Question

我创建了这样的代码片段来测试nonlocal：

def main():
    increased = 1
    def try_stat():
        nonlocal increased
        unchanged = 1
        unchanged += 1
        increased += 1
        print(f"unchanged = {unchanged} and increased = {increased}.") 

    for count in range(1, 4):
        print(f"Here comes iteration {count}")
        try_stat()
main()

运行它并获取：

$ python loc_stat.py
Here comes iteration 1
unchanged = 2 and increased = 2.
Here comes iteration 2
unchanged = 2 and increased = 3.
Here comes iteration 3
unchanged = 2 and increased = 4.

请注意，在声明非局部变量时，应处理两个步骤：

声明外部范围已增加= 1
将其重新声明为非本地。

为什么不直接将语法设计为

def main():
    # increased = 1
    def try_stat():
        nonlocal increased = 1

C使用带有块范围的静态设计来实现它

#include <stdio.h>
void try_stat(void);

int main(void)
{
    for (int count = 1; count <=3; count++)
    {
        printf("Here comes iteration %d:\n", count);
        try_stat();
    }

    return 0;
}

void try_stat(void)
{
    int unchanged = 1;
    static int increased = 1;

    printf("unchanged = %d and increased = %d\n", ++unchanged, ++increased);
}

编译并随附

$ ./a.out
Here comes iteration 1:
unchanged = 2 and increased = 2
Here comes iteration 2:
unchanged = 2 and increased = 3
Here comes iteration 3:
unchanged = 2 and increased = 4

此解决方案似乎更直观，并且Cpython是用C编写的，

选择块范围内的当前非本地模式而不是C的本机静态的原因是什么？

Answer 1

两者。：-）

它们不是等效的。与nonlocal变量不同，static变量在此块之外可见。

实际上，我认为您要做的实际上是一个类成员变量。 C不是OO语言，static变量用于封装两次调用之间的持久状态。 Python为此具有类和属性self。

非本地模式，而不是具有块范围的C的本机静态

1 个答案: