javascript - 在这个简单的代码中，为什么函数内部的提示值被忽略了？

Question

在下面的代码中:

我也尝试过传递whatThrow参数。我似乎没有发现什么问题。

var ranThrow = ["empty", "rock", "paper", "scissors"];
var ranNum = Math.random();
var postRanNum = (ranNum * 3) + 1;
var roundPostRanNum = Math.floor(postRanNum);
var compThrow = ranThrow[roundPostRanNum];
var whatThrow = prompt("Rock, Papper or Scissors?", "rock");

var rpsGame = function () {

  if (whatThrow === "rock" && compThrow === "rock")  {
      return "You tie!";
  }
  else if (whatThrow === "rock" && compThrow === "paper") {
      return "You lose!";
  }
  else if (whatThrow === "rock" && compThrow === "scissor") {
      return "You win!";
  }
  else {
      return "Error";
  }
};
rpsGame();

console.log("The computer threw" + " " + compThrow);

Answer 1

该值在函数内部不会被忽略，而是调用函数的代码会忽略函数返回的内容。

如果显示返回值，您将看到该函数使用该值:

console.log(rpsGame());