b*d
2 楼
int i = 1;
int j = 2;
int k = i +++ j;
k?
i?
j?
int j = 2;
int k = i +++ j;
k?
i?
j?
v*d
3 楼
...
b*d
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?
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*d
5 楼
怎么换另一个形象啊 我换衣服了啊
怎么还是这个
怎么还是这个
w*g
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?
【在 b*****d 的大作中提到】
: int i = 1;
: int j = 2;
: int k = i +++ j;
: k?
: i?
: j?
v*d
7 楼
。。。
l*8
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"
when it encounters a sequence of "i+++j" it will generate four tokens:
(1) identifier "i"
(2) operator "++"
(3) operator "+"
(4) identifier "j"
v*d
9 楼
，，
o*o
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;
【在 b*****d 的大作中提到】
: another one.
: k = i+ +j;
v*d
11 楼
晕 不会用
b*d
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?
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*d
13 楼
now
b*d
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"
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*o
15 楼
１０分。系统有延迟吧，没关系的。
【在 v***d 的大作中提到】
: now
【在 v***d 的大作中提到】
: now
t*t
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.
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*y
17 楼
10分
【在 v***d 的大作中提到】
: ,,,,,
【在 v***d 的大作中提到】
: ,,,,,
b*d
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.
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*i
19 楼
somehow remind me of 茴香豆的茴字有几种写法, hehe
l*8
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".
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*g
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?
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*e
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".
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*8
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.
a compiler... So yes, it is 99.9% useless.
b*d
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.
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.
z*e
25 楼
C/C++是LL(1)或者LR(1)，所以左边优先。
【在 b*****d 的大作中提到】
: puzzle...
: how about i +++j,
: the( ++j ) should be recognized as a token right?
【在 b*****d 的大作中提到】
: puzzle...
: how about i +++j,
: the( ++j ) should be recognized as a token right?
d*n
26 楼
who will actually write this?
maybe some tester?
maybe some tester?
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