{"resolved":true,"match_type":"uuid","source_type":"original_book","book_id":"38270854-596f-4a26-8183-db8c968c6020","title":"编码 隐匿在计算机软硬件背后的语言","author":null,"edition":null,"publisher":null,"year":null,"isbn":"9787121181184","requested_by":"printed_page","book":{"source_type":"original_book","book_id":"38270854-596f-4a26-8183-db8c968c6020","id":"38270854-596f-4a26-8183-db8c968c6020","uuid":"38270854-596f-4a26-8183-db8c968c6020","title":"编码 隐匿在计算机软硬件背后的语言","normalized_title":"编码隐匿在计算机软硬件背后的语言","author":null,"edition":null,"publisher":null,"year":null,"isbn":"9787121181184","isbns":["9787121181184","0123456789","6044921875","3039331000","0000000000","0000000000000"],"parsed":true,"ai_readable":true,"document_type":"book","page_count":438,"navigation_url":"https://workshop.closeclaw.top/ai/book/38270854-596f-4a26-8183-db8c968c6020","title_url":"https://workshop.closeclaw.top/ai/title/%E7%BC%96%E7%A0%81%20%E9%9A%90%E5%8C%BF%E5%9C%A8%E8%AE%A1%E7%AE%97%E6%9C%BA%E8%BD%AF%E7%A1%AC%E4%BB%B6%E8%83%8C%E5%90%8E%E7%9A%84%E8%AF%AD%E8%A8%80","fixed_page_url_template":"https://workshop.closeclaw.top/ai/book/38270854-596f-4a26-8183-db8c968c6020/page/{printed_page}","path_api_url_template":"https://workshop.closeclaw.top/api/ai/book/38270854-596f-4a26-8183-db8c968c6020/page/{printed_page}","read_url":"https://workshop.closeclaw.top/api/ai/read/38270854-596f-4a26-8183-db8c968c6020","page_url":"https://workshop.closeclaw.top/api/ai/books/38270854-596f-4a26-8183-db8c968c6020/page","search_url":"https://workshop.closeclaw.top/api/ai/books/38270854-596f-4a26-8183-db8c968c6020/search"},"page":{"printed_page":152,"pdf_page":168,"physical_index":168,"physical_index_base":0,"printed_page_status":"recognized","page_sha256":"07db3f4d9648307430d7bbf6858eb72fee2b9efbafce732095db5f0eb039b2d1"},"chapter":"编码隐匿在计算机软硬件背后的语言","content":"我们把这两个数颠倒位置后再做一遍。在十进制中，减法题目对应于：\n\n$$\n\\frac { - 2 5 3 } { 3 7 3 }\n$$\n\n而用二进制表示为：\n\n$$\n\\frac { - 1 1 1 1 1 0 } { ? ? ? 2 ? 2 ? 2 }\n$$\n\n第一步用11111111减去减数，得到对1的补数：\n\n$$\n\\frac { - 1 1 1 1 1 1 0 1 } { 0 0 0 0 0 1 0 }\n$$\n\n第二步，将减数对1的补数与被减数相加：\n\n$$\n\\frac { 1 0 1 1 0 0 0 0 } { 1 0 1 1 0 0 1 0 }\n$$\n\n现在我们要用某种方法在结果中减去11111111。当减数小于被减数的时候，我们将结果加1再减100000000来完成计算。但是你无法在不借位的情况下做到这一点。所以，我们先用11111111减去所得结果：\n\n$$\n\\frac { - 1 0 1 1 0 0 1 0 } { 0 1 0 0 1 1 0 1 }\n$$\n\n这里又一次用到了将各位取反来求得结果的方法，但是这个结果是77，而真正的答案应该是-77。\n\n到这里，我们已经有了足够的条件来改造上一章所搭建的加法器，并让它像实现加法一样来实现减法运算。为了不让问题太复杂，这个新的加/减法器只执行在减数小于被减数的减法操作，即结果为正数的操作。\n\n该加法器的核心是由逻辑门集成的8位全加器。\n\n编码—隐匿在计算机软硬件背后的语言\n\n152","previous_page":151,"next_page":153}