addl eax ebx ecx addl 0 times 13 edi 4 esi addl esp edx add

addl %eax, %ebx, %ecx addl 0 times 13 (, %edi, 4), %esi addl (%esp), (%edx) addl %eax, $4 Which choice corresponds to ALL the invalid instructions? 1, 2 and 4 1, 3 and 4 1, 2 and 3 1 only 1, 2, 3 and 4

Solution

Answer:

1 only

Because add1 takes only two operators not more than two. Since in the first instruction , add1 take more than two operands , hence invald.