loaded question解释

本文探讨了既定观点问题的概念,通过具体例子展示了如何在对话中利用这种技巧,并引用了历史实例来说明其在实际场景中的运用。文章特别提到了在新西兰体罚公投和美国国家篮球协会总裁对电台主持人的提问中使用既定观点问题的例子。

详细网址http://zh.wikipedia.org/wiki/%E6%97%A2%E5%AE%9A%E8%A7%80%E9%BB%9E%E5%95%8F%E9%A1%8C

既定观点问题(英语:loaded question[1]),有时也称复合问题(英语:complex question;拉丁语:plurium interrogationum),系在问题中掺入默认的观点,对方一旦回答,就等同承认这些观点是真的。如果回答者并未同意该默认,仅是因正面回答问题而被迫承认,以其回答证成默认则属谬误。
例一
甲:你还在抽烟吗?
乙:没有。
甲:看吧!你果然抽过烟!
甲的问题已默认乙抽过烟,乙无论回答“是”或“否”,皆无异于承认以前抽过烟。 如果乙根本没抽过烟,最安全的做法是回答“我从未抽过烟。”或干脆不回答。
例二
“你不打老婆了吗?”
若正面回答,等同承认有老婆,且以前曾经打过老婆。
例三
“为什么私人企业总是比公营企业有效率?”
若正面回答,等同承认“私人企业总是比公营企业有效率”是事实。
历史实例[编辑]
2009年新西兰体罚公投时曾问以下问题:
"Should a smack as part of good parental correction be a criminal offence in New Zealand?"(中译:做为好的家长管教方法的一部份的掴打该被禁止吗?)
慕来‧艾多里吉(Murray Edridge)指称这问题本身“带有既定观点且模棱两可”,并宣称道“这问题已预先假定掴打是做好的好家长管教方法的一部份”。[2]
2012年6月13日,美国国家篮球协会总裁大卫·斯特恩曾向电台主持人吉姆‧罗姆(Jim Rome)问道:“你不再打你老婆了吗?”借此以表达他对罗姆的访谈的感觉。

### Question B1: Add-and-Shift Multiplier #### Part a: Circuit Diagram of the Data Path **Circuit Diagram:** ``` +-----------------+ | 4-bit Adder | +--------+--------+ | | | | v v +--------+--------+ | 9-bit Accumulator (ACC) | | (5 MSB, 4 LSB) | +--------+--------+ | v +--------+--------+ | 4-bit Multiplier (M) | +-----------------+ ``` - **4-bit Adder:** Adds the multiplicand (A) and the current value in the lower 4 bits of the accumulator. - **9-bit Accumulator (ACC):** Stores the intermediate results. The upper 5 bits are initially cleared, and the lower 4 bits are loaded with the multiplier. - **4-bit Multiplier (M):** Stores the multiplier. #### Part b: State Graph for the Control Circuit **State Graph:** ``` S0 (Initial) -> S1 (Load) -> S2 (Add) -> S3 (Shift) -> S4 (Check Completion) | | +-----------------------------------+ ``` - **S0 (Initial):** Wait for the start signal (St = 1). - **S1 (Load):** Load the multiplier into the lower 4 bits of the accumulator and clear the upper 5 bits. - **S2 (Add):** If the least significant bit (LSB) of the multiplier is 1, add the multiplicand to the accumulator. - **S3 (Shift):** Shift the entire 9-bit accumulator right by one position. - **S4 (Check Completion):** Check the counter value. If the counter reaches the maximum value (indicating all shifts are done), set the Done signal and return to S0. Otherwise, go back to S2. #### Part c: Table for Multiplication Process Given: - Multiplicand (A) = 12 (1100 in binary) - Multiplier (M) = 10 (1010 in binary) **Table:** | Time | State | Counter | Product Register | St | M | K | Load | Ad | Sh | Done | |--|------------------|----|---|---|------|----|----|------| | t0 | S0 | 00 | 000000000 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | | t1 | S0 | 00 | 000000000 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | | t2 | S1 | 00 | 000001010 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | | t3 | S2 | 00 | 000001010 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | | t4 | S3 | 01 | 000000101 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | | t5 | S2 | 01 | 000000101 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | | t6 | S3 | 10 | 000000010 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | | t7 | S2 | 10 | 000000010 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | | t8 | S3 | 11 | 000000001 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | | t9 | S2 | 11 | 000000001 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | | t10 | S3 | 11 | 000000000 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | **Explanation:** - **t1:** Start signal (St = 1) is detected, and the multiplier is loaded into the lower 4 bits of the accumulator. - **t2:** The initial state after loading the multiplier. - **t3:** The first addition is performed since the LSB of the multiplier is 1. - **t4:** The result is shifted right. - **t5-t10:** The process continues until the counter reaches its maximum value (11), indicating all shifts are complete. The Done signal is set to 1, and the multiplication process is complete.
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