示例
Demonstrate that Set#<=> implements subset partial ordering rather than total ordering, returning nil for incomparable sets and causing Array#sort to raise ArgumentError.
sha256:72cc01c64895659755d1fa04f846f49f48adf91a79aee92dd664dde5076fe8ed
PUBLISHED
L3_CONTRACT_PASS
MIT-0
执行证据
分开显示声明环境和签名验证运行,明确样本的证明范围。
证据依据签名契约通过
验证回执1
验证级别L3_CONTRACT_PASS
声明的环境
- 执行上下文
- ruby
- 操作系统
- linux
- 架构
- x64
- 运行时
- ruby
- 语言
- ruby
- 包管理器
- bundler
验证运行环境
- 执行上下文
- ruby 3
- 操作系统
- linux debian · glibc
- 架构
- x64
- 运行时
- ruby 3
- 语言
- ruby
- 包管理器
- bundler
- 执行方式
- container · docker
CONTAINER_RUN · compile:SKIPPED · contract:PASS · load:PASS · resolve:PASS · rubygems@1 · 2026-08-17
案例
- 目标
- Demonstrate that Set#<=> implements subset partial ordering rather than total ordering, returning nil for incomparable sets and causing Array#sort to raise ArgumentError. HOW
- 包
-
set 1.1.3
- 环境
- ruby
- 创建时间
- 2026-08-17T07:23:19Z
常见的想当然
Set#<=> provides a total ordering across any sets so arrays of sets can be sorted with Array#sort.
这是本样本作者记下的、开发者或模型在此处通常会有的预期。下面的契约才是真正运行过的东西。
契约
- Set#<=> returns nil when neither set is a subset of the other, which causes Array#sort on incomparable sets to raise ArgumentError.
- Set#<=> returns -1, 0, or 1 only when sets are in a proper subset, equal, or proper superset relationship.
- Array#sort on an array of sets succeeds only when all elements can be ordered by subset inclusion.
文件
- Gemfile
- Gemfile.lock
- NOTES.md
- csx.json
- test/contract.rb
下载源代码构件 (tar.gz)
原始种子者
csx-seed
验证回执
- ruby 3 · linux debian/x64 · docker · CONTAINER_RUN · compile:SKIPPED · contract:PASS · load:PASS · resolve:PASS · rubygems@1 · 2026-08-17 · ed25519:d91480838ac982c9