CODEWITHSUNDEEP
cbit-shiftbitmaskmachine-code
KunaiBunsen
KunaiBunsen

Reputation: 55

How do i mask the bits i dont want from my machinecode?

So I have a txt file that I'm taking numbers in like 02324020 which is the machine code from the MIPS instruction: add $t0, $s1, $s2

When it's in binary and cut into sections of 6,5,5,5,5,6 the numbers taken are: 0,17,18,8,0,20

This gives the idea of the instruction as R[rd] = R[rs]+R[rt] and the number from 26-31 is opcode 0 meaning R format, rd=8, rs=17, rt=18

I'm trying to mask the bits so I only see the bits between 21-25 and then receive the number that they are: 0,17,18,8,0,20 because shifting to the right: printf("\n%x\n", code >>26); gives me the opcode 0 and tells me its the R format.

So now I'm trying to find the 17 but I don't know what to place after & 0b in:

printf("\n%x\n", code >>21 & 0b____);

I could really use an example to find the remaining numbers from my number 02324020

This is my code so far:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char* argv[])
{

    unsigned int code = 0;
    FILE *fp;
    fp = fopen(argv[1], "r+");
    int result = fscanf(fp, "%x", &code);
    printf("\n hex-1: %08X, decimal-1: %d \n", code, code);

    printf("\n%x\n", code >>26);
    printf("\n%x\n", code >>21 & 0x0b00000011111); //what can i do to change this
    while (result != EOF)
    {
        result = fscanf(fp, "%x", &code);
        printf("\n read hex: %08X, decimal: %d \n", code, code);
        if(feof(fp))
        {
            puts("EOF");
            break;
        }//if
    }//while
}

Upvotes: 1

Views: 106

Answers (1)

frslm
frslm

Reputation: 2978

To define a binary number, you'd use 0b___ (as you've mentioned in your answer), not 0x0b0___.

To extract the number you need, you'd have to bit-shift code so that the bits you're interested in are at the right-most end. Then OR only those bits you're interested in. Putting this together, you have:

printf("\n%x\n", code >> 26);
printf("\n%x\n", code >> 21 &  0b11111);
printf("\n%x\n", code >> 16 &  0b11111);
printf("\n%x\n", code >> 11 &  0b11111);
printf("\n%x\n", code >>  6 &  0b11111);
printf("\n%x\n", code       & 0b111111);

Upvotes: 1

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