mips

问题 I am trying to print binary digits of any integer input and store them in the array starting at the last index. Then I am trying to print it from the array. .data prompt: .asciiz "Enter an int: " errorLarge: .asciiz "Error Value to large CANNOT be held in 16 bit" errorSmall: .asciiz "Error Value is to small CANNOT be held in 16 bits" # 64bytes =512 bits created (1 int =4 bytes):: (16 int =64 bytes) array: .space 64 newLine: .asciiz "\n" .globl main .text main: li $v0,4 la $a0,prompt syscall

问题 I have the following data segment .data a: .byte 0x11 .align 1 b: .word 0x22334455 Assuming that address "a" is 0x10010000, then the expected address for the word at b is 0x10010002, but MARS stores the word at 0x10010004, ignoring the explicit ".align" directive. By the way, I used MARS MIPS simulator (Version 4.5 on a MacBook Pro) to assemble the above code. Therefore, my question is: Is this a bug, or is it expected that the behavior of MARS differs from SGI's 1992 documentation for MIPS

问题 I have the following data segment .data a: .byte 0x11 .align 1 b: .word 0x22334455 Assuming that address "a" is 0x10010000, then the expected address for the word at b is 0x10010002, but MARS stores the word at 0x10010004, ignoring the explicit ".align" directive. By the way, I used MARS MIPS simulator (Version 4.5 on a MacBook Pro) to assemble the above code. Therefore, my question is: Is this a bug, or is it expected that the behavior of MARS differs from SGI's 1992 documentation for MIPS

问题 B[8] = A[i–j]; how do you use the arrays while trying to convert into MIPS the above C expression? i thought something like : lw $t0, 16 ($s7) sub $t1, $s3, $s4 lw $t2, $t1 ($s6) considering that $s6 holds A $s7 --- B $s3 ---- i and $s4 -- j but am not sure about how to do the calculationg to actually get on A [i-j] any suggestion? 回答1: Well, first off, 16($s7) is wrong. Assuming that B and A are arrays of 32-bit integers, the offset is 8 * 4 = 32 . Next, simply add $t1 and $s6 and load from

问题 My professor gave a video that looks like this: In the lower right, he wrote $ra at location 124 while the $sp is at 128 which implies that the first sw $ra, 4($sp) instruction stores the $ra value at a location 4 bytes less than the $sp . But my book does it differently: and The image implies that the lw instruction stores it at locations larger, more positive numbers than the $sp . So which is right? Does lw and sw offset numbers refer to numbers higher or lower than the $sp ? 回答1: You are

问题 This is an assignment. the idea is to prompt the user to: input the first positive integer input any one of the operations /,*,-,+ input the second positive integer Then, call do_math which calls one of: do_add if operator is +, add the 2 inputs without using add or sub mips instruction and the return the result in a register. return an error condition if there is an overflow. do_sub if operator is -, subtract the second integer from the first without using sub or add mips instruction (you

问题 Followup to this I figured out what went wrong in the linked post. The Sigma0 and Sigma1 boxes weren't doing what they should: modulo 2 addition of the rotated versions of either A or E. Basically, it should work like this: Notice that if the number of bits in a position is even the result is 0, if odd it's 1. To do this I made a truth table and karnaugh maps: LINK I simplified the resulting expression to !A!BC + A(B XNOR C) And I tried to apply it to Sigma0 and Sigma1. But it's not working.

问题 From the man pages of OS161 : Synopsis #include <unistd.h> #include <fcntl.h> int open(const char *filename, int flags); int open(const char *filename, int flags, mode_t mode); How the standard c library function open is defined: int open(const char *filename, int flags, ...); The declaration: /* * Definition for each syscall. * All we do is load the syscall number into v0, the register the * kernel expects to find it in, and jump to the shared syscall code. * (Note that the addiu instruction

问题 I'm using a MIPS cross compiler on my Linux machine which works great. Now I need to compile the same application on Windows. I'm searching the web for some prebuilt MIPS cross compiler (with toolchain) for MS-Windows, but without success. Since I'm not sure how to do so, I'm asking if someone knows such prebuilt toolchain? or some guide how to convert my Linux toolchain for windows if it's possible? Thanks. 回答1: You should use Codescape MIPS SDK. 回答2: It looks like Mentor has taken away the

问题 In MIPS, a 32-bit jump instruction consists of 6-bits for the opcode and 26-bits for the target (destination) address that we want to set the program counter to. However, it is possible to set the program counter to a 28-bit target address. How is this possible if we can only fit 26-bits in the jump instruction? 回答1: Instructions on MIPS are always 4-byte aligned, so the low 2 bits of any valid target address can only be zero. Thus, the 26 bits specified in jump instructions are always