英文:
How to return a value from if-statement into a function golang
问题
我在从Python转到Golang后遇到了一个问题。在Python中，如果在函数内部的if语句中声明了一个变量，那么该变量对于函数/方法来说是可见的，只要该语句在函数内部。
当我尝试在Golang中做类似的事情时，我无法在if语句之外使用该变量。
这个错误（"Unresolved reference '...'"）将在IDE中出现，并且代码将无法编译。
我的问题是，为什么会发生这种情况？
英文:
I had a problem when I started to use Golang after Python. In Python, a variable that is declared inside an if-statement will be visible for a function/method if the statement is inside the function.
from pydantic import BaseModel
class sometype(BaseModel):
"""
A model describes new data structure which will be used
"""
sometype1: str
def someaction(somedata:sometype):
"""
do some action
:param somedata: a sometype instance
:return:
"""
print("%s" % somedata.sometype1 )
def somefunc(somedata:int, somebool:bool, anydata:sometype):
"""
It is a function
:param somedata: some random int
:param somebool: thing that should be True (else there will be an error)
:param anydata: sometype instance
:return:
"""
if somebool==True:
somenewdata=anydata
someaction(somenewdata)
if __name__=="__main__":
print("some")
thedata :sometype = sometype(sometype1="stringtypedata")
somefunc(1, True, thedata)
An IDE only can warn you ("Local variable '...' might be referenced before assignment") that this could not be referenced in some cases (to be precise - there will be no variable named "somenewdata" if the "somebool" be False).
When I tried to do something similar in Go - I couldn't use the variable outside if-statement.
// main package for demo
package main
import "fmt"
//sometype organizes dataflow
type sometype struct {
sometype1 string
}
//someaction does action
func someaction(somedata sometype) {
fmt.Printf("%v", somedata)
}
//somefunc is a function
func somefunc(somedata int, somebool bool, anydata sometype) {
if somebool == true {
somenewdata = anydata
}
someaction(somenewdata)
}
func main() {
fmt.Println("some")
thedata := sometype{"stringtype"}
somefunc(1, true, thedata)
}
This error ("Unresolved reference "..."") will appear in IDE and the code will not compile.
My question was - why does that happening?
答案1
得分: -3
我在这个问题上遇到了困难，因为我没有意识到在 if 函数内部使用的变量对于函数来说是不可见的。
答案很简单 - 在这种情况下，你不需要返回值，只需要填充值即可。所以，在函数内部的 if 语句之前引入它，它将对函数和语句都可见。
// 用于演示的主要包
package main
import "fmt"
// sometype 组织数据流
type sometype struct {
sometype1 string
}
// someaction 执行动作
func someaction(somedata sometype) {
fmt.Printf("%v", somedata)
}
// somefunc 是一个函数
func somefunc(somedata int, somebool bool, anydata sometype) {
// 引入变量
var somenewdata sometype
if somebool == true {
// 用数据填充变量
somenewdata = anydata
}
someaction(somenewdata)
}
func main() {
fmt.Println("some")
thedata := sometype{"stringtype"}
somefunc(1, true, thedata)
}
希望对你有帮助！
英文:
I struggled with this problem as I didn't realise that it was implied that the variable which is used inside if-function is not visible for a function.
The answer is simple - you don't need to return the value in this case as you should just fill the value. So, you need to introduce it before the if-statement inside the function, and it will be visible for both: the function and the statement.
// main package for demo
package main
import "fmt"
//sometype organizes dataflow
type sometype struct {
sometype1 string
}
//someaction does action
func someaction(somedata sometype) {
fmt.Printf("%v", somedata)
}
//somefunc is a function
func somefunc(somedata int, somebool bool, anydata sometype) {
//introduce the variable
var somenewdata sometype
if somebool == true {
//fill the variable with data
somenewdata = anydata
}
someaction(somenewdata)
}
func main() {
fmt.Println("some")
thedata := sometype{"stringtype"}
somefunc(1, true, thedata)
}
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