右移操作员 [英] shift right operator

问题描述

main（）

{

int a = 100;

a = a>> 32;

printf（"％d"，a）;

}


它打印a"只有100 ....但我期待a = 0 .....

有人可以告诉我原因吗？


bye

Deepak \

解决方案

DeepaK KC写道：

main（）
{
int a = 100;
a = a>> 32;
printf（"％d"，a）;
}
<它打印a只有100 ....但我期待a = 0 .....
有人可以告诉我原因吗？

它''表现未定义的行为。 （例如，有些机器

只有5位移位数。）


如果你打开编译器警告怎么办？

-

Chris" electric hedgehog" Dollin

DeepaK KC写道：

main（）
{
int a = 100;
a = a>> 32;
printf（"％d"，a）;
}
它打印a只有100 ....但我期待a = 0 .....
有人可以告诉我原因吗？

再见
Deepak \

你的程序的固定版本：

#include< stdio.h>

int main（void）

{

int a = 100;

a = a>> 32;

printf（"％d \\ \\ n"，a）;

返回0;

}


temp（558）

gcc -Wall -o foo foo.c

foo.c：在函数`main''中：

foo.c：6：警告：右移count> =类型的宽度

temp（559）

main()
{
int a =100;
a = a>>32;
printf(" %d", a);
}

it prints "a" as 100 only....but I am expecting a = 0.....
can some one tell me the reason?

bye
Deepak\

解决方案

DeepaK K C wrote:

main()
{
int a =100;
a = a>>32;
printf(" %d", a);
}

it prints "a" as 100 only....but I am expecting a = 0.....
can some one tell me the reason?

It''s a manifestation of undefined behaviour. (Some machines
only have a 5-bit shift count, for example.)

What happens if you turn your compiler warnings up?
--
Chris "electric hedgehog" Dollin

DeepaK K C wrote:

main()
{
int a =100;
a = a>>32;
printf(" %d", a);
}

it prints "a" as 100 only....but I am expecting a = 0.....
can some one tell me the reason?

bye
Deepak\

Fixed version of your program:
#include <stdio.h>
int main(void)
{
int a = 100;
a = a>>32;
printf(" %d\n", a);
return 0;
}

temp(558)

gcc -Wall -o foo foo.c
foo.c: In function `main'':
foo.c:6: warning: right shift count >= width of type
temp(559)

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