javascript - 为什么此代码未引发错误？

这是我的代码(Javascript):

function getNumbers(){
    var numberString = document.getElementById("numbers").value;
    var actualNumbers = [];
    var flowIndex = 0;
    for(let i = 0; i < numberString.length; i++){
        if(numberString.charAt(i) != " " && numberString.charAt(i) != "\," && numberString.charAt(i) != "\t"){
            actualNumbers[flowIndex] = parseInt(numberString.charAt(i));
            flowIndex++;
        }
        else continue;
    }
    return actualNumbers;
}

function division(){
    try{
        var answer = getNumbers()[0] / getNumbers()[1];
        if(getNumbers()[1] == 0)
            throw "bad";
    }
    catch(error){
        throw error.description + "Division by zero error";
    }
    finally{
        return answer;
    }
}

我有一个函数getNumbers()，它返回一个包含array[0] = 1和array[1] = 0的数组。现在，我想在array[1] == 0时抛出一个异常“bad”。但是，既没有引发try异常，也没有引发catch异常，但是finally子句正在运行。问题是什么？

注意:除以零时，不会引发异常，而是答案是Infinity。 getNumbers()正常工作。

最佳答案

异常被抛出，但是您通过执行以下操作来抑制该异常:

finally {
    return answer;
}

finally子句拥有最终决定权。如果从中使用return，则可以抑制该异常并使该函数正常完成。

解决该问题的一种方法是删除finally子句，然后将return answer;放入try中。

只是FWIW，还有一些其他注释作为注释:

function getNumbers(){
    var numberString = document.getElementById("numbers").value;
    var actualNumbers = [];
    var flowIndex = 0;

    // You might consider splitting the string into an array of one-character strings so you
    // aren't constantly calling a method (`charAt`), like this:
    //     `var chars = numberString.split("");`
    // Then index into `chars`

    // N.B. `let` is an ES2015 (ES6) feature not all JavaScript engines have as it's new;
    // the rest of your code is using the older `var`
    // --v
    for (let i = 0; i < numberString.length; i++){
        // No need to escape the comma --------------------------------v
        if(numberString.charAt(i) != " " && numberString.charAt(i) != "\," && numberString.charAt(i) != "\t"){
            actualNumbers[flowIndex] = parseInt(numberString.charAt(i));
            flowIndex++;
        }
        // No need for the `else continue;` at all
        else continue;
        // In the above, you regularly call `charAt` four times when once would have been sufficient.
        // You might also consider a `switch`
    }
    return actualNumbers;
}

function division(){
    try{
        // Rather than calling `getNumbers` three separate times, call it once and remember its return value
        // doing the calculation should be AFTER checking [1] for 0, not before
        var answer = getNumbers()[0] / getNumbers()[1];
        if(getNumbers()[1] == 0)
            throw "bad";                    // Recommend using Error, e.g.: `throw new Error("bad")`
        // Move the `return answer;` here
    }
    catch(error){
        // You've thrown a string, it doesn't have a `description` property
        // Separately: Why throw something above, just to catch it here and throw something else?
        throw error.description + "Division by zero error";
    }
    // Remove the finally
    finally{
        return answer;
    }
}

还是FWIW，我可能要负责从input(甚至在调用division的东西中)而不是division中从getNumbers获取值，并使用/\d/.test(...)来测试字符是否为数字，因为有很多不是" "，","或"\t"的非数字。并且一旦我们知道它们是数字，就可以使用+ch而不是parseInt进行转换(但是使用此输入，这只是一种样式选择[嗯，有性能暗示，但是99.99％的时间都没关系])。

所以也许:

function getNumbers(str) {
    var numbers = [];
    str.split("").forEach(function(ch) {
        if (/\d/.test(ch)) {
            numbers.push(+ch);
        }
    });
    return numbers;
}

function division() {
    var numbers = getNumbers(document.getElementById("numbers").value);
    if (numbers[1] == 0) {
        throw new Error("Division by zero error");
    }
    return numbers[0] / numbers[1];
}

或者使用更简洁的备用getNumbers，但通过输入进行更多循环(通常无关紧要):

function getNumbers(str) {
    return str.split("")
        .filter(function(ch) { return /\d/.test(ch); })
        .map(function(ch) { return +ch; });
}

function division() {
    var numbers = getNumbers(document.getElementById("numbers").value);
    if (numbers[1] == 0) {
        throw new Error("Division by zero error");
    }
    return numbers[0] / numbers[1];
}