Translate the following C code to MIPS.
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. Assume that the elements of the arrays A and B are 4-byte words.
B[8] = A[i] + A[j];

1) From the very first line we could see that $t0 points to the fifth byte of an A array. If A is a 32-bit value array, this means $t0 refers to the array's second element-not the fourth or fifth element.

2) It collects the value in the $t1 register from the third line, and retains it into the memory position to which $t0 refers.

3) The fourth line gets the second component A and places it into the $t0 register (that also formerly kept the element's address).

In C the corresponding code is:-f= A[1] + A[1];

the answer is d underlined.

d?

explanation: