How to convert C code to MIPS assembly?

Question

I'm currently writing a program for a class that requires me to convert a C program to MIPS assembly. The premise is to take two hex values and see how many 1's match up in binary. For instance, here is the C code:

// Count the number of items that match the pattern (have '1' bits in 
// the same place as the pattern)
// pu32_a points to the first element of the array
// u32_n is the number of elements in the array
// u32_pat is the pattern to match

uint32_t pmatch(uint32_t* pu32_a, uint32_t u32_n, uint32_t u32_pat) {

  uint32_t u32_result;

  u32_result = 0;
  while(u32_n != 0) {
    if((*pu32_a & u32_pat) == u32_pat) u32_result++;
    pu32_a++;
    u32_n--;
  }

return u32_result;

}

uint32_t au32_k[] = {0x80000000, 0xFFFFFFFF, 0x0000A5AA, 0xF0F0F0F0};
uint32_t u32_count;

main() {

  u32_count = pmatch(au32_k, 4, 0x0000000F0);
  printf("Number of matches is: %d \n", u32_count);

}  

And here is my current assembly code:

#
#
# au32_k = {0x80000000, 0xFFFFFFFF,
#       0x0000A5AA, 0xF0F0F0F0}
#
#

         .data
au32_k:  .space 16
u32_n:   .word  4
u32_pat: .word  240
         .text

main:

    # Store values in $s# registers

    addi $s0, $zero, 2147483648
    addi $s1, $zero, 4294967295
    addi $s2, $zero, 42410
    addi $s3, $zero, 4042322160

    # Index = $t0
    # This loads each of the values into an array
    # starting at index $t0.

    addi $t0, $zero, 0
    sw   $s0, au32_k($t0)
    addi $t0, $t0, 4
    sw   $s1, au32_k($t0)
    addi $t0, $t0, 4
    sw   $s2, au32_k($t0)
    addi $t0, $t0, 4
    sw   $s3, au32_k($t0)

    # Reset pointer to 0

    addi $t0, $t0, -12

    # Storing values in argument registers for subroutine

    la $a0, au32_k
    lw $a1, u32_n
    lw $a2, u32_pat
    jal pmatch

# This is where I get completely lost.

pmatch:

    addi $s0, $zero, 0
    addi $s1, $zero, 0

    while:

        beqz $a1, while_end

        if_start:

    while_end:

I am completely lost on this part in the C code:

if((*pu32_a & u32_pat) == u32_pat) u32_result++;

I have no clue on how to convert this to assembly, and quite frankly, have no clue how this counts the number of matching 1 bits in two numbers.

Answer 1

I'm not going to do your whole assignment for you but here's what the code does.

if((*pu32_a & u32_pat) == u32_pat) u32_result++;

The first significant chunk is this:

(*pu32_a & u32_pat)

It involves de-referencing a pointer variable named pu32_a . This is then ANDed with another variable named u32_pat . The next chunk is as follows.

 == u32_pat) u32_result++;

This chunk compares result of the AND operation to the variable named u32_pat . If it is equal the variable named u32_result is incremented by one.