变量隐藏Accidental Variable Shadowing

6.5 — Variable shadowing (name hiding) | Learn C++ https://www.learncpp.com/cpp-tutorial/variable-shadowing-name-hiding/

Go的50度灰：Golang新开发者要注意的陷阱和常见错误 | 鸟窝 https://colobu.com/2015/09/07/gotchas-and-common-mistakes-in-go-golang/

6.5 — Variable shadowing (name hiding)

BY ALEX ON JANUARY 3RD, 2020 | LAST MODIFIED BY NASCARDRIVER ON APRIL 25TH, 2020

Each block defines its own scope region. So what happens when we have a variable inside a nested block that has the same name as a variable in an outer block? When this happens, the nested variable “hides” the outer variable in areas where they are both in scope. This is called name hiding or shadowing.

Shadowing of local variables

#include <iostream>

int main()
{ // outer block
    int apples { 5 }; // here's the outer block apples

    { // nested block
        // apples refers to outer block apples here
        std::cout << apples << '\n'; // print value of outer block apples

        int apples{ 0 }; // define apples in the scope of the nested block

        // apples now refers to the nested block apples
        // the outer block apples is temporarily hidden

        apples = 10; // this assigns value 10 to nested block apples, not outer block apples

        std::cout << apples << '\n'; // print value of nested block apples
    } // nested block apples destroyed

    std::cout << apples << '\n'; // prints value of outer block apples

    return 0;
} // outer block apples destroyed

　　

If you run this program, it prints:

5
10
5

In the above program, we first declare a variable named apples in the outer block. This variable is visible within the inner block, which we can see by printing its value (5). Then we declare a different variable (also named apples) in the nested block. From this point to the end of the block, the name apples refers to the nested block apples, not the outer block apples.

Thus, when we assign value 10 to apples, we’re assigning it to the nested block apples. After printing this value (10), nested block apples is destroyed. The existence and value of outer block apples is not affected, and we prove this by printing the value of outer block apples (5).

Note that if the nested block apples had not been defined, the name apples in the nested block would still refer to the outer block apples, so the assignment of value 10 to apples would have applied to the outer block apples:

#include <iostream>

int main()
{ // outer block
    int apples{5}; // here's the outer block apples

    { // nested block
        // apples refers to outer block apples here
        std::cout << apples << '\n'; // print value of outer block apples

        // no inner block apples defined in this example

        apples = 10; // this applies to outer block apples

        std::cout << apples << '\n'; // print value of outer block apples
    } // outer block apples retains its value even after we leave the nested block

    std::cout << apples << '\n'; // prints value of outer block apples

    return 0;
} // outer block apples destroyed

　　

The above program prints:

5
10
10

When inside the nested block, there’s no way to directly access the shadowed variable from the outer block.

Shadowing of global variables

Similar to how variables in a nested block can shadow variables in an outer block, local variables with the same name as a global variable will shadow the global variable wherever the local variable is in scope:

#include <iostream>
int value { 5 }; // global variable

void foo()
{
    std::cout << "global variable value: " << value << '\n'; // value is not shadowed here, so this refers to the global value
}

int main()
{
    int value { 7 }; // hides the global variable value until the end of this block

    ++value; // increments local value, not global value

    std::cout << "local variable value: " << value << '\n';

    foo();

    return 0;
} // local value is destroyed

　　

This code prints:

local variable value: 8
global variable value: 5

However, because global variables are part of the global namespace, we can use the scope operator (::) with no prefix to tell the compiler we mean the global variable instead of the local variable.

#include <iostream>
int value { 5 }; // global variable

int main()
{
    int value { 7 }; // hides the global variable value
    ++value; // increments local value, not global value

    --(::value); // decrements global value, not local value (parenthesis added for readability)

    std::cout << "local variable value: " << value << '\n';
    std::cout << "global variable value: " << ::value << '\n';

    return 0;
} // local value is destroyed

　　

This code prints:

local variable value: 8
global variable value: 4

Avoid variable shadowing

Shadowing of local variables should generally be avoided, as it can lead to inadvertent errors where the wrong variable is used or modified. Some compilers will issue a warning when a variable is shadowed.

For the same reason that we recommend avoiding shadowing local variables, we recommend avoiding shadowing global variables as well. This is trivially avoidable if all of your global names use a “g_” prefix.

Best practice

Avoid variable shadowing.

Go的50度灰：Golang新开发者要注意的陷阱和常见错误 | 鸟窝 https://colobu.com/2015/09/07/gotchas-and-common-mistakes-in-go-golang/

然的变量隐藏Accidental Variable Shadowing

短式变量声明的语法如此的方便（尤其对于那些使用过动态语言的开发者而言），很容易让人把它当成一个正常的分配操作。如果你在一个新的代码块中犯了这个错误，将不会出现编译错误，但你的应用将不会做你所期望的事情。

package main
import "fmt"
func main() {
    x := 1
    fmt.Println(x)     //prints 1
    {
        fmt.Println(x) //prints 1
        x := 2
        fmt.Println(x) //prints 2
    }
    fmt.Println(x)     //prints 1 (bad if you need 2)
}

　　

即使对于经验丰富的Go开发者而言，这也是一个非常常见的陷阱。这个坑很容易挖，但又很难发现。

你可以使用 vet命令来发现一些这样的问题。 默认情况下， vet不会执行这样的检查，你需要设置-shadow参数：
go tool vet -shadow your_file.go。