相比较快比无符号整数签署 [英] Faster comparing signed than unsigned ints

问题描述

可能重复：结果
  <一href=\"http://stackoverflow.com/questions/4712315/performance-of-unsigned-vs-signed-integers\">performance无符号VS符号整数

我已经读的地方，它是一个很小的有点快于 x86_64的比较在<$ C $签署整数 C> C / C ++ 相比，无符号整数，例如： 的for（int我... ）是（UINT我...）比更快。

I have read somewhere that it's a tiny bit faster on x86_64 to compare signed ints in C/C++ compared to unsigned ints, e.g. for (int i...) is "faster" than for (uint i...).

是这样吗？为什么是真的吗？我知道其中的差别是超小，但无论如何。

Is that true? Why is that true? I know the difference is super small, but anyway.

推荐答案

您最好引用来源为这种说法，这是它的脸可笑了。

You'd better cite a source for such a claim, which is ridiculous on the face of it.

我们正在谈论的x86_64，这意味着现代的处理器。这些的ALU将完成整数加法，减法，和/或在单个时钟周期比较（当然高速缓存未命中将需要更长的时间，但只对数据的大小和存储器布局取决于，未签名岬）。或甚至更低，用SIMD协处理器

We're talking about x86_64, which means modern processors. These ALUs will complete integer addition, subtraction, and/or comparison in a single clock cycle (cache misses will of course take longer, but depend only on the size and memory layout of the data, not signed-ness). Or even less, with the SIMD coprocessor.

我最好更容易相信有两种类型的比较之间的微小功率差，但不产生速度差。

I'd more likely to believe that there's a slight power difference between the two types of comparison, but not a speed difference.

现在，它可能是对于特定的编译器，定位于x86_64平台时code代是相较于其他一个数据类型更糟。但是，这将是一个非常特殊的情况下，并不太可能适用于所有x86_64的编译器。仍然，我怀疑缓存效果或后台进程是影响性能的测量（即测量时间每个进程花费将由上下文切换无效缓存受到影响，甚至性能计数器）。

Now, it is possible that for a particular compiler, code generation is worse for one data type vs the other when targeting the x86_64 platform. But that would be a very specialized case, and not likely to apply to all x86_64 compilers. And still, I'd suspect cache effects or background processes were affecting the performance measurement (even performance counters that measure time spent per-process will be affected by a context switch invalidating caches).

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