How to convert C++ program to Mips assembly language in MARs assembler?

Question

C++ program to be converted to mips assembly language

    #include <iostream>
    using namespace std;

    int main() {

int a = 1; 
int b = 1; 
    int array[a];
   while ( a < 10)
    {   
     array[a] = b + a;
 cout << array[a] << endl; // print elements in array //check values in array
        a +=1;
    }
      system("pause");  
    return 0;
    }

Mips assembly language program

    .data
     str: .ascii "abcdef"
     array: .space 20
    .text
    main: 
     li $s0, 1 # a = 1
     li $s2, 1 # b = 1         
     loop:
     la  $t1, array
     slti $t0, $s0, 3   # t0<- 1 if a < 3
     beq $t0, $zero, exit
     sll $t0, $s0, 2    # t1<- 4*a
     add $t1, $t1, $t0  # new base addr
     add $t2, $s2, $s0  # t2<- a+b   
     sw  $t1, 0($t2)    # D[a]=a+b
     addi $s0, $s0, 1   # a = a +1
     j loop             # goes to loop: label 
     exit:
     li $v0, 10 # v0<- (exit)
     syscall 

I've tried lw, sw, lb, sb to put the value of register $t2 in the array, but I continue to get an error when I single step run program in Mars compiler

Updated Mips assembly language program

     .data
     str: .ascii "abcdef"
     .align 2
     array: .space 40
     .text
     main: 
     li $s0, 1 # a = 1
     li $s2, 1 # b = 1
     loop:
     la  $t1, array
     slti $t0, $s0, 3   # t0<- 1 if a < 3
     beq $t0, $zero, exit
     sll $t0, $s0, 2   # t1<- 4*a
     addu $t2, $t1, $t0 # new base addr
     add $s1, $s0, $s2 # s1<- a+b
     sw  $s1, 0($t2)   # D[a]=a+b

     li  $v0, 1        # load appropriate system call code into register $v0;
               # code for printing integer is 1
     lw $a0, 0($t0)
     addiu $t0, $t0, 4
     syscall           # call operating system to perform print operation

     addi $s0, $s0, 1  # a = a +1
     j loop            # goes to loop: label          
     exit:
     li $v0, 10 # v0<- (exit)
     syscall

I run this program and I get this error: "address out of range 0x00000004" I want to print the values of my array to check if it is right.

Answer 1

There are several errors in your code. Regarding the storing of items in the array, you have to add the base address of the array with the index (multiplied by 4) to get the address of the item to be stored.

Assuming that $s0 holds a and $s2 holds b, to store D[a] = a + b you would:

la $t1, array
sll $t0, $s0, 2
addu $t2, $t1, $t0  # $t2 is &D[a]
add $s1, $s0, $s2    # $s1 = a + b
sw $s1, 0($t2)  # D[A] = a + b

Note that you didn't reserve enough memory to hold 10 items in array D, assuming an int is 32 bits-wide, then each item is 4 bytes long, therefore you should reserve 40 bytes...

You should also ensure that the array is properly aligned at a word boundary. To do that, you can instruct the assembler to do the alignment for you with the .align 2 directive, e.g:

.align 2
array: .space 40

Answer 2

intN;
cout<<"Enter the array size: ";
cin>>N; //size must be less than MAX_SIZE

int one[MAX_SIZE]; 
int two[MAX_SIZE]; 
cout<<"Ente the elements of array one:"<<endl;
for(inti = 0; i<N;i++)
    cin>>one[i];

cout<<"Ente the elements of array two:"<<endl;
for(inti = 0; i<N;i++)
    cin>>two[i];

int result = compare(one,two,N);

if(result == 0)
    cout<<"Array one is Equal to array two"<<endl;
elseif(result==-1)
    cout<<"Array one is less than array two"<<endl;
elseif(result==1)
    cout<<"Array one is greater than array two"<<endl;