英文:
Does my gofmt work wrongly or I don't understand something?
问题
我猜我的gofmt工作不像它应该的那样，我是对的吗？
原始文件：
package main
import "fmt"
func main() {
fmt.Printf("hello, world\n")
}
然后我执行了：
gofmt -r 'h -> H' -w "hello.go"
之后文件的内容：
package H
import "fmt"
func H() {
H
}
英文:
I suppose that my gofmt works not how it's supposed to, am I right ?
Original file:
package main
import "fmt"
func main() {
fmt.Printf("hello, world\n")
}
Then I did:
gofmt -r 'h -> H' -w "hello.go"
Content of the file after:
package H
import "fmt"
func H() {
H
}
答案1
得分: 5
据我理解，gofmt的工作方式与其作者预期的可能不同于你的期望。
文档中提到：
> 模式和替换都必须是有效的Go表达式。在模式中，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替换中的相同标识符。
由于模式中只有一个小写字母，它匹配所有子表达式，然后用H替换它们。让我们进一步看一个例子，考虑以下代码：
package main
import "fmt"
func compare(a, b int) {
if a + b < a * b {
fmt.Printf("hello, world\n")
}
}
经过相同的gofmt命令后，上述代码变成了：
package H
import "fmt"
func H(H, H H) {
if H+H < H*H {
H
}
}
如果这不是你想要的结果，那么你应该使用更具体的模式表达式。
英文:
Presumably gofmt works as its authors intended, which might be different from what you expected.
The documentation says:
> 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.
As you have only a single lowercase letter in the pattern, it matches all sub-expressions. And then replaces them with H. Let's take your example further, consider this:
package main
import "fmt"
func compare(a, b int) {
if a + b < a * b {
fmt.Printf("hello, world\n")
}
}
After the same gofmt command the above code becomes:
package H
import "fmt"
func H(H, H H) {
if H+H < H*H {
H
}
}
If this is not what you want, then you should use a more specific pattern expression.
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