C++ Bug probably scope problem

Applications Programming
1

Hello,
I have some difficulty to understand the scope of this program adapted from C++ book of D.S. Malik I am reading.
I put the header and the main program as attachment to save space. My problem is the output:

$ ./prog1009_die
Line 4: Not yet rolled, die1 gets default number: 1
Line 5: Not yet rolled, die2 gets default number: 1

Line 6: After first rolling die1: 4
Line 7: After first rolling die2: 5
Line 8: The sum of the numbers rolled by the dice is: 9

Line 6-2: After second rolling die1: 2
Line 7-2: After second rolling die2: 4

Line 9: After rolling again (third time), 
die1 (die1.roll()): 5
die2 (die2.roll()): 4
the sum of the numbers rolled [die1.getNum() + die2.getNum()] is: 8
the sum of the numbers rolled [die1.roll() + die2.roll()] is: 8

Line 9-2: After rolling again (third time)
die1 (die1.getNum()): 5
die2 (die2.getNum()): 4
the sum of the numbers rolled [die1.getNum() + die2.getNum()] is: 9

I was expecting Line 9 part would be the same result as Line 9-2 part, but the sum of the two numbers is not right. It is difficult for me to understand why. The problem may be related to die::roll() and die::getNum() functions in this case, but I am not sure.

int die::roll()
{
    num = rand() % 6 + 1;
    return num;
}

int die::getNum() const
{
    return num;
}

Also I am aware the rand() function here so each time the number will change. Can somebody elaborate it for me? Thanks a lot!

2

You rolled again for sum so the sum is for that roll. Do get for sum.

1 Like
3

Thanks!
Just to confirm: Did you mean this part in the source code?:

Line 9 ... ... <<"\nthe sum of the numbers rolled [die1.roll() + die2.roll()] is: " << die1.roll() + die2.roll() << endl;

causing roll() anther time?

Wait a minute! Yes, roll() was called again, then Line 9-2 the sum of the numbers rolled [die1.getNum() + die2.getNum()] should print the same as Line 9 the sum of the numbers rolled [die1.roll() + die2.roll()] , Right?
2) Does getNum() const; play some trick here?
I was confused with the scope of getNum() and roll() here. Thanks again!

4

You have a big long cout calling multiple things which have side-effects on each other, and the order they are done is ambiguous -- remember this is a compiled, heavily optimized language, not a one-at-a-time string interpreter. Re-imagine your code like this:

myfunction(die1.getnum(), die2.getnum(), die1.roll(), die2.roll());

When written that way, whether roll() gets called before or after getnum() is more obviously ambiguous.

Split it into two cout statements to avoid the side-effects.

It has nothing to do with the const. It just means that calling that function has no side-effects. It doesn't guarantee that all other functions have no side effects however.

1 Like
5

Almost got your point. I re-wrote the code as:

cout << "Line 9: After rolling again (third time), ";
cout << "\ndie1 (die1.roll()): " << die1.roll();    //Line 9 - a1
cout << "\ndie2 (die2.roll()): " << die2.roll();    //Line 9 - a2
cout << "\nthe sum of the numbers rolled [die1.getNum() + die2.getNum()] is: " << die1.getNum() + die2.getNum();         //Line 9 - c1
cout << "\nthe sum of the numbers rolled [die1.roll() + die2.roll()] is: " << die1.roll() + die2.roll() << endl;          //Line 9 - c2

Line 9-c1 die1.getNum() + die2.getNum() was from previous rolling.
At Line 9-c2 die1.roll() and die2.roll() gets rolled again when they are combined together as die1.roll() + die2.roll() . That caused the confusion.
In turn, Line 9-c2 prints out the same as next Line 9-2 because die1.getNum()+ die2.getNum() are from the roll() of Line 9-c2.

Without your suggestion, I could not figure this out.
Thanks again!

Yifang

6

If roll() returns void then people will not be confused when rolling and when examining with get(). Compiled code will be same or smaller and faster.