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【Hackathon 5th No.34】 为 Paddle 新增 bitwise_right_shift / bitwise_right_shift_ / bitwise_left_shift / bitwise_left_shift_ API (update RFC) #788
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@xuxinyi389 补充了更详细的描述,对比了竞品的具体实现,pr中bitwise_functors.h的实现和这一部分相对应,辛苦review~
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具体行为定义:(`n_bits`表示数据类型存储位数,例如int8的`n_bits`为8,uint16的`n_bits`为16;当`y`小于0时为“未定义行为”,等效于位移超过最大位数溢出) | ||
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+ 算术位移 |
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注意到一种情形,比如x为有符号数,比如x=-45(int8),无论是算术移位还是逻辑移位,其左移2位为76,左移3位为-104。这种有符号数左移出现时正时负的情形,可能会让人迷惑,RFC中还可以详细描述下不同情形下,对于有符号数x的符号位的处理逻辑,相应的例子和描述也可以更新到API文档中。
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注意到一种情形,比如x为有符号数,比如x=-45(int8),无论是算术移位还是逻辑移位,其左移2位为76,左移3位为-104。这种有符号数左移出现时正时负的情形,可能会让人迷惑,RFC中还可以详细描述下不同情形下,对于有符号数x的符号位的处理逻辑,相应的例子和描述也可以更新到API文档中。
嗯嗯好,这种属于溢出,位移溢出的结果都是不可控无意义的,我说明一下~
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补充下竞品对于未定义操作的的实际处理方式,看是否和当前现有设计方案是一致的
+ 有符号数时,例如`int8 x=-100`,补码为`1001,1100`,最高位为符号位,仅需要右移7位,所有的`int8`就都会变成`1111,1111`,即`-1`; | ||
+ 无符号数时候,例如`uint8 x=200`,存储为`1100,1000`,八位均表示数值大小,需要右移8位才可以将所有的`uint8`变为`0000,0000`,即`0`; | ||
+ 当`b`位负数这一未定义行为时,同样等效于右移无穷位,与移动`max_shift`等效,有符号数变为`-1`,无符号数变为`0` | ||
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错别字,“位“应该为”为“。对于未定义的行为怎么处理,应该列出竞品实际是如何处理的,处理方式是否一致,这些都是要说清楚的。
@xuxinyi389 应该比较详细了,哪里有疏漏的话我再补充一些,辛苦review~ |
@@ -145,7 +145,9 @@ void rshift_kernel_cuda(TensorIteratorBase& iter) { | |||
+ 算术右移时,`max_shift`需要考虑最大的移动距离,有符号数最高位为符号位,故表示数值的位数实际上会少一位。 | |||
+ 有符号数时,例如`int8 x=-100`,补码为`1001,1100`,最高位为符号位,仅需要右移7位,所有的`int8`就都会变成`1111,1111`,即`-1`; | |||
+ 无符号数时候,例如`uint8 x=200`,存储为`1100,1000`,八位均表示数值大小,需要右移8位才可以将所有的`uint8`变为`0000,0000`,即`0`; | |||
+ 当`b`位负数这一未定义行为时,同样等效于右移无穷位,与移动`max_shift`等效,有符号数变为`-1`,无符号数变为`0` | |||
+ 当`b`位负数这一未定义行为时,同样等效于右移无穷位,与移动`max_shift`等效(从代码中可以看到,`b<0`和`b>=max_shift`是在同一个`if`判断中),只要满足两个条件中任意一个,则使得有符号数变为`-1`,无符号数变为`0`。 |
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“当‘b’位负数这一未定义行为时“,有个错别字“位“应该为”为“
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“当‘b’位负数这一未定义行为时“,有个错别字“位“应该为”为“
sry,更正了~
LGTM |
补充和更新api设计细节
原RFC:
PR: