### Chapter 1

Problem 1, web page 37Explain each of the following terms in your own words.Translator.Interpreter.Virtual machine.You are watching: Principle of equivalence of hardware and software

**Answer:**A translator converts program in one language into programs in another, usually lower level language. Example, a Java compiler converts Java resource code into Java Byte code.An interpreter carries out a program, indict by instruction. It an initial converts an accuse in one languge right into a set of lower-level instructions and executes those.A virtual an equipment is a conceptual machine. That does not exist except as a regime in part (usually) higher level language. The SimpComp simulation program is an example of a virtual machine. That there have the right to be virtual computer systems demonstrates the an essential equivalence that hardware and also software.Problem 8, page 38In what sense are hardware and software equivalent? not equivalent?

**Answer:**Hardware and software are functionally equivalent. Any function done through one can, in principle, be done by the other. They are not identical in the feeling that to do a computer system run the lowest level have to be hardware. There have to be part embodied (physical) maker executing logic operations as a basis for software. There is additionally a difference in performance. Hardware will constantly execute operations faster than tantamount software.

### Appendix A - Binary Numbers

Problem 7. Do the complying with calculations ~ above 8-bit, two"s match numbers. 00101101 11111111 00000000 11110111 +01101111 +11111111 -11111111 -11110111 --------- --------- --------- ---------**Answer:**10011100 11111110 00000001 00000000The an initial calculation outcomes in one overflow. The subtractions room done by creating the two"s enhance of the subtrahend and also then adding it to the minuend.Problem 9. Consider the following enhancement problems for 3-bit binary numbers in two"scomplement. Because that each sum, state:whether the sign little of the an outcome is 1whether the low-order 3 bits room 0whether one overflow occurred. 000 000 111 100 100 +001 +111 +110 +111 +100 ---- ---- ---- ---- ----A: The very first row listed below gives the answer v a lug out indigenous thethree little bit register shown by CO below that bit. Consider a flags register with N (negative), Z (zero) and also V (overflow)flags. Climate the 2nd row of number (under "flags") to represent the state the the flags NZV after the operation.

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001 111 1 101 1 011 1 000 CO CO COflags NZV NZV NZV NZV NZV 000 100 100 001 011

### Appendix B - Floating point Numbers

Problem 2. Convert the adhering to IEEE single-precision floating-point number from hex to decimal.42E48000H3F880000H00800000HC7F00000H**A:**composing 42E48000H in binary we get:0 10000101 11001001000000000000000(a) The unbiased exponent is 133 - 127 = 6 so we have actually 1.11001001 x 26 which is 114.25.Similarly us get:(b) 1.0625(c) 2-126(d) -122,880Close this window to return to the food site.