NSpec和多个befores

Question

我正在使用NSpec，我对前面的示例感到困惑：

void they_are_loud_and_emphatic()
{
    //act runs after all the befores, and before each spec
    //declares a common act (arrange, act, assert) for all subcontexts
    act = () => sound = sound.ToUpper() + "!!!";
    context["given bam"] = () =>
    {
        before = () => sound = "bam";
        it["should be BAM!!!"] = 
            () => sound.should_be("BAM!!!");
    };
}
string sound;

它可以工作，但当我进行下一次更改时：

void they_are_loud_and_emphatic()
{
    //act runs after all the befores, and before each spec
    //declares a common act (arrange, act, assert) for all subcontexts
    act = () => sound = sound.ToUpper() + "!!!";
    context["given bam"] = () =>
    {
        before = () => sound = "b";
        before = () => sound += "a";
        before = () => sound += "m";
        it["should be BAM!!!"] = 
            () => sound.should_be("BAM!!!");
    };
}
string sound;

字符串声音只有“M！”。当我调试代码时，它只调用前面的最后一个。也许我不理解这个理论，但我相信所有的lambdas都在'act‘和'it’之前运行。出什么问题了？

Answer 1

我使用下面的语法和工作(方法之前是外部的，上下文中是内部的)：

    void they_are_loud_and_emphatic()
    {
        act = () => sound = sound.ToUpper() + "!!!";
        context["given bam"] = () =>
        {
            before = () =>
            {
                sound = "b";
                sound += "a";
                sound += "m";
            };

            it["should be BAM!!!"] = () => sound.should_be("BAM!!!");
        };
    }

    string sound;

Answer 2

即使在前面的例子中它是递增的，但是对于每个规范再次运行之前的操作将会停止。

void they_are_loud_and_emphatic(){
act = () => sound = sound.ToUpper() + "!!!";
 context["given bam"] = () =>
{
before = () => sound = "b";   //now sound is B!!!
before = () => sound += "a";  //now sound is A!!!
before = () => sound += "m";  //now sound is M!!!
it["should be BAM!!!"] = 
() => sound.should_be("BAM!!!");  // when this line is runing ,sound is"M!!!"
};
}
string sound;