The subsequent problems deal with translating from C to MIPS. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3, and $s4, respectively. Suppose that the base address of the arrays A and B are in registers $s6 and $s7, respectively.
Convert to MIPS the subsequent expression.
B[8] = A[i-j]
I have the answer already but i do not understand it; I need expert to guide through it step by step. thanks!