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i +++ j

i +++ j# Programming - 葵花宝典

v*d2008-10-15 07:10

1 楼

,,,,,

b*d2008-10-15 07:10

2 楼

int i = 1;
int j = 2;
int k = i +++ j;
k?
i?
j?

v*d2008-10-15 07:10

3 楼

...

b*d2008-10-15 07:10

4 楼

another one.
k = i+ +j;

【在 b*****d 的大作中提到】

: int i = 1;
: int j = 2;
: int k = i +++ j;
: k?
: i?
: j?

v*d2008-10-15 07:10

5 楼

怎么换另一个形象啊我换衣服了啊
怎么还是这个

w*g2008-10-15 07:10

6 楼

这个东西我觉得不同的编译器可能会给出不同的结果，未必有正确解。

【在 b*****d 的大作中提到】

: int i = 1;
: int j = 2;
: int k = i +++ j;
: k?
: i?
: j?

v*d2008-10-15 07:10

7 楼

。。。

l*82008-10-15 07:10

8 楼

The parser/tokenizer will always try to match the longest string. Therefore
when it encounters a sequence of "i+++j" it will generate four tokens:
(1) identifier "i"
(2) operator "++"
(3) operator "+"
(4) identifier "j"

v*d2008-10-15 07:10

9 楼

，，

o*o2008-10-15 07:10

10 楼

before wasting time, think what you may achieve from this kind of tricks

【在 b*****d 的大作中提到】

: another one.
: k = i+ +j;

v*d2008-10-15 07:10

11 楼

晕不会用

b*d2008-10-15 07:10

12 楼

In fact, it is defined in C++ standard.
The reason is i++ has higher precedence than ++i.
Surprisingly.
so, i +++ j is
(i++)+j rather than
i+ (++j).
VS 2008 gives the correct answer.

【在 b*****d 的大作中提到】

: int i = 1;
: int j = 2;
: int k = i +++ j;
: k?
: i?
: j?

v*d2008-10-15 07:10

13 楼

now

b*d2008-10-15 07:10

14 楼

that's totally true.
so i++j is wrong, but i+ +j is okay.

Therefore

【在 l***8 的大作中提到】

: The parser/tokenizer will always try to match the longest string. Therefore
: when it encounters a sequence of "i+++j" it will generate four tokens:
: (1) identifier "i"
: (2) operator "++"
: (3) operator "+"
: (4) identifier "j"

H*o2008-10-15 07:10

15 楼

１０分。系统有延迟吧，没关系的。

【在 v***d 的大作中提到】

: now

t*t2008-10-15 07:10

16 楼

it is defined in C++ standard, but your reason is wrong
it is defined in the tokenization process.

【在 b*****d 的大作中提到】

: In fact, it is defined in C++ standard.
: The reason is i++ has higher precedence than ++i.
: Surprisingly.
: so, i +++ j is
: (i++)+j rather than
: i+ (++j).
: VS 2008 gives the correct answer.

k*y2008-10-15 07:10

17 楼

10分

【在 v***d 的大作中提到】

: ,,,,,

b*d2008-10-15 07:10

18 楼

puzzle...
how about i +++j,
the( ++j ) should be recognized as a token right?

【在 t****t 的大作中提到】

: it is defined in C++ standard, but your reason is wrong
: it is defined in the tokenization process.

w*i2008-10-15 07:10

19 楼

somehow remind me of 茴香豆的茴字有几种写法, hehe

l*82008-10-15 07:10

20 楼

>> how about i +++j
No, "i +++j" is still tokenized as "i", "++", "+", "j".
This has nothing to do with operator precedence.
To force the parser to recognize (++j), you need to put a space between the 1st and 2nd
"+", like this: "i+ ++j".

g*g2008-10-15 07:10

21 楼

Why do you care, I'll make sure whoever writes this code
will get fired if I am the boss.

【在 b*****d 的大作中提到】

: int i = 1;
: int j = 2;
: int k = i +++ j;
: k?
: i?
: j?

m*e2008-10-15 07:10

22 楼

Wow...
I think it's good to know stuff, but this is very close to useless knowledge
to me.

the 1st and 2nd

【在 l***8 的大作中提到】

: >> how about i +++j
: No, "i +++j" is still tokenized as "i", "++", "+", "j".
: This has nothing to do with operator precedence.
: To force the parser to recognize (++j), you need to put a space between the 1st and 2nd
: "+", like this: "i+ ++j".

l*82008-10-15 07:10

23 楼

You probably will never care about this problem unless your job is to write
a compiler... So yes, it is 99.9% useless.

b*d2008-10-15 07:10

24 楼

强人。
Thank you.

write

【在 l***8 的大作中提到】

: You probably will never care about this problem unless your job is to write
: a compiler... So yes, it is 99.9% useless.