Question

我与QUnit坐在一起,我 one了一件事。

我对 Chrome进行了这一简单检验:

deepEqual(new RegExp(), /(?:)/);

我假定它将通过,因为:new RegExp( 回归代码>/(?:)/ in the Development Console。 <代码>new RegExp() ==/(?)/ for RegExps,但toString(功能既是回报又是are<>。

我认为,字面/非物字句会起作用,但是,由于这一试样,情况并非如此:

deepEqual(new RegExp(" "), / /);

因此,在以下试验中,第一个试验失败:

test("test", function() { deepEqual(new RegExp(), /(?:)/); // fail deepEqual(new RegExp(" "), / /); // pass equal(new RegExp().toString(), /(?:)/.toString()); // pass });

Therefore, could someone point me in the right direction as to why the first test fails please?

Answer 1

www.un.org/Depts/DGACM/index_spanish.htm 简短回答: 。在字面上,它有<代码>(?)/,而就物体而言,它只是空洞。当您填写/和new RegExp(>时,source property is the same (两者都是用一个空间-character做的说明)。

。如果你看着Qunit的消息来源,你看着这个法典的范畴:

"regexp": function (b, a) {
    return QUnit.objectType(b) === "regexp" &&
        a.source === b.source && // the regex itself
        a.global === b.global && // and its modifers (gmi) ...
        a.ignoreCase === b.ignoreCase &&
        a.multiline === b.multiline;
};

你们可以看看,如何使用本法典来区分源参数(它只是将每一浮游动物的特性加以输出,并检验其平等):

function eq(x, y) {
   console.log("x.source:", " " + x.source + " ", "y.source:", " " + y.source + " ", "===:", x.source === y.source);
   console.log("x.global:", x.global, "y.global:", y.global, "===:", x.global === y.global);
   console.log("x.ignoreCase:", x.ignoreCase, "y.ignoreCase:", y.ignoreCase, "===:", x.ignoreCase === y.ignoreCase);
   console.log("x.multiline:", x.multiline, "y.multiline:", y.multiline, "===:", x.multiline === y.multiline);
}

When you call this with eq(/(?:)/, new RegExp());, you get:

x.source:  (?:)  y.source:    ===: false
x.global: false y.global: false ===: true
x.ignoreCase: false y.ignoreCase: false ===: true
x.multiline: false y.multiline: false ===: true

当你打电话到<条码>以下(///,新编号(“”)时,请上。

x.source:     y.source:     ===: true
x.global: false y.global: false ===: true
x.ignoreCase: false y.ignoreCase: false ===: true
x.multiline: false y.multiline: false ===: true

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