有人可以告诉我为什么吗？ （一元前后增量运算符 [英] Can someone tell me why? (Unary pre and post increment operator

问题描述

请评估以下片段：


int a = 5;

int b;

b = a ++ + a;

Edit1-> Text = b;

a = 5;

b = a ++ + ++ a;

Edit2-> Text = b;


你能告诉我为什么分配给Edit1-> Text的b值是10

而值b分配给Edit2->文字是12？


如果可以的话，为什么不记录？


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Please evaluate the following snippet:

int a=5;
int b;
b = a++ + a;
Edit1->Text = b;
a = 5;
b = a++ + ++a;
Edit2->Text = b;

Can you tell me why the value of b assigned to Edit1->Text is 10
while the value of b assigned to Edit2->Text is 12?

And if you can, why isn''t it documented?

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推荐答案

Andreas Sheriff发布：

Andreas Sheriff posted:

请评估以下片段：

int a = 5;


定义一个名为a的变量类型为int的并将其值设置为5.

int b;


定义变量：Name =" b"。 Type =" int"。值=没有特定值

b = a ++ + a;


" b"最终可能会有两个可能的值，或者：


10

或


11


因为标准并没有规定是否a ++或a或a。将首先完成。


变量a最终得到值6.

Edit1-> Text = b;



文本为10或11.


a = 5;



将a的值设为5.


b = a ++ + ++ a;



" b"获得：


12


或者是5 + 7或6 + 6。


" a"变成7.


Edit2-> Text = b;



文本设置为12.


你能告诉我为什么分配给Edit1-> Text的b的值分配给Edit2-> Text的b的值是12？

Please evaluate the following snippet:

int a=5;
Defines a variable called "a" of the type "int" and sets its value to 5.
int b;
Defines variable: Name = "b". Type = "int". Value = no particular value
b = a++ + a;
"b" could end up with two possible values here, either:

10

or

11

as the Standard doesn''t dictate whether "a++" or "a" will be done first.

The variable "a" ends up with the value 6.
Edit1->Text = b;

The text will be either 10 or 11.

a = 5;

Sets "a"''s value to 5.

b = a++ + ++a;

"b" gets:

12

either by "5 + 7" or "6 + 6".

"a" becomes 7.

Edit2->Text = b;

Text is set to 12.

Can you tell me why the value of b assigned to Edit1->Text is 10
while the value of b assigned to Edit2->Text is 12?

欢迎你！

-JKop

You''re welcome!
-JKop

" Andreas Sheriff" < SP ********* @ iion.com>在消息中写道

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"Andreas Sheriff" <sp*********@iion.com> wrote in message
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Lj.229505@fed1read03...

请评估以下代码段：

int a = 5;
int b;
b = a ++ + a;
Edit1-> Text = b;
a = 5;
b = a ++ + ++ a;
Edit2-> Text = b;

你能告诉我为什么分配给Edit1-> Text的b的值是10
而分配给Edit2-> Text的b的值是12？

如果可以的话，为什么不记录？

- <警告：
请勿回复此电子邮件
仅在此新闻组回复我


这就是我的想法：

int a = 5;
int b;
b = a ++ + a;
1）b = 5 + 5

2）a ++在完成任务后完成，a = 6

Edit1-> Text = b;
a = 5;
b = a ++ + ++ a;
1）++ a在作业之前完成，a = 6

2）b = 6 + 6

3）a ++在作业完成后完成， a = 7

编辑2 - >文字= b;

Please evaluate the following snippet:

int a=5;
int b;
b = a++ + a;
Edit1->Text = b;
a = 5;
b = a++ + ++a;
Edit2->Text = b;

Can you tell me why the value of b assigned to Edit1->Text is 10
while the value of b assigned to Edit2->Text is 12?

And if you can, why isn''t it documented?

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Here''s what I think happens:
int a=5;
int b;
b = a++ + a; 1) b = 5 + 5
2) a++ is done after the assignment, a = 6
Edit1->Text = b;
a = 5;
b = a++ + ++a; 1) ++a is done before the assignment, a = 6
2) b = 6 + 6
3) a++ is done after the assignment, a = 7
Edit2->Text = b;

PKH

PKH

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