英文:
Using the gofmt refactoring tool to rename a global variable
问题
我正在尝试使用gofmt工具根据这篇博客文章来重构基于Go的代码，我有一个简单的示例：
package main
import (
"fmt"
)
var v = 12
func main() {
fmt.Println(v)
}
我试图将变量v重命名为m，并应用以下命令：
gofmt -r 'v -> m' -w main.go
重构后的代码看起来是这样的（有问题）：
package m
import (
"fmt"
)
var m = m
func m() {
m
}
我在这里漏掉了什么？
英文:
I'm experimenting the gofmt tool capabilities for refactoring [tag:go] code based on this blog post, I have this trivial example:
package main
import (
"fmt"
)
var v = 12
func main() {
fmt.Println(v)
}
I'm trying to rename the v variable to mapplying this recipe:
gofmt -r 'v -> m' -w main.go
The code after the refactoring looks (broken) like:
package m
import (
"fmt"
)
var m = m
func m() {
m
}
What am I missing here?
答案1
得分: 17
你尝试的方法存在问题，gofmt手册中有说明：
使用-r标志指定的重写规则必须是以下形式的字符串：
pattern -> replacementpattern和replacement都必须是有效的Go表达式。在pattern中，单个小写字母标识符用作通配符，匹配任意子表达式；这些表达式将被替换为replacement中的相同标识符。
（已添加强调）
如果你有var vee = 12并使用-r vee -> foo，一切都会正常。但是，使用v -> m时，v -> m会匹配每个Go表达式，将其标识为v并用m替换。
英文:
There is a problem with what you're trying, the gofmt manual states:
>The rewrite rule specified with the -r flag must be a string of the form:
>
> pattern -> replacement
>
>Both pattern and replacement must be valid Go expressions. In the pattern, single-character lowercase >identifiers serve as wildcards matching arbitrary sub-expressions; those expressions will be substituted for the same identifiers in the replacement.
(highlighting added)
If you had var vee = 12 and used -r vee -> foo everything would be fine. With v -> m however,
v -> m matches every Go expression, identifies it as v and replaces it by m.
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