与if / else语句一起使用

Question

I am creating a function that should add the sum of three cards for both the dealer and the player. The values should then be evaluated and what is given back from the function should be a recommendation, "safe", "stop", "black jack" or "busted". There will be two different functions evaluating the player´s cards and the dealer´s. Please look at the code I have written. I think I have figured it out codewise, I just have one question.

function printResult(p1, p2, p3, d1, d2, d3) {
  var x = p1 + p2 + p3; 
  var ab = "player:" + " " + player(x);
  var y = d1 + d2 + d3;
  var de = "dealer:" + " " + dealer(y);
  return ab + ", " + de;
 };

function player(p1, p2, p3) {
  var summa = p1 + p2 + p3
  var result;
  if (summa < 21)
  result = "safe";
  else if (summa = 21)
    result = "black jack";
  else
    result = "busted";

  return result;
}

function dealer(d1, d2, d3) {
  var summa = d1 + d2 + d3;
  var dealer;
  if (summa < 17)
  dealer =  "safe";
else if (summa >= 17 && summa <= 20)
  dealer = "stop";
else if (summa == 21)
  dealer = "black jack";
else 
  dealer = "busted";

return dealer; 
}

When I print printResult(4, 8, 1, 3, 8, 4) it gives me a faulty recommendation (the player gets busted when it should get safe).

However, if I change the code to the following, then it works.

function printResult(p1, p2, p3, d1, d2, d3) {
  var x = p1 + p2 + p3; 
  var ab = "player:" + " " + player(p1, p2, p3);
  var y = d1 + d2 + d3;
  var de = "dealer:" + " " + dealer(d1, d2, d3);
  return ab + ", " + de;
 };

I don´t understand why there is a difference in the result? Ps, in the latter piece of code I understand that I do not have to have x and y as variables, they can be removed.

Thanks in advance!

Answer 1

There is a difference:

var x = p1 + p2 + p3; 
var ab = "player: " + player(x);

You're calling player with one parameter, x , that contains the sum of your variables. This is valid code, but the other parameters will now be undefined . To illustrate, your function with actually look something like:

function player(p1, p2, p3) {
  var summa = p1 + p2 + p3 //p1 is x, p2 is undefined, p3 is undefined
  ...
  else
    result = "busted";
  return result;
}

Lets say x = 2 . When you try to add 2 + undefined + undefined JavaScript will return NaN (Not A Number) and this is what the variable summa will be set to. The only way out from that function is now through "busted" since the other conditions of your if-else are not met (only the else).

This can be solved by:

var summa = (p1 + p2 + p3) || (p1 + p2) || p1;

This basically means that you only add the parameters that were passed to the function, ie that exists.

Answer 2

Javascript allow you to not specify d2 & d3 parameters to your player function but if they aren't there they would have the value undefined and x + undefined + undefined is NaN (Special value for Not a Number)

If you want to allow both ways to call your function you can add a check at the top that convert undefined to 0 :

function player(p1, p2, p3) {
  p1 = p1 || 0;
  p2 = p2 || 0;
  p3 = p3 || 0;
  var summa = p1 + p2 + p3
  var result;
  if (summa < 21)
  result = "safe";
  else if (summa = 21)
    result = "black jack";
  else
    result = "busted";

  return result;
}

This syntax works because undefined is 'falsey' so the value of 0 will be used, 0 is also 'falsey' but it's not important because it will be replaced by 0 and all other values will be kept as they are 'truthy'