绑定不适用于特定零件 [英] Binds not working on a specific part

问题描述

因此，在程序的前面，我使用了该行

So earlier in my program, I use the line

l.bind("<Button-1>",lambda e: getSide(i))

当我单击该元素时，它可以正常工作.

and when I click on the element, it works fine.

但是，稍后我会使用该行

However, later I use the line

l.bind("<Button-1>",lambda e: sortby(x))

用于其他本地对象. getSide是一个存根，它显示绑定时定义的i的值. sortby是一个Quicksort(出于调试目的)在开始时打印x的值.奇怪的是，尽管getSide返回正确的值，但sortby却没有.

for a different local object. getSide is a stub that prints the value of i defined when binding. sortby is a Quicksort that (for debugging purposes) prints the value of x at the start. The curious thing is that while getSide returns the correct value, sortby does not.

getSide返回i，而sortby打印len(column)-1，即要绑定的最后一个Label.

getSide returns i, whereas sortby prints len(column)-1, i.e the last Label to be bound.

推荐答案

问题是您正在循环内创建这些绑定.执行此操作时，必须捕获"在lambda函数的参数列表中使用的所有值:

The problem is that you are creating these bindings inside a loop. When you do this, you must "capture" any values you use inside a lambda function in its argument list:

for x in range(0,len(columns)):
...
    l.bind("<Button-1>",lambda e, x=x: sortby(x))
#                                 ^^^

这是因为lambda函数中包含的表达式是在调用时而不是定义时求值的.因此，sortby(x)中的x将始终引用循环中x保留的最后一个值.

This is because the expression contained inside a lambda function is evaluated at call-time, not definition-time. So, the x in sortby(x) will always refer to the last value held by x in the loop.

但是，默认参数是在定义时求值的.因此，执行x=x可以确保x引用循环内x的当前值.

Default arguments however are evaluated at definition-time. Thus, doing x=x ensures that x refers to the current value of x inside the loop.

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