英文:
Return values of function as input arguments to another
问题
如果我有以下代码:
func returnIntAndString() (i int, s string) {...}
func doSomething(i int, s string) {...}
然后我可以成功执行以下操作:
doSomething(returnIntAndString())
然而,假设我想在doSomething
中添加另一个参数,像这样:
func doSomething(msg string, i int, s string) {...}
当我像这样调用它时,Go 在编译时会报错:
doSomething("message", returnIntAndString())
错误信息为:
main.go:45: multiple-value returnIntAndString() in single-value context
main.go:45: not enough arguments in call to doSomething()
有没有办法解决这个问题,或者我应该放弃,将returnIntAndString
的返回值赋值给一些引用,并像doSomething(msg, code, str)
这样传递消息和这些值?
英文:
If I have
func returnIntAndString() (i int, s string) {...}
And I have:
func doSomething(i int, s string) {...}
Then I can do the following successfully:
doSomething(returnIntAndString())
However, let's say I want to add another argument to doSomething like:
func doSomething(msg string, i int, s string) {...}
Go complains when compiling if I call it like:
doSomething("message", returnIntAndString())
With:
main.go:45: multiple-value returnIntAndString() in single-value context
main.go:45: not enough arguments in call to doSomething()
Is there a way to do this or should I just give up and assign the return values from returnIntAndString
to some references and pass msg and these values like doSomething(msg, code, str)
?
答案1
得分: 18
在规范中这里有详细描述。它要求内部函数对所有参数返回正确的类型。不允许在返回多个值的函数中使用额外的参数。
作为特例,如果函数或方法g的返回值数量与另一个函数或方法f的参数数量相等,并且每个返回值都可以分别赋值给f的参数,则调用f(g(g的参数))将按顺序将g的返回值绑定到f的参数上。调用f除了调用g之外不能包含其他参数,并且g必须至少有一个返回值。如果f有一个最终的...参数,则该参数将被赋予g的返回值,这些返回值在分配了常规参数之后仍然存在。
func Split(s string, pos int) (string, string) {
return s[0:pos], s[pos:]
}
func Join(s, t string) string {
return s + t
}
if Join(Split(value, len(value)/2)) != value {
log.Panic("test fails")
}
如果不满足这些特定条件,则需要分别赋值返回值并调用函数。
英文:
It's described here in the spec. It requires the inner function to return the correct types for all arguments. There is no allowance for extra parameters along with a function that returns multiple values.
> As a special case, if the return values of a function or method g are
> equal in number and individually assignable to the parameters of
> another function or method f, then the call f(g(parameters_of_g)) will
> invoke f after binding the return values of g to the parameters of f
> in order. The call of f must contain no parameters other than the call
> of g, and g must have at least one return value. If f has a final ...
> parameter, it is assigned the return values of g that remain after
> assignment of regular parameters.
>
> func Split(s string, pos int) (string, string) {
> return s[0:pos], s[pos:]
> }
>
> func Join(s, t string) string {
> return s + t
> }
>
> if Join(Split(value, len(value)/2)) != value {
> log.Panic("test fails")
> }
If those specific conditions are not met, then you need to assign the return values and call the function separately.
答案2
得分: 0
我有同样的问题。我能想到的最好的解决方案是创建我所需的额外参数的类型或结构,并为它们编写方法,就像这样:
package main
import (
"fmt"
)
type Message string
type MessageNumber struct {
Message string
Number int
}
func testfunc() (foo int, bar int) {
foo = 4
bar = 2
return
}
func (baz Message) testfunc2(foo int, bar int) {
fmt.Println(foo, bar, baz)
}
func (baz MessageNumber) testfunc3(foo int, bar int) {
fmt.Println(foo, bar, baz.Number, baz.Message)
}
func main() {
Message("the answer").testfunc2(testfunc())
MessageNumber{"what were we talking about again?", 0}.testfunc3(testfunc())
fmt.Println("Done. Have a day.")
}
输出结果如下:
4 2 the answer
4 2 0 what were we talking about again?
Done. Have a day.
英文:
I had the same question. The best solution I could come up with is creating types or structs for my desired extra parameters and writing methods for them like this:
package main
import (
"fmt"
)
type Message string
type MessageNumber struct {
Message string
Number int
}
func testfunc() (foo int, bar int) {
foo = 4
bar = 2
return
}
func (baz Message) testfunc2(foo int, bar int) {
fmt.Println(foo, bar, baz)
}
func (baz MessageNumber) testfunc3(foo int, bar int) {
fmt.Println(foo, bar, baz.Number, baz.Message)
}
func main() {
Message("the answer").testfunc2(testfunc())
MessageNumber{"what were we talking about again?", 0}.testfunc3(testfunc())
fmt.Println("Done. Have a day.")
}
The output looks like this:
user@Frodos-Atari-MEGA-STE:~/go/test$ go run main.go
4 2 the answer
4 2 0 what were we talking about again?
Done. Have a day.
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