Kết quả của chương trình sẽ là gì?

The code defines two functions, `f` and `g`, and a `main` function. `f` takes an integer reference and an integer value, increments both, and returns their sum. Because x is passed as reference any changes to x in function f will affect the original variable in main. `g` takes two integer values, increments both, and returns their sum. Because the values are passed by value not reference changes will only occur within function g, and will not affect the original variables in main. The `main` function initializes two integers, `a` and `b`, prints their initial values, calls `f`, prints the updated values of `a` and `b`, calls `g`, and then prints the values of `a` and `b` again. The initial print statement will print `3, 3`. Function f increments a from 3 to 4. It increments b locally in f to 4. The returned value of 4+4=8 is printed to console. The next print statements show that a now equals 4 but b is still 3 because it was passed by value to f. Function g gets copies of a and b. It adds 2 to each internally and returns 6+5=11 which is then printed to console. The final print statements show that a and b are still 4 and 3 since g only had local copies of the variables. So, 3,3 8 4,3 11 4,3 is printed to console. Since no answer matches this, the correct answer is "Không có đáp án nào ở trên".