{"kind":"task","effective_mode":"full","benchmark":{"kind":"benchmark","effective_mode":"full","slug":"humaneval-plus","formal_name":"HumanEval+","introduction":"HumanEval+ keeps the 164 original HumanEval problems and multiplies their tests by roughly 80×. It exists because the original suite was loose enough to pass implementations that were actually wrong.","introduction_ja":"","introduction_en":"","category":"Category not supplied","task_count":null,"acquisition_status":"Acquisition status not supplied","official_url":"https://github.com/evalplus/evalplus","indexing_mode":"noindex","profile":{"resources":[],"task_format":"","scoring":"","metric":"","size":"","answer_access":"","license":"","citation":"","maintainer":"","released":"","why_hard":"","related":[]}},"task_id":"e8f83e30-16b8-5927-adb2-a53b420f46e5","task_key":"default--test--e8f83e30-16b8-5927-adb2-a53b420f46e5","task_revision_id":"2","upstream_id":"","short_description":"def truncate_number(number: float) -> float:","config":"default","split":"test","body":"{\"entry_point\":\"truncate_number\",\"prompt\":\"\\n\\ndef truncate_number(number: float) -> float:\\n    \\\"\\\"\\\" Given a positive floating point number, it can be decomposed into\\n    and integer part (largest integer smaller than given number) and decimals\\n    (leftover part always smaller than 1).\\n\\n    Return the decimal part of the number.\\n    >>> truncate_number(3.5)\\n    0.5\\n    \\\"\\\"\\\"\\n\"}","display_format":"code","language":"","answer_status":"published","assets":[],"source_url":"https://github.com/evalplus/evalplus","history":"initial import","indexing_mode":"noindex","subproblems":[],"grids":[]}