为什么函数的整数返回值与该整数的解释不同

Question

为什么这样做有效？

def nextSquare():
    i = 1;

    # An Infinite loop to generate squares
    while True:
        yield i*i
        i += 1  # Next execution resumes
                # from this point

# Driver code to test above generator
# function
for num in nextSquare():
    if num > 100:
        break
    print(num)

但不是这个？

def nextSquare():
    i = 1;

    # An Infinite loop to generate squares
    while True:
        yield i*i
        i += 1  # Next execution resumes
                # from this point

# Driver code to test above generator
# function
for num in 1:
    if num > 100:
        break
    print(num)

nextsquare每次返回一个整数运行所以nextsquare第一次returns和

有什么区别

for num in nextSquare():

和

for num in 1:

Answer 1

An integer is returned each time nextsquare is run

对了，nextSquare应该是“运行”在前。你称它为 nextSquare()，它 returns 是一个可迭代对象，然后 for 循环遍历它，像这样：

retval = nextSquare()
it = iter(retval)
num = next(it)
while True:
    # loop body
    try:
        num = next(it)
    except StopIteration:
        break

但是，调用 next(1) 没有意义，因为迭代单个数字没有意义 - 循环迭代 collections 或类似“collection-like”的东西。一个数字肯定不像 collection，所以你不能遍历它。

错误消息正是这样说的：

>>> for _ in 1:
...  ...
... 
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'int' object is not iterable

为什么函数的整数返回值与该整数的解释不同

Why is integer returnvalue of function not interpreted the same as that integer

for-loop

yield

python-3.x