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];

answer: internal fragmentation is the difference between the (fixed) size of a memory hole and the data within in that memory. for example, if a 1,024 byte block of memory holds 800 bytes of data, then the 224 bytes difference is considered lost to internal fragmentation.

external fragmentation occurs when the sum of memory fragments is large enough to satisfy a request, but the holes are non-contiguous and the request cannot be satisfied. this can occur when memory is dynamically allocated and freed.

explanation:

deudtgc hdb d gd gsyjs bgwgsvsvd bsbeve,r3hfnfb. dv c cncnd.

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