[英]Why are x and z evaluating differently if “x=(date<<7)>>12” and {y=date<<7;z=y>>12;}?
It is really frustrating.What is the possible reason that (date<<7)>>12
is giving a different result from y>>12
where y
is date<<7
?.I should add that the latter is working correctly as I intend,but the first is not.What's the difference?I can see none. 这真是令人沮丧。可能的原因是(date<<7)>>12
给出了与y>>12
不同的结果,其中y
是date<<7
?。我应该补充一点,后者正在我正常工作打算,但第一个不是。有什么区别?我看不到。
#include<stdio.h>
int main()
{
unsigned short date=5225,x,y,z;
x=(date<<7)>>12;
printf("Month is %hu\n",x);
y=date<<7;
z=y>>12;
printf("Month is %hu\n",z);
}
OUTPUT OUTPUT
Month is 163
Month is 3
In C, all integer computations are promoted to at least int
1 . 在C语言中,所有整数计算至少提升为int
1 。 So 所以
x = (date << 7) >> 12
= (5225 << 7 /* result as int */) >> 12
= 668800 >> 12
= 163
after the computation is complete, we truncate the result back to an unsigned short
to get 163. 计算完成后,我们将结果截断回unsigned short
以获得163。
In the second case, the y
forced the result to be truncated to be an unsigned short
, so 在第二种情况下, y
强制将结果截断为unsigned short
,因此
y = (unsigned short) (date << 7)
= (unsigned short) 668800
= 13440
z = y >> 12
= 3
1 : C11 §6.5.7/3: "The integer promotions are performed on each of the operands. The type of the result is that of the promoted left operand."; 1 :C11§6.5.7/ 3:“整数提升是对每个操作数执行的。结果的类型是提升后的左操作数的类型。” §6.3.1.1/2: "If an int
can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int
; otherwise, it is converted to an unsigned int
. These are called the integer promotions ." §6.3.1.1/ 2:“如果int
可以表示原始类型的所有值(由宽度限制,对于位域),则该值将转换为int
;否则,它将转换为unsigned int
这些被称为整数促销 。“
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