Question

Translate the following MIPS code to C. 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.

addi \)t0, \(s6, 4

add \)t1, \(s6, \)0

sw \(t1, 0(\)t0)

lw \(t0, 0(\)t0)

add \(s0, \)t1, $t0

Short answer

The C code:

A [1] = A[0];

f = & A[0] + A[1];

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 addi  $t0,   $s6,  4 MIPS code to C

addi $\$t0,\$s6,4$

$$\begin{gathered} \$t0=\$s6+4 \\ \$t0=\&A\left[1\right] \end{gathered}$$

Convert add  $t1, $s6, $0 MIPS code to C

$$\begin{gathered} add\$t1,\$s6,\$0 \\ \$t1=\$s6+\$0 \\ \$t1=\$s6 \\ \$t1=\&A\left[0\right] \end{gathered}$$

Convert sw $t1, 0($t0) MIPS code to C

$$\begin{gathered} sw\$t1,0(\$t0) \\ Memory[\$t0+0]=\$t1 \\ Memory[\&A0]=\$t1 \end{gathered}$$

A[0] will have the address of A[1]

Convert lw $t0, 0($t0) MIPS code to C

$$\begin{gathered} lw\$t0,0(\$t0) \\ \$t0=Memory[\$t0+0] \\ \$=Memory[\$t0] \\ =Memory[\&A[1\left]\right] \end{gathered}$$

Convert add  $s0, $t1, $t0 MIPS code to C

add$\$s0,\$t1,\$t0$

$$\begin{gathered} \$s0=\$t1+\$t0 \\ =\&A\left[0\right]+A\left[1\right] \\ f=\&A\left[0\right]+A\left[1\right] \end{gathered}$$