由于延缓计算,所以最终计算时,i已经变成了3.

Bard:

he result of calling f1(), f2(), and f3() will be 9, 9, and 9. This is because the f() functions are closures, which means that they capture the value of i when they are created. In this case, i is 3 when the f() functions are created, so they will always return 9.

Here is a breakdown of what happens in the code:

• The count() function creates a list called fs.
• The count() function iterates over the range 1 to 3.
• For each iteration, the count() function defines a function called f.
• The f() functions are closures, which means that they capture the value of i when they are created.
• The f() functions return the value of i squared.
• The count() function appends each f() function to the fs list.
• The count() function returns the fs list.
• The f1, f2, and f3 variables are assigned the values of the first three elements in the fs list.
• When f1(), f2(), and f3() are called, they return the value of i squared, which is 9.

修改下:

def count():

    fs = []

    for i in range(1, 4):

        def f(x=i):

             return x*x

        fs.append(f)

    return fs

f1, f2, f3 = count()

每次循环创建一个函数,在每个函数创建时将 i 的值 (1,2,3) 分别赋值给每个函数 f1, f2,  f3 的单独的局部变量 x (f1的x=1,f2的x=2,f3的x=3)

Bard:

The result of calling f1(), f2(), and f3() will be 1, 4, and 9. This is because the f() functions are closures, which means that they capture the value of i when they are created. In this case, i is set to 1, 2, and 3 when the f() functions are created, so they will return 1, 4, and 9 respectively.

Here is a breakdown of what happens in the code:

• The count() function creates a list called fs.
• The count() function iterates over the range 1 to 3.
• For each iteration, the count() function defines a function called f.
• The f() functions are closures, which means that they capture the value of i when they are created.
• The f() functions return the value of x squared, where x is set to the value of i when the function is called.
• The count() function appends each f() function to the fs list.
• The count() function returns the fs list.
• The f1, f2, and f3 variables are assigned the values of the first three elements in the fs list.
• When f1(), f2(), and f3() are called, they return the value of x squared, which is 1, 4, and 9 respectively.

GPT:

f1, f2, f3 = count() 是一种多重赋值的方式。在这个例子中，count() 函数返回了一个包含三个函数的列表，通过多重赋值将列表中的函数分别赋给了变量 f1、f2 和 f3。

具体来说，count() 函数返回了一个包含三个闭包函数的列表 fs。每个闭包函数都有对应的索引 i，但是在闭包函数内部访问 i 时，实际上是引用了外部作用域中的变量 i。由于循环结束后 i 的值为 3，因此每个闭包函数在被调用时都返回了 3 的平方。

这意味着，通过 f1()、f2() 和 f3() 调用这三个函数时，它们都会返回 9，因为在创建这些函数时，循环已经结束并且 i 的值为 3。

如果想要每个函数分别返回 1、4、9，可以通过在闭包函数中使用默认参数的方式解决：