a)25b)6c)Compilation errord)Undefined behaviorCorrect answer is option 'A'. Can you explain this answer?
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