输出取决于编译器 [英] does the output depends on the compiler

问题描述

大家好


这个命令的输出取决于编译器。


int val = 1234;

int * ptr =& val;

printf（"％d％d％d"，val，* ptr ++，++ * ptr）;


O / p

i期待1234,1234,1235。

但我得到了1235,1235,1235

推荐答案

sunny写道：

sunny wrote:

嗨所有


这个命令的输出取决于编译器。


int val = 1234;

int * ptr =& val;

printf（"％d％d％d"，val，* ptr ++，++ * ptr）;


O / p

i期待1234,1234,1235。

但我得到了1235,1235,1235

Hi All

does the output of this command depend on the compiler.

int val=1234;
int* ptr=&val;
printf("%d %d %d",val,*ptr++,++*ptr);

O/p
i was expecting 1234,1234,1235.
but i got 1235, 1235, 1235

sunny写道：

sunny wrote:

嗨所有


输出这个命令取决于编译器。


int v al = 1234;

int * ptr =& val;

printf（"％d％d％d"，val，* ptr ++，++ * ptr） ;


O / p

i期待1234,1234,1235。

但我得到了1235,1235,1235

Hi All

does the output of this command depend on the compiler.

int val=1234;
int* ptr=&val;
printf("%d %d %d",val,*ptr++,++*ptr);

O/p
i was expecting 1234,1234,1235.
but i got 1235, 1235, 1235

sunny写道：

sunny wrote:

大家好


此命令的输出是否依赖于编译器。

Hi All

does the output of this command depend on the compiler.

int val = 1234;

int * ptr =& val;

printf （％d％d％d，val，* ptr ++，++ * ptr）;


O / p

i期待1234,1234 ，1235。

但我得到1235,1235,1235

int val=1234;
int* ptr=&val;
printf("%d %d %d",val,*ptr++,++*ptr);

O/p
i was expecting 1234,1234,1235.
but i got 1235, 1235, 1235

您的代码有未定义的行为。你正在改变ptr，并且

阅读它不止一次在一个序列点之间，如果你想让程序正常运行，你就不应该这样做。


它也是未定义的功能的顺序参数是

评估。

Your code has undefined behaviour. You''re altering ptr, and
"reading" it more than once between a sequence point, which
you should not do if you want the program to behave properly.

It''s also undefined in which order arguments to functions are
evaluated.

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