“进行mod整洁”与“进行构建”的行为

时间:2019-08-21 11:56:38

标签: go go-modules

假设我有一个具有以下结构的项目:

+-- app/
+-- otherstuff/
+-- test/
+-- go.mod
+-- go.sum
+-- main.go

通过运行go.mod,可以确保go mod tidy不包含未使用的依赖项:

# 1) Verify that no dependency containing the name "modern-go" is found on go.mod
$ grep 'modern-go' go.mod
<-- Empty (nothing found) (OK)


# 2) Run "go mod tidy", verify that nothing changes (i.e. go.mod is already clean)
$ go mod tidy -v
unused github.com/modern-go/concurrent
unused github.com/modern-go/reflect2
<-- messages above are displayed, but go.mod did not change


# 3) Verify that go.mod did not change
$ grep 'modern-go' go.mod
<-- Empty (nothing found) (OK)

现在,如果我运行go buildgo.mod得到更新:

# 4) Run "go build"
$ go build


# 5) go.mod was updated by "go build":
$ grep 'modern-go' go.mod
    github.com/modern-go/concurrent v0.0.0-20180306012644-bacd9c7ef1dd // indirect
    github.com/modern-go/reflect2 v1.0.1 // indirect

我不知道这是怎么回事。我希望go mod tidy会保持go.mod处于干净状态,因此运行go build不会改变它。

有什么想法吗?

1 个答案:

答案 0 :(得分:0)

您期望的行为通常是正确的。

Go 1.13在此区域中进行了一些修复。我建议尝试一下:

https://twitter.com/golang/status/1164276194524762113