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If a 32-bit integer overflows, can we use a 40-bit structure instead of a 64-bit long one?
---------
#Q:
If, say, a 32-bit integer is overflowing, instead of upgrading int to long, can we make use of some 40-bit type if we need a range only within 240, so that we save 24 (64-40) bits for every integer?
If so, how?
I have to deal with billions and space is a bigger constraint.
#A:
##Yes, but...
It is certainly possible, but it is usually nonsensical (for any program that doesn't use billions of these numbers):
```C
#include <stdint.h> // don't want to rely on something like long long
struct bad_idea
{
uint64_t var : 40;
};
```
Here, var will indeed have a width of 40 bits at the expense of much less efficient code generated (it turns out that "much" is very much wrong -- the measured overhead is a mere 1-2%, see timings below), and usually to no avail. Unless you have need for another 24-bit value (or an 8 and 16 bit value) which you wish to pack into the same structure, alignment will forfeit anything that you may gain.
In any case, unless you have billions of these, the effective difference in memory consumption will not be noticeable (but the extra code needed to manage the bit field will be noticeable!).
###Note:
The question has in the mean time been updated to reflect that indeed billions of numbers are needed, so this may be a viable thing to do, presumed that you take measures not to lose the gains due to structure alignment and padding, i.e. either by storing something else in the remaining 24 bits or by storing your 40-bit values in structures of 8 each or multiples thereof).
Saving three bytes a billion times is worthwhile as it will require noticeably fewer memory pages and thus cause fewer cache and TLB misses, and above all page faults (a single page fault weighting tens of millions instructions).
While the above snippet does not make use of the remaining 24 bits (it merely demonstrates the "use 40 bits" part), something akin to the following will be necessary to really make the approach useful in a sense of preserving memory -- presumed that you indeed have other "useful" data to put in the holes:
```C
struct using_gaps
{
uint64_t var : 40;
uint64_t useful_uint16 : 16;
uint64_t char_or_bool : 8;
};
```
Structure size and alignment will be equal to a 64 bit integer, so nothing is wasted if you make e.g. an array of a billion such structures (even without using compiler-specific extensions). If you don't have use for an 8-bit value, you could also use an 48-bit and a 16-bit value (giving a bigger overflow margin).
Alternatively you could, at the expense of usability, put 8 40-bit values into a structure (least common multiple of 40 and 64 being 320 = 8*40). Of course then your code which accesses elements in the array of structures will become much more complicated (though one could probably implement an operator[] that restores the linear array functionality and hides the structure complexity).
Update:
Wrote a quick test suite, just to see what overhead the bitfields (and operator overloading with bitfield refs) would have. Posted code (due to length) at gcc.godbolt.org, test output from my Win7-64 machine is:
```TXT
Running test for array size = 1048576
what alloc seq(w) seq(r) rand(w) rand(r) free
-----------------------------------------------------------
uint32_t 0 2 1 35 35 1
uint64_t 0 3 3 35 35 1
bad40_t 0 5 3 35 35 1
packed40_t 0 7 4 48 49 1
Running test for array size = 16777216
what alloc seq(w) seq(r) rand(w) rand(r) free
-----------------------------------------------------------
uint32_t 0 38 14 560 555 8
uint64_t 0 81 22 565 554 17
bad40_t 0 85 25 565 561 16
packed40_t 0 151 75 765 774 16
Running test for array size = 134217728
what alloc seq(w) seq(r) rand(w) rand(r) free
-----------------------------------------------------------
uint32_t 0 312 100 4480 4441 65
uint64_t 0 648 172 4482 4490 130
bad40_t 0 682 193 4573 4492 130
packed40_t 0 1164 552 6181 6176 130
```
What one can see is that the extra overhead of bitfields is neglegible, but the operator overloading with bitfield reference as a convenience thing is rather drastic (about 3x increase) when accessing data linearly in a cache-friendly manner. On the other hand, on random access it barely even matters.
These timings suggest that simply using 64-bit integers would be better since they are still faster overall than bitfields (despite touching more memory), but of course they do not take into account the cost of page faults with much bigger datasets. It might look very different once you run out of physical RAM (I didn't test that).
------
via:[stackoverflow](http://stackoverflow.com/questions/27705409/if-a-32-bit-integer-overflows-can-we-use-a-40-bit-structure-instead-of-a-64-bit/27705562#27705562)
作者:[Damon][a][Michael Kohne][b]
译者:[译者ID](https://github.com/译者ID)
校对:[校对者ID](https://github.com/校对者ID)
本文由 [LCTT](https://github.com/LCTT/TranslateProject) 原创翻译,[Linux中国](http://linux.cn/) 荣誉推出
[a]:http://stackoverflow.com/users/572743/damon
[b]:http://stackoverflow.com/users/5801/michael-kohne

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如果使用32位整型会溢出那么是否可以使用一个40位结构体代替64位长整型
---------
#问题
假如说使用32位的整型会溢出在不考虑使用长整型的情况下如果我们只需要表示2的40次方范围内的数是否可以利用某些40位长的数据类型来表示呢这样的话每个整型数就可以节省24位的空间。
如果可以,该怎么做?
需求是:我现在必须处理数以亿计的数字,所以在存储空间上受到了很大的限制。
#回答
##可以是可以,但是……
这种方法的确可行,但这么做通常没什么意义(因为几乎没有程序需要处理多达十亿的数字):
```C
#include <stdint.h> // 不要考虑使用long long类型
struct bad_idea
{
uint64_t var : 40;
};
```
在这里变量var占据40位大小但是以生成代码时拥有非常低的运行效率来换取的事实证明“非常”二字言过其实了——测试中程序开销仅仅增加了1%到2%正如下面的测试时间所示而且这么做通常没什么用。除非你还需要保存一个24位的值或者是8位、16位的值这样你皆可以它们放到同一个结构中。不然的话因为对齐内存地址产生的开销会抵消这么做带来的好处。
在任何情况下,除非你是真的需要保存数以亿计的数字,否则这样做给内存消耗带来的好处是可以忽略不计的(但是为了处理这些位字段的额外代码量是不可忽略的!)。
###说明:
在此期间这个问题已经被更新了是为了说明实际上确实有需要处理数以亿计数字的情况。假设采取某些措施来防止因为结构体对齐和填充抵消好处比如在后24位中存储其它的内容或者使用多个8位来存储40位那么这么做就变得有意义了。
如果有十亿个数每个数都节省三个字节的空间那么这么做就非常有用了。因为使用更小的空间存储要求更少的内存页也就会产生更少的cache和TLB不命中和内存缺页单个缺页会产生数以千万计的指令译者注直译是这样但语义说不通
尽管上面提到的情况不足以充分利用到剩余的24位它仅仅使用了40位部分如果确实在剩余位中放入了有用的数据那么使用类似下面的方法会使得这种思路就管理内存而言显得非常有用。
```C
struct using_gaps
{
uint64_t var : 40;
uint64_t useful_uint16 : 16;
uint64_t char_or_bool : 8;
};
```
结构体大小和对齐长度等于64位整型的大小所以只要使用得当就不会浪费空间比如对一个保存10亿个数的数组使用这个结构不考虑使用指定编译器的扩展。如果你不会用到一个8位的值那么你可以使用一个48位和16位的值giving a bigger overflow margin
或者以牺牲可用性为代价把8个64位的值放入这样的结构体中或者使用40和64的组合使得其和满足320。当然在这种情况下通过代码去访问数组结构体中的元素会变得非常麻烦尽管一种方法是实现一个operator[]在功能上还原线性数组,隐藏结构体的复杂性)。
更新:
我写了一个快速测试工具只是为了获得位字段的开销以及伴随位字段引用的重载操作。由于长度限制将代码发布在gcc.godbolt.org上在本人64位Win7上的测试结果如下
```TXT
运行测试的数组大小为1048576
what alloc seq(w) seq(r) rand(w) rand(r) free
-----------------------------------------------------------
uint32_t 0 2 1 35 35 1
uint64_t 0 3 3 35 35 1
bad40_t 0 5 3 35 35 1
packed40_t 0 7 4 48 49 1
运行测试的数组大小为16777216
what alloc seq(w) seq(r) rand(w) rand(r) free
-----------------------------------------------------------
uint32_t 0 38 14 560 555 8
uint64_t 0 81 22 565 554 17
bad40_t 0 85 25 565 561 16
packed40_t 0 151 75 765 774 16
运行测试的数组大小为134177228
what alloc seq(w) seq(r) rand(w) rand(r) free
-----------------------------------------------------------
uint32_t 0 312 100 4480 4441 65
uint64_t 0 648 172 4482 4490 130
bad40_t 0 682 193 4573 4492 130
packed40_t 0 1164 552 6181 6176 130
```
我们看到位字段的额外开销是微不足道的但是当以友好的方式线性访问数据时伴随位字段引用的操作符重载产生的开销则相当显著大概有3倍。在另一方面随机访问产生的开销则无足轻重。
这些时间表明简单的使用64位整型会更好因为它们在整体性能上要比位字段好尽管占用更多的内存但是显然它们并没有考虑随着数据集增大带来的缺页开销。一旦程序内存超过RAM大小结果可能就不一样了未亲自考证
------
via:[stackoverflow](http://stackoverflow.com/questions/27705409/if-a-32-bit-integer-overflows-can-we-use-a-40-bit-structure-instead-of-a-64-bit/27705562#27705562)
作者:[Damon][a][Michael Kohne][b]
译者:[KayGuoWhu](https://github.com/KayGuoWhu)
校对:[校对者ID](https://github.com/校对者ID)
本文由 [LCTT](https://github.com/LCTT/TranslateProject) 原创翻译,[Linux中国](http://linux.cn/) 荣誉推出
[a]:http://stackoverflow.com/users/572743/damon
[b]:http://stackoverflow.com/users/5801/michael-kohne