如何使这个忽略偶数？ [英] How do I make this ignore even numbers?

问题描述

底部的通话功能中存在24会导致错误。

The presence of 24 in the call function at the bottom causes an error.

def sum_odd_ints(list):
    add = 0
    for n in range(len(list)):
        if type(n) == int and int(n % 2 == 1):
            add += list[n]
    print(add)
    return add



sum_odd_ints(['cat', 3, 3.0, 7, 24, 'dog'])

推荐答案

你可以试试

Could you try

if type(n) == int and int(n % 2) == 1:
   add += int(n)

代替？



n 不是数组中的索引，它是从数组枚举中提取的值。这会产生巨大的差异。



你也可以改变测试奇怪的方式：

instead ?

n is not an index in your array, it is the value extracted from the enumeration of the array. That makes a huge difference.

You can also change the way you test the oddity:

if type(n) == int and int (n & 1) == 1:
   add += int(n)

第一行有一个错位的括号。 [/ Edit]

There was a misplaced parenthesis on the first line. [/Edit]

您需要使用调试器运行程序并在程序中检查 n 崩溃。

由于编译器可能会出现问题。

You need to run your program with a debugger and inspect n when the program crash.
your may have a problem because of compiler.

if type(n) == int and int(n % 2) == 1:
   add += int(n)

某些编译器不优化如果表达式，即使第一部分错误，也要评估第二部分。

Some compiler don't optimize the if expression, which make than even if the first part is wrong, the second part be evaluated.

if type(n) == int:
    if int(n % 2) == 1:
       add += int(n)

这种方式确保仅在第一部分为整数时才评估第二部分。

This way you ensure that second part is evaluated only when first part is integer.

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