Question

For the following C statement, what is the corresponding MIPS assembly code? Assume that the variables f, g, h, i, and j are assigned to registers \(s0, \)s1, \(s2, \)s3, and \(s4, respectively. Assume that the base address of the arrays A and B are in registers \)s6 and $s7, respectively.

$\mathbf{B}\mathbf{\left[}\mathbf{8}\mathbf{\right]}\mathbf{=}\mathbf{A}\mathbf{[}\mathbf{i}\mathbf{-}\mathbf{j}\mathbf{]}$

Short answer

MIPS assembly code is

sub$\$t1,\$s3,\$s4$

sll$\$t2,\$t1,2$

add$\$t3,\$s6,\$t2$

lwrole="math" localid="1655312386210" $\$t1,0\left(t3\right)$

sw$\$t1,32\left(\$s7\right)$

Step-by-step solution

Define MIPS assembly code

MIPS stands for Microprocessor without interlocked pipeline. To write MIPS assembly code we should know about Registers and instructions format each MIPS instruction is of 32 bits.

 Convert C code into MIPS assembly code

C code:$B\left[8\right]=A\left[i-j\right]$;

Now compiled MIPS code

MIPS code	comment
sub$\$t1,\$s3,\$s4$	$\$t1=i-j$, i is assigned to register and j is assigned to register$\$s4$
sll$\$t2,\$t1,2$	shift left by a constant number of bits
add$\$t3,\$s6,\$t2$	$\$t3=$the address of$A\left[i-j\right]$
lw$\$t1,0\left(t3\right)$	$\$t1=A\left[i-j\right]$copy from memory to register
sw$\$t1,32\left(\$s7\right)$	$B\left[8\right]=A\left[i-j\right]$copy from register to memory

Here, 32 is an offset and$\$s7$is a base register.