如果缺少其余参数，为什么此函数不会引发错误？
showStatistics("Mark Teixeira", "New York Yankees", "1st Base");
这是定义的函数：
function showStatistics(name, team, position, average, homeruns, rbi) {
document.write("<p><strong>Name:</strong> " + arguments[0] + "<br />");
document.write("<strong>Team:</strong> " + arguments[1] + "<br />");
if (typeof arguments[2] === "string") {
document.write("<strong>Position:</strong> " + position + "<br />");
}
if (typeof arguments[3] === "number") {
document.write("<strong>Batting Average:</strong> " + average + "<br />");
}
if (typeof arguments[4] === "number") {
document.write("<strong>Home Runs:</strong> " + homeruns + "<br />");
}
if (typeof arguments[5] === "number") {
document.write("<strong>Runs Batted In:</strong> " + rbi + "</p>");
}
}
任何未传递的参数将undefined在函数内部显示。在JavaScript中，没有方法重载。
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