10

我有以下有效的功能

function sum ()
{
    var total = 0,
        num = 0,
        numArgs = arguments.length;

    if (numArgs === 0) {
        throw new Error("Arguments Expected");
    }

    for(var c = 0; c < numArgs; c += 1) {
        num = arguments[c];
        if (typeof(num) !== "number") {
            throw new Error("Only number are allowed but found", typeof (num));
        }
        total += num;

    }

    return total;

}


sum(2, "str"); // Error: Only number are allowed but found "string"

茉莉花规格文件如下:

describe("First test; example specification", function () {
    it("should be able to add 1 + 2", function (){
        var add = sum(1, 2);
        expect(add).toEqual(3);
    });
    it("Second Test; should be able to catch the excption 1 +'s'", function (){
        var add = sum(1, "asd");
        expect(add).toThrow(new Error("Only number are allowed but found", typeof("asd")));
    });
});

第一个测试效果很好,第二个测试失败了。
我应该如何处理中的预期错误Jasmine

4

1 回答 1

16

正如这个问题中所讨论的,您的代码不起作用,因为您应该传递一个函数对象来期望,而不是调用 fn() 的结果

    it("should be able to catch the excption 1 +'s'", function (){
//        var add = sum(1, "asd");
        expect(function () {
            sum(1, "asd");
        }).toThrow(new Error("Only number are allowed but found", typeof ("asd")));
    });
于 2012-04-19T17:17:35.503 回答