设置int64_t中的所有位

From following code I expect to set all bits in x to 1, but somehow only first 32 bits are set:

从下面的代码我希望将x中的所有位设置为1,但不知何故只设置了前32位:

int64_t x = 0xFFFFFFFF;
x<<32;
x|=0xFFFFFFFF;

Note: printing x after each line results in 4294967295 (32 lower bits set to 1). Also, tried using numeric_limits<int64_t>::min() with no success. My question is how to set all bits in x? Using RHEL5.5.

注意:在每行之后打印x会产生4294967295(32位低位设置为1)。此外,尝试使用numeric_limits :: min()但没有成功。我的问题是如何设置x中的所有位?使用RHEL5.5。

Thx

3 个解决方案

#1

x<<32 calculates the result of shifting x left by 32 bits and does nothing with the value. You want to use x <<= 32 instead.

x << 32计算将x向左移位32位的结果,并且不对该值进行任何操作。你想用x << = 32代替。

#2

Why not int64_t x = -1? or uint64_t x = ~0?

为什么不int64_t x = -1?或uint64_t x = ~0?

#3

This will work:

这将有效:

int64_t x = ~0LL;   (iner

or

int64_t x = -1LL;

You may get away with not having the LL, but not guaranteed - depends on the compiler.

你可能没有LL,但没有保证 - 取决于编译器。