前/后增量蓝调 [英] pre/post increment blues
问题描述
代码：
int aaa = 100;
printf（"％d％d％d \ n"， - aaa， aaa，aaa - ）;
printf（"％d \ n"，aaa）;
打印：
99 100 100
有人可以解释这背后的逻辑吗？
di *********** *@gmail.com 写道：
代码：
int aaa = 100;
printf（"％d％d％d \ n"， - aaa，aaa，aaa--）;
printf（"％d \ n"， aaa）;
打印：
99 100 100
98
有人可以解释这背后的逻辑吗？
是的。空无一人。它是未定义的行为。
di * ***********@gmail.com 写道：
代码：
int aaa = 100;
printf（"％d％d％d \ n"， - aaa，aaa，aaa--）;
printf （％d \ n，aaa）;
打印：
99 100 100
98
有人可以解释这背后的逻辑吗？
这背后没有逻辑。在语句中多次修改
变量是未定义的行为。阅读编译器
（可能/希望）输出的警告。任何事情都可能发生。
请阅读：
http://www.parashift.com/c++-faq-lit...html#faq-39.15
-
rbh
On Thu，2007年6月14日19:12:28 -0700，" di在con.lang.c ++中，************ @ gmail.com"
< di ********** @ gmail.comwrote：
代码：
int aaa = 100;
printf（"％d％ d％d \ n"， - aaa，aaa，aaa--）;
printf（"％d \ nn"，aaa）;
打印：
99 100 100
98
有人可以解释这背后的逻辑吗？
这背后没有逻辑。你写了一个声明，它修改了一个对象的值两次，并另外读取了另一个没有参与计算新值的时间。所有没有
之间的任何序列点。
行为未定义，C ++标准未指定或
care会发生什么。
-
杰克克莱恩
主页： http://JK-Technology.Com
常见问题解答
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The code:
int aaa = 100;
printf("%d %d %d\n", --aaa, aaa, aaa--);
printf("%d\n", aaa);
prints:
99 100 100
98
could somebody explain the logic behind this?
di************@gmail.com wrote:The code:
int aaa = 100;
printf("%d %d %d\n", --aaa, aaa, aaa--);
printf("%d\n", aaa);
prints:
99 100 100
98
could somebody explain the logic behind this?Yes. There is none. It''s undefined behaviour.
di************@gmail.com wrote:
The code:
int aaa = 100;
printf("%d %d %d\n", --aaa, aaa, aaa--);
printf("%d\n", aaa);
prints:
99 100 100
98
could somebody explain the logic behind this?There is no logic behind this. It is undefined behaviour to modify a
variable more than once in a statement. Read the warnings the compiler
(presumably/hopefully) output. Anything could happen.
Please read this:
http://www.parashift.com/c++-faq-lit...html#faq-39.15
--
rbh
On Thu, 14 Jun 2007 19:12:28 -0700, "di************@gmail.com"
<di**********@gmail.comwrote in comp.lang.c++:
The code:
int aaa = 100;
printf("%d %d %d\n", --aaa, aaa, aaa--);
printf("%d\n", aaa);
prints:
99 100 100
98
could somebody explain the logic behind this?There is no logic behind this. You wrote a statement that modifies
the value of an object twice, and additionally reads the value another
time not involved in the calculation of the new value. All without
any sequence points in between.
The behavior is undefined, and the C++ standard does not specify or
care what happens.
--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://c-faq.com/
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++
http://www.club.cc.cmu.edu/~ajo/docs/FAQ-acllc.html
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