mips

问题 Let's say I have $t0 , and I'd like to divide its integer contents by two, and store it in $t1 . My gut says: srl $t1, $t0, 2 ... but wouldn't that be a problem if... say... the right-most bit was 1? Or does it all come out in the wash because the right-most bit (if positive) makes $t0 an odd number, which becomes even when divided? Teach me, O wise ones... 回答1: Use instruction sra: Shift right arithmetic !! sra $t1, $t0, 1 Divides the content of $t0 by the first power of 2. Description:

来源： https://stackoverflow.com/questions/37153381/why-is-my-mips-base-converter-printing-out-the-values-from-a-previous-loop-after

来源： https://stackoverflow.com/questions/37153381/why-is-my-mips-base-converter-printing-out-the-values-from-a-previous-loop-after

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来源： https://stackoverflow.com/questions/38045665/store-certain-bits-from-one-register-into-certain-bits-of-another-register

来源： https://stackoverflow.com/questions/38045665/store-certain-bits-from-one-register-into-certain-bits-of-another-register

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来源： https://stackoverflow.com/questions/23123106/mips-i-instruction-immediate-field