如何在对模块进行单元测试时模拟导入的子例程

Question

考虑一个模块,它导出一个连接到Internet并返回结果的子例程:

unit module A;

sub download is export {
     "result from internet"  # Not the actual implementation, obviously.
}

另一个导入并调用该子例程的模块:

use A;  # imports &download into this lexical scope
unit module B;

sub do-something is export {
     download().uc ~ "!!"   # Does something which involves calling &download
}

现在我想为模块编写单元测试B.
但我不希望测试真正连接到互联网; 我希望他们使用download由我的测试脚本控制的子程序的模拟版本:

use Test;
plan 2;

use B;

my $mock-result;
my &mock-download = -> { $mock-result }

# ...Here goes magic code that installs &mock-download
# as &download in B's lexical scope...

$mock-result = "fake result";
is do-something(), "FAKE RESULT!!", "do-something works - 1";

$mock-result = "foobar";
is do-something(), "FOOBAR!!", "do-something works - 2";

问题是缺少魔法代码来覆盖子download...

在Perl 5中,我认为使用glob赋值可以很容易地实现这一点,或者在Sub :: Override或Test :: MockModule的帮助下更好.

但是在Perl 6中,模块的词法范围在B完成编译后关闭,因此在测试脚本运行时不再能够修改(如果我错了,请纠正我).所以这种方法似乎不可行.

那么如何在Perl 6中解决这个任务呢？
即如何编写单元测试B::do-something,而不让它调用真实A::download？

Answer 1

最简单的方法可能是使用https://docs.perl6.org/language/functions#Routines中wrap描述的方法,但前提条件是阻止内联的pragma.您需要在模块A中:use soft;use soft;

unit module A;
use soft;

sub download is export {
    "result from internet";
}

模块B:

unit module B;
use A;

sub do-something is export {
    download.uc ~ "!!";
}

和测试脚本:

use Test;
use A;
use B;

&download.wrap({
    "mock result";
});

is do-something, "MOCK RESULT!!", "mock a 'use'd sub";
# ok 1 - mock a 'use'd sub