为什么在这种情况下会丢弃该值？ [英] Why does the value get discarded in this case?

问题描述

当我有：


int main（无效）

{

int x = 256;

x>> 8;


printf（"值为：％d \ n"，x）;

返回0 ;

}


我得到：


[cdalten @ localhost~] $ gcc -g -Wall seq .c -o seq

seq.c：在函数''main''：

seq.c：6：warning：语句无效

[cdalten @ localhost~] $ ./seq

值为：256

然而，当我将x从x>> 8更改为x ++时/>

#include< stdio.h>


int main（无效）

{

int x = 256;

x ++;


printf（"值为：％d \ n"，x）;

返回0;

}


我得到：


[cdalten @ localhost~] $ gcc -g -Wall seq.c -o seq

[cdalten @ localhost~] $ ./seq

值为：257

问题是，为什么x>> 8丢弃了值得

走了，但x ++不是吗？

解决方案

gcc -g -Wall seq.c -o seq

seq.c：在函数''main''中：

seq.c：6：warning：语句无效

[cdalten @ localhost~]

./ seq

值为：256

然而，当我改变时x从x>> 8到x ++


#include< stdio.h>


int main（无效）

{

int x = 256;

x ++;


printf（"值为：％d \ n"，x）;

返回0;

}


我得到：

[cdalten @ localhost~]

gcc -g -Wall seq.c -o seq

[cdalten @ localhost~]

When I have:

int main(void)
{
int x = 256;
x>>8;

printf("The value is: %d\n", x);
return 0;
}

I get:

[cdalten@localhost ~]$ gcc -g -Wall seq.c -o seq
seq.c: In function ''main'':
seq.c:6: warning: statement with no effect
[cdalten@localhost ~]$ ./seq
The value is: 256
However, when I change x from x>>8 to x++

#include <stdio.h>

int main(void)
{
int x = 256;
x++;

printf("The value is: %d\n", x);
return 0;
}

I get:

[cdalten@localhost ~]$ gcc -g -Wall seq.c -o seq
[cdalten@localhost ~]$ ./seq
The value is: 257

The question is, how come something like x>>8 discards the value right
away, but x++ doesn''t?

解决方案

gcc -g -Wall seq.c -o seq
seq.c: In function ''main'':
seq.c:6: warning: statement with no effect
[cdalten@localhost ~]

./seq
The value is: 256
However, when I change x from x>>8 to x++

#include <stdio.h>

int main(void)
{
int x = 256;
x++;

printf("The value is: %d\n", x);
return 0;
}

I get:

[cdalten@localhost ~]

gcc -g -Wall seq.c -o seq
[cdalten@localhost ~]

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