意外的答案，编译错误？ [英] Unexpected answer, compiler bug?

问题描述

我运行下面的程序。它应该打印2但实际打印

out 1.发生了什么？我应该报告编译器的错误吗？


谢谢。

void main（）

{

int a = 1;

a = a ++;

printf（"％d"，a）;

}

解决方案

Port Pirie写道：

我运行下面的程序。它应该打印2但实际打印

out 1.发生了什么？我应该报告编译器的错误吗？


谢谢。


void main（）

{

int a = 1;

a = a ++;

printf（"％d"，a）;

}

您已调用未定义的行为三次。首先，无效的主要原因不是适当的原型。正确的形式是int main（void）或int main（int，

char **）。其次，您访问一个对象，在序列点内修改它的值两次

。第三，你正在调用一个可变函数

，没有原型范围。


请通过以下链接阅读常见问题：


< http://www.c-faq.com/>

< http://www.clc-wiki.net/>

在文章< fk ********** @ aioe.orgPort Pirie< no *** @ nospam.invalidwrote：

>我运行下面的程序。它应该打印2但实际上它打印出来1.发生了什么？我应该报告编译器的错误吗？

void main（）

int a = 1;

a = a ++;

printf（"％d"，a）;
}

您问过一个常见问题3.1的变体，3.2和3.3。

请参阅< http://c-faq.com/expr/index.html>。另见问题11.12a至

11.15 at< http://c-faq.com/ansi/index.html> ;.

-

In-Real-Life：风河系统Chris Torek

美国犹他州盐湖城（40°39.22''N，111°50.29''W）[phone]

电子邮件：忘了它 http：// web .torek.net / torek / index.html

由于垃圾邮件发送者，阅读电子邮件就像在垃圾中搜索食物一样。

< BLOCKQUOTE>"桑托什" < sa ********* @ gmail.comschrieb im Newsbeitrag

news：fk ********** @ registered.motzarella.org ...

Port Pirie写道：

>

我运行程序下面。它应该打印2但实际上它打印出来1.发生了什么？我应该报告编译器的错误吗？

谢谢。

void main（）
{
int a = 1;
a = a ++;
printf（"％d"，a）;
}

您已调用未定义的行为三次。

实际四次

首先无效主要不是

正确原型。正确的形式是int main（void）或int main（int，

char **）。其次，您访问一个对象，在序列点内修改它的值两次

。第三，你正在调用一个可变函数

而没有原型范围。

和第四个你不会以一个\\\


再见结束你的输出，Jojo

Hi,

I ran the program below. It should print out 2 but actually it prints
out 1. What''s going on? Should I report a bug against the compiler?

Thanks.
void main()
{
int a=1;
a=a++;
printf("%d", a);
}

解决方案

Port Pirie wrote:

Hi,

I ran the program below. It should print out 2 but actually it prints
out 1. What''s going on? Should I report a bug against the compiler?

Thanks.
void main()
{
int a=1;
a=a++;
printf("%d", a);
}

You have invoked undefined behaviour thrice. Firstly void main is not
the proper prototype. Proper forms are int main(void) or int main(int,
char**). Secondly you access an object to modify it''s value twice
within a sequence point. Thirdly you are calling a variadic function
with no prototype in scope.

Please read the FAQ at the following link:

<http://www.c-faq.com/>
<http://www.clc-wiki.net/>

In article <fk**********@aioe.orgPort Pirie <no***@nospam.invalidwrote:

>I ran the program below. It should print out 2 but actually it prints
out 1. What''s going on? Should I report a bug against the compiler?

void main()
{
int a=1;
a=a++;
printf("%d", a);
}

You have asked a variant of Frequently Asked Questions 3.1, 3.2, and 3.3.
See <http://c-faq.com/expr/index.html>. See also questions 11.12a through
11.15 at <http://c-faq.com/ansi/index.html>.
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22''N, 111°50.29''W) [phone]
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.

"santosh" <sa*********@gmail.comschrieb im Newsbeitrag
news:fk**********@registered.motzarella.org...

Port Pirie wrote:

>Hi,

I ran the program below. It should print out 2 but actually it prints
out 1. What''s going on? Should I report a bug against the compiler?

Thanks.
void main()
{
int a=1;
a=a++;
printf("%d", a);
}

You have invoked undefined behaviour thrice.

Four times actually

Firstly void main is not
the proper prototype. Proper forms are int main(void) or int main(int,
char**). Secondly you access an object to modify it''s value twice
within a sequence point. Thirdly you are calling a variadic function
with no prototype in scope.

and fourth you don''t end your output with a \n

bye, Jojo

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