我正在寻找一种生成前缀位掩码的便携式方法,该前缀具有为@Component
@Order(SwaggerPluginSupport.SWAGGER_PLUGIN_ORDER + 1000)
public class SwaggerIncludeMissingNicknameIntoUniqueIdReader implements OperationBuilderPlugin {
@Override
public void apply(OperationContext context) {
Optional<ApiOperation> methodAnnotation = context.findControllerAnnotation(ApiOperation.class);
Operation operationBuilder = context.operationBuilder().build();
String uniqueId = operationBuilder.getUniqueId().replaceAll("Using(GET|POST|PUT|DELETE)", "");
// If nickname exists, populate the value of nickname annotation into uniqueId
String fillId = methodAnnotation.transform(ApiOperation::nickname).or(uniqueId);
context.operationBuilder().uniqueId(fillId);
context.operationBuilder().codegenMethodNameStem(fillId);
}
@Override
public boolean supports(DocumentationType delimiter) {
return SwaggerPluginSupport.pluginDoesApply(delimiter);
}
}
(或64或任意整数类型的位宽)设置的前n
位。
示例:
0 <= n <= 32
如果我们忽略案例prefix_bitmask(0) = 0b00000000000000000000000000000000u
prefix_bitmask(4) = 0b00000000000000000000000000001111u
prefix_bitmask(32) = 0b11111111111111111111111111111111u
或n == 0
,有两种方法已经可以起作用:
n == 32
// "constructive": set only the required bits
uint32_t prefix_mask1(int i) { return (uint32_t(1) << i) - 1; }
// "destructive": shift unneeded bits out
uint32_t prefix_mask2(int i) { return ~uint32_t(0) >> (32 - i); }
失败32且prefix_mask1
失败0都是因为大于整数类型的移位是未定义的行为(因为允许CPU仅使用移位大小的最低5位)。
是否有一种“规范”的方法可以解决此问题而无需分支?
答案 0 :(得分:4)
((uint32_t) 1 << i/2 << i-i/2) - 1
。
以上方法可以将uint32_t
替换为任何无符号类型。并且不需要其他更改。需要知道类型中的位数b
和掩码m
= 2 b
-1的其他选项包括:
((uint32_t) 1 << (i & m)) - 1 - (i >> b)
(来自supercat)
和:
((uint32_t) i >> b) ^ 1) << (i & m)) - 1
(来自Matt Timmermans的建议)。
答案 1 :(得分:3)
这可以通过使用prefix_mask2
想法和算术移位来完成,以准备正确的模式来完成,总共有3条指令(假定CPU中的移位计数为模数字宽):
// minimal instruction dependency (2 cycles), but requires large constant
// that some architectures have trouble generating
uint32_t prefix_mask2a(int i) {
return ((int32_t) (i + (0x80000000 - 32))) >> ((i ^ 31) & 31);
}
// 3 cycles
uint32_t prefix_mask2b(int i) {
return (uint32_t) ((int32_t) -i >> 31) >> (-i & 31);
}
答案 2 :(得分:2)
您可以将uint32_t
投射到更多位上,进行移位,然后再转换回去:
uint32_t prefix_mask(int i) {
return UINT32_MAX & ((UINT64_C(1) << i) - 1);
}
答案 3 :(得分:0)
我认为它很便携
#define PREFIX(type, n) (type)(((sizeof(type) * CHAR_BIT - (n)) == sizeof(type) * CHAR_BIT) ? ((type)0) : (!(sizeof(type) * CHAR_BIT - (n)) ? (~(type)(0)) : ((~(type)(0)) << (sizeof(type) * CHAR_BIT - n))))
#define POSTFIX(type, n) (type)(((sizeof(type) * CHAR_BIT - (n)) == sizeof(type) * CHAR_BIT) ? ((type)0) : (!(sizeof(type) * CHAR_BIT - (n)) ? (~(type)(0)) : ((~(type)(0)) >> (sizeof(type) * CHAR_BIT - n))))
#define TEST_TYPE unsigned long long
void printbin(TEST_TYPE x)
{
TEST_TYPE mask = (TEST_TYPE)1 << (sizeof(x) * CHAR_BIT - 1);
while(mask)
{
printf("%d", !!(x & mask));
mask >>= 1;
}
}
int main()
{
for(int x = 0; x <= sizeof(TEST_TYPE) * CHAR_BIT; x++)
{
printbin(PREFIX(TEST_TYPE, x)); printf("\n");
}
printf("\n");
for(int x = 0; x <= sizeof(TEST_TYPE) * CHAR_BIT; x++)
{
printbin(POSTFIX(TEST_TYPE, x)); printf("\n");
}
}