mips

问题 I'm trying to get my feet wet with MIPS assembly language using the MARS simulator. My main problem now is how do I initialize a set of memory locations so that I can access them later via assembly language instructions? For example, I want to initialize addresses 0x1001000 - 0x10001003 with the values 0x99, 0x87, 0x23, 0x45. I think this can be done in the data declaration (.data) section of my assembly program but I'm not sure of the syntax. Is this possible? Alternatively, in the .data

问题 I'm writing strncpy in MIPS, but I'm having trouble null-terminating the string. If i do not null-terminate it myself, the string goes on and on. I have tried sb $__ 0($0) but that does not seem to work... $a0 = pointer to destination array $a1 = source string $a2 = number of characters to copy strncpy: add $t1 $zero $zero #counter beq $a2 $0 done # if num chars to copy is 0, return. j cpyLoop cpyLoop: beq $t1 $a2 done # if counter == num to copy, end lb $t2 0($a1) # load the character beq

问题 I'm writing strncpy in MIPS, but I'm having trouble null-terminating the string. If i do not null-terminate it myself, the string goes on and on. I have tried sb $__ 0($0) but that does not seem to work... $a0 = pointer to destination array $a1 = source string $a2 = number of characters to copy strncpy: add $t1 $zero $zero #counter beq $a2 $0 done # if num chars to copy is 0, return. j cpyLoop cpyLoop: beq $t1 $a2 done # if counter == num to copy, end lb $t2 0($a1) # load the character beq

问题 I'm writing strncpy in MIPS, but I'm having trouble null-terminating the string. If i do not null-terminate it myself, the string goes on and on. I have tried sb $__ 0($0) but that does not seem to work... $a0 = pointer to destination array $a1 = source string $a2 = number of characters to copy strncpy: add $t1 $zero $zero #counter beq $a2 $0 done # if num chars to copy is 0, return. j cpyLoop cpyLoop: beq $t1 $a2 done # if counter == num to copy, end lb $t2 0($a1) # load the character beq

问题 How do you take in an input of a floating number in MIPS? I have tried using: li.s $f0, 6 syscall But I just keep getting that there is an error with the line. 回答1: li $v0, 6 syscall //the float value that is read will be in $f0 register 回答2: you cannot load-immediate a number into float registers li $t0, 6 # load-immediate 6 into an int register mtc1 $t0, $f0 # copies the bit pattern "...110". It is NOT 6.0!! cvt.s.w. $f12, $f0 # convert word to (single) float. $f12 now contains 6.0 回答3: You

问题 How do you take in an input of a floating number in MIPS? I have tried using: li.s $f0, 6 syscall But I just keep getting that there is an error with the line. 回答1: li $v0, 6 syscall //the float value that is read will be in $f0 register 回答2: you cannot load-immediate a number into float registers li $t0, 6 # load-immediate 6 into an int register mtc1 $t0, $f0 # copies the bit pattern "...110". It is NOT 6.0!! cvt.s.w. $f12, $f0 # convert word to (single) float. $f12 now contains 6.0 回答3: You

问题 Does assembly for MIPS read every label? ignore the task and syntax, I just put something together quick. add reg3, reg1, $zero add reg1, reg1, reg2 beq reg1, reg3, BRANCH1 #reg2 contents are zero bne reg1, $zero, BRANCH2 #reg1 doesn't equal zero BRANCH1: add returnReg, reg1, $zero BRANCH2: add returnReg, reg2, $zero jr returnAddress would this read line-by-line, including the labels, unless they were jumped over? For instance, would BRANCH1 be executed every single time, unless the contents

问题 I have written some code which opens and reads from a file, input.txt, in a loop. I now want to take each word from the first file, and write it to a new file like so. If file 1 says "This is my file", File 2 would say "This /n is /n my /n file" each on a separate line. I have written a label which opens the file to be written to, now I am unsure of what I should write in order to access each word from my input.txt. file_open: li $v0, 13 la $a0, output_file_name # output.txt li $a1, 1 li $a2,

问题 I'm learning MIPS by myself with sources found online* and see a lot of people doing addi $reg, $0, N to store a numerical value N inside a register $reg , and am wondering why is this so ubiquitous when li $reg N would do. What am I missing ? *(so don't have exhaustive while accessible explanations of what I learn) 回答1: Assembler has: directives — .text , .data , .align 2 , others labels — main: regular instructions — addi $a0, $0, 1 and beq $a0, $a1, label pseudo instructions — li $a0, 1

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: