[英]Bitwise shift operation in C on uint64_t variable
I have the following sample code: 我有以下示例代码:
uint64_t x, y;
x = ~(0xF<<24);
y = ~(0xFF<<24);
The result would be: 结果将是:
x=0xfffffffff0ffffff
y=0xfffff
Can anyone explain the difference? 有人可以解释这个区别吗? Why x is calculated over 64 bit and y only on 32?
为什么x计算超过64位而y只计算在32位?
The default operation is 32 bit. 默认操作是32位。
x=~(0xf<<24);
This code could be disassembled into the following steps: 此代码可以反汇编为以下步骤:
int32_t a;
a=0x0000000f;
a<<=24; // a=0x0f000000;
a=~a; // a=0xf0ffffff;
x=(uint64_t)a; // x = 0xfffffffff0ffffff;
And, 和,
y = ~(0xFF<<24);
int32_t a;
a=0x000000ff;
a<<=24; // a=0xff000000;
a=~a; // a=0x00ffffff;
x=(uint64_t)a; // x = 0x000000000ffffff;
Because 0x0f << 24
is a positive number when viewed as an int
, it's sign-extended to a positive number, ie to 0x00000000_0f000000
(the underscore is just for readability, C does not support this syntax). 因为
0x0f << 24
在被视为int
时是正数,所以它被符号扩展为正数,即到0x00000000_0f000000
(下划线只是为了可读性,C不支持这种语法)。 This is then inverted into what you're seeing. 然后将其转换为您所看到的内容。
0xff << 24
on the other hand is negative, so it's sign-extended differently. 另一方面,
0xff << 24
是负数,因此它的符号扩展方式不同。
Other posters have shown why it does this. 其他海报已经说明了为什么会这样做。 But to get the expected results:
但要获得预期的结果:
uint64_t x, y;
x = ~(0xFULL<<24);
y = ~(0xFFULL<<24);
Or you can do this (I don't know if this is is any slower than the above though): 或者你可以这样做(我不知道这是否比上面的慢):
uint64_t x, y;
x = ~(uint64_t(0xF)<<24);
y = ~(uint64_t(0xFF)<<24);
Then: 然后:
x = 0xfffffffff0ffffff
y = 0xffffffff00ffffff
You have undefined behavior in your program so anything might happen. 您的程序中有未定义的行为,因此可能发生任何事情。
int
, which is equivalent to signed int
. int
,相当于signed int
。 On this particular platform, int
is apparently 32 bits. int
显然是32位。 int
. int
。 int
so no implicit type promotions take place. int
因此不会发生隐式类型的提升。 The result of the << operation is therefore also a (signed) int
. int
。 int
as a non-negative value, so everything is ok. int
,因此一切正常。 int
! int
! Here, undefined behavior is invoked and anything might happen. ISO 9899:2011 6.5.7/4: ISO 9899:2011 6.5.7 / 4:
"The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros."
“E1 << E2的结果是E1左移E2位位置;空位用零填充。” /--/
/ - /
"If E1 has a signed type and nonnegative value, and E1 × 2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.
“如果E1具有带符号类型和非负值,并且E1×2E2在结果类型中可表示,那么这就是结果值;否则,行为是未定义的。
So the expression 0xFF<<24 can't be used. 因此不能使用表达式0xFF << 24。 The program is free to print any garbage value after that.
该程序可以随后打印任何垃圾值。
But if we ignore that one and focus on 0x0F<24: 但如果我们忽略那一个并专注于0x0F <24:
int
. int
。 The ~operator is applied to this. Bugs like this is why the coding standard MISRA-C contains a number of rules to ban sloppy use of integer literals in expression like this. 像这样的错误是为什么编码标准MISRA-C包含许多规则来禁止在这样的表达式中使用整数文字。 MISRA-C compliant code must use the
u
suffix after each integer literal (MISRA-C:2004 10.6) and the code is not allowed to perform bitwise operations on signed integers (MISRA-C:2004 12.7). 符合MISRA-C的代码必须在每个整数文字后使用
u
后缀(MISRA-C:2004 10.6),并且不允许代码对有符号整数执行按位运算(MISRA-C:2004 12.7)。
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