将C程序转换为汇编代码

Question

How exactly do I convert this C program into assembly code? I am having a hard time understanding this process or how to even start it. I am new to this. Any help would be appreciated!

while(a!=b){
     if(a > b){
        a = a - b;
       }
        else{
          b = b - a;
    }
  }
   return a;
   }

Side Note: Assume two positive integers a and b are already given in register R0 and R1.
Can you leave comments explaining how you did it?

Answer 1

If you are using gcc , you can get the assembly as gcc -S -o as ac if your source code is ac . If you are using Visual Studio, you can get it when you debug by selecting the "disassembly" window. Here is the output of Visual studio (I named the subrountine/function called "common" that's why "common" appears):

    while(a!=b){
    003613DE  mov         eax,dword ptr [a]  
    003613E1  cmp         eax,dword ptr [b]  
    003613E4  je          common+44h (0361404h)  
         if(a > b){
    003613E6  mov         eax,dword ptr [a]  
    003613E9  cmp         eax,dword ptr [b]  
    003613EC  jle         common+39h (03613F9h)  
            a = a - b;
    003613EE  mov         eax,dword ptr [a]  
    003613F1  sub         eax,dword ptr [b]  
    003613F4  mov         dword ptr [a],eax  
         }
         else{
    003613F7  jmp         common+42h (0361402h)  
             b = b - a;
    003613F9  mov         eax,dword ptr [b]  
    003613FC  sub         eax,dword ptr [a]  
    003613FF  mov         dword ptr [b],eax  
        }
      }
    00361402  jmp         common+1Eh (03613DEh)  
       return a;
    00361404  mov         eax,dword ptr [a]  
    }

Here variable a is saved in memory initially and so is b ( dword ptr [b] ).

Answer 2

The professor that taught me system programming used what he called 'atomic-C' as a stepping stone between C and assembly. The rules for atomic-C are (to the best of my recollection):

1. only simple expressions allowed, ie a = b + c; is allowed a = b + c + d; is not allowed because there are two operators there.
2. only simple boolean expressions are allowed in an if statement, ie if (a < b) is allowed but if (( a < b) && (c < d)) is not allowed.
3. only if statements, no else blocks.
4. no for / while or do-while is allowed, only goto's and label's

So, the above program would translate into;

 label1:
     if (a == b) 
         goto label2;

     if (a < b)
         goto label4;

     a = a - b;
     goto label3;

 label4:
     b = b - a;

 label3:
     goto label1; 

 label2:
     return a;

I hope I got that correct...it has been almost twenty years since I last had to write atomic-C. Now assuming the above is correct, lets start converting some of the atomic-C statements into MIPS (assuming that is what you are using) assembly. From the link provided by Elliott Frisch, we can almost immediately translate the subtraction steps:

a = a - b     becomes R0 = R0 - R1 which is: SUBU R0, R0, R1
b = b - a     becomes R1 = R1 - R0 which is: SUBU R1, R1, R0

I used unsigned subtraction due to both a and b being positive integers.

The comparisons can be done thusly:

if(a == b) goto label2 becomes if(R0 == R1) goto label2 which is: beq R0, R1, L2?

The problem here is that the third parameter of the beq op-code is the displacement that the PC moves. We will not know that value till we are done doing the hand assembly here.

The inequality is more work. If we leave of the pseudo code instructions, we first need to use the set on less than op-code which put a one in destination register if the first register is less than the second. Once we have done that, we can use the branch on equal as described above.

if(a < b)              becomes    slt R2, R0, R1  
    goto label4                   beq R2, 1, L4?        

Jumps are simple, they are just j and then the label to jump to. So,

goto label1 becomes j label1

Last thing we have to handle is the return. The return is done by moving the value we want to a special register V0 and then jumping to the next instruction after the call to this function. The issue is MIPS doesn't have a register to register move command (or if it does I've forgotten it) so we move from a register to RAM and then back again. Finally, we use the special register R31 which holds the return address.

return a     becomes   var = a      which is SW R0, var
                       ret = var    which is LW var, V0
                       jump RA      which is JR R31

With this information, the program becomes. And we can also adjust the jumps that we didn't know before:

           L1:
 0x0100        BEQ R0, R1, 8
 0x0104        SLT R2, R0, R1                 ; temp = (a < b)  temp = 1 if true, 0 otherwise
 0x0108        LUI R3, 0x01                   ; load immediate 1 into register R3
 0x010C        BEQ R2, 1, 2                   ; goto label4         
 0x0110        SUBU R0, R0, R1                ; a = a - b
 0x0114        J L3                           ; goto label3
           L4:
 0x0118        SUBU R1, R1, R0                ; b = b - a;
           L3:
 0x011C        J L1                           ; goto lable1
           L2:
 0x0120        SW R0, ret                     ; move return value from register to a RAM location
 0x0123        LW ret, V0                     ; move return value from RAM to the return register.
 0x0124        JR R31                         ; return to caller

It has been almost twenty years since I've had to do stuff like this (now a days, if I need assembly I just do what others have suggested and let the compiler do all the heavy lifting). I am sure that I've made a few errors along the way, and would be happy for any corrects or suggestions. I only went into this long-winded discussion because I interpreted the OP question as doing a hand translation -- something someone might do as they were learning assembly.

cheers.

Answer 3

I've translated that code to 16-bit NASM assembly:

loop:
    cmp ax, bx
    je .end;        if A is not equal to B, then continue executing. Else, exit the loop
    jg greater_than;    if A is greater than B...

    sub ax, bx;     ... THEN subtract B from A...

    jmp loop;       ... and loop back to the beginning!

.greater_than:
    sub bx, ax;     ... ELSE, subtract A from B...

    jmp loop;       ... and loop back to the beginning!

.end:
    push ax;        return A

I used ax in place of r0 and bx in place of r1

Answer 4

ORG 000H                   // origin
MOV DPTR,#LUT              // moves starting address of LUT to DPTR
MOV P1,#00000000B          // sets P1 as output port
MOV P0,#00000000B          // sets P0 as output port
MAIN: MOV R6,#230D         // loads register R6 with 230D
      SETB P3.5            // sets P3.5 as input port
      MOV TMOD,#01100001B  // Sets Timer1 as Mode2 counter & Timer0 as Mode1 timer
      MOV TL1,#00000000B   // loads TL1 with initial value
      MOV TH1,#00000000B   // loads TH1 with initial value
      SETB TR1             // starts timer(counter) 1
BACK: MOV TH0,#00000000B   // loads initial value to TH0
      MOV TL0,#00000000B   // loads initial value to TL0
      SETB TR0             // starts timer 0
HERE: JNB TF0,HERE         // checks for Timer 0 roll over
      CLR TR0              // stops Timer0
      CLR TF0              // clears Timer Flag 0
      DJNZ R6,BACK
      CLR TR1              // stops Timer(counter)1
      CLR TF0              // clears Timer Flag 0
      CLR TF1              // clears Timer Flag 1
      ACALL DLOOP          // Calls subroutine DLOOP for displaying the count
      SJMP MAIN            // jumps back to the main loop
DLOOP: MOV R5,#252D
BACK1: MOV A,TL1           // loads the current count to the accumulator
       MOV B,#4D           // loads register B with 4D
       MUL AB              // Multiplies the TL1 count with 4
       MOV B,#100D         // loads register B with 100D
       DIV AB              // isolates first digit of the count
       SETB P1.0           // display driver transistor Q1 ON
       ACALL DISPLAY       // converts 1st digit to 7seg pattern
       MOV P0,A            // puts the pattern to port 0
       ACALL DELAY
       ACALL DELAY
       MOV A,B
       MOV B,#10D
       DIV AB              // isolates the second digit of the count
       CLR P1.0            // display driver transistor Q1 OFF
       SETB P1.1           // display driver transistor Q2 ON
       ACALL DISPLAY       // converts the 2nd digit to 7seg pattern
       MOV P0,A
       ACALL DELAY
       ACALL DELAY
       MOV A,B             // moves the last digit of the count to accumulator
       CLR P1.1            // display driver transistor Q2 OFF
       SETB P1.2           // display driver transistor Q3 ON
       ACALL DISPLAY       // converts 3rd digit to 7seg pattern
       MOV P0,A            // puts the pattern to port 0
       ACALL DELAY         // calls 1ms delay
       ACALL DELAY
       CLR P1.2
       DJNZ R5,BACK1       // repeats the subroutine DLOOP 100 times
       MOV P0,#11111111B
       RET

DELAY: MOV R7,#250D        // 1ms delay
 DEL1: DJNZ R7,DEL1
       RET

DISPLAY: MOVC A,@A+DPTR    // gets 7seg digit drive pattern for current value in A
         CPL A
         RET
LUT: DB 3FH                // LUT starts here
     DB 06H
     DB 5BH
     DB 4FH
     DB 66H
     DB 6DH
     DB 7DH
     DB 07H
     DB 7FH
     DB 6FH
END

Answer 5

http://ctoassembly.com

Try executing your code here. Just copy it inside the main function, define a and b variables before your while loop and you are good to go.

You can see how the code is compiled to assembly with a fair amount of explanation, and then you can execute the assembly code inside a hypothetical CPU.

Answer 6

Although this is compiler's task but if you want to make your hands dirty then look at godbolt

This is great compiler explorer tool let you convert your C/C++ code into the assembly line by line.

If you are a beginner and wants to know "How C program converts into the assembly?" then I have written a detailed post on it here .