声明64位变量用C [英] Declaring 64-bit variables in C

问题描述

我有一个问题。

  uint64_t中VAR = 1; //这是... 000000 00001吧？
 

在我的code这个作品：

  VAR ^（1 <<;＆LT; 43）
 

但它是如何知道1应该在64位？应该不是我写这篇文章呢？

  VAR ^（（uint64_t中）1 LT;＆LT; 43）
 

解决方案

当你认为，1是一个纯签署 INT （这可能你的平台上为32位宽2的补运算），所以是43，所以任何机会<$​​ C $ C> 1＆LT;＆LT; 43 导致溢出：在事实，如果两个参数的数据类型的 INT 运营规则规​​定，其结果将是一个 INT 以及

不过，在C符号整数溢出是未定义行为，所以原则上任何的行可能发生。你的情况，可能是编译器发出code来执行在64位寄存器，移位，所以的靠运气的它似乎工作;让你应该用你写的第二种形式，或者，可以通过指定有保证的，正确的结果 1 为无符号长长文字使用 ULL 后缀（无符号长长是保证的至少 64位）。

  VAR ^（1ULL＆LT;＆LT; 43）
 

I have a question.

uint64_t var = 1; // this is 000000...00001 right?

And in my code this works:

var ^ (1 << 43)

But how does it know 1 should be in 64 bits? Shouldn’t I write this instead?

var ^ ( (uint64_t) 1 << 43 )

解决方案

As you supposed, 1 is a plain signed int (which probably on your platform is 32 bit wide in 2's complement arithmetic), and so is 43, so by any chance 1<<43 results in an overflow: in facts, if both arguments are of type int operator rules dictate that the result will be an int as well.

Still, in C signed integer overflow is undefined behavior, so in line of principle anything could happen. In your case, probably the compiler emitted code to perform that shift in a 64 bit register, so by luck it appears to work; to get a guaranteed-correct result you should use the second form you wrote, or, in alternative, specify 1 as an unsigned long long literal using the ull suffix (unsigned long long is guaranteed to be at least 64 bit).

var ^ ( 1ULL << 43 )

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