# 01111 binary system

The 01111 binary system will consist of short-answer questions and writing your own code. A significant portion will consist of writing code more than half the pointsso you should be prepared to do this.

The review below, in addition to Lab 1Lab 2and Lab 3are all fair game for the exam. For the **01111 binary system,** this includes both code you've written and questions you've answered.

Between this review and the first three labs, this information is intended to be comprehensive; there will be no material on the exam which isn't touched by either the labs or this review. If you're pushed for time, my personal recommendation is to spend the majority of your time studying what you wrote for your labs, both code and questions. You'll find the review below to be largely though not entirely redundant with that information.

In 5 bits, what is the most negative value representable in signed form, using two's complement? Express your answer in both binary and decimal.

In 5 bits, what is the most positive value representable in signed form, using two's complement? In 5 bits, what is the most negative value representable 01111 binary system unsigned form? You cannot represent a negative value in unsigned form; unsigned form means that all the bits 01111 binary system used for representing positive values. In 5 bits, what is the most positive value representable 01111 binary system unsigned form? 01111 binary system you are given the following 4-bit binary number, shown in two's complement: Is this information important in knowing what the value of the number is, in decimal?

That is, do you need to know if it's signed or unsigned to say what the decimal value is? Why or why not? First of all, we know that it's signed because it's in two's complement. However, even we hadn't been told that it's in two's complement and only been told it's a binary number, we wouldn't need to be told if it's signed or not 01111 binary system the left-most bit is a 0, meaning it will be positive a number regardless of whether the number was signed or unsigned.

It is equal to 6 decimal in either case. Like the previous question, we know it's signed because it's in two's complement. If we hadn't been told it was two's complement, then because of the left-most bit being a 1, we would be unable to determine the value of the number unless more information was given. This is because the left-most bit would make it a negative number if the number was signed, and a different positive number if the number was unsigned.

Suppose that a binary addition was performed in a processor, and the processor set the carry bit at the end 01111 binary system the computation. What does this mean? That is, what is the significance of the carry bit being set after an addition is performed? After addition, the carry bit being set means that the result of the addition did not fit into the 01111 binary system of bits alloted. For example, adding two unsigned 4-bit binary numbers 5 and Therefore, the result of the additon is actually with a carry set the carry corresponding to the bit that would have been in position 4 if we had a 5 bits available.

Suppose that a binary addition was performed in a processor, and the processor set the overflow bit at the end of the computation. That is, what is the significance of the overflow bit being set after an addition is performed? After addition, the overflow 01111 binary system being set means that the result of adding two same-signed numbers resulted in a **01111 binary system** of a different 01111 binary system just like in decimal, if you add two positive numbers, you must get a 01111 binary system number, and if you add two negative numbers, you must get a negative number.

For example, here we add two signed 4-bit binary numbers 5 and 7: Both numbers were positive, but we got a negative result, thus we 01111 binary system overflow.

Generally, the carry bit is ignored when performing signed arithmetic. The carry bit is used to signify that we do not have 01111 binary system bits to represent the result of a computation using unsigned arithmetic, and doesn't correspond to anything interesting in a signed setting. Generally, the overflow bit is ignored when performing unsigned arithmetic.

The overflow bit is set when the two inputs have the same sign, but the output lacks this same sign. In an unsigned setting, because there are no such things as negative numbers, this distinction is meaningless. In MIPS, what is the difference between the add and addu instructions? Other than that, they are identical. In generating the assembly code for line 3, MIPS compilers will generally use the addu instruction, which is for unsigned addition.

This 01111 binary system not 01111 binary system bug in the 01111 binary system, and result will, in fact, end up holding the correct result. There are two questions here:.

Why does addu return the correct result, despite being intended for unsigned addition? The addu instruction exists to allow programmers to add numbers and ignore any overflow that occurs. It does not mean that the operands of the instructions are automatically made unsigned; they will remain negative if they are already negative. Thus, the computation produces the correct result, even when using addu.

For more details, see the "Elaboration" paragraph in your book on page What feature of 01111 binary system allows compilers to use addu here instead of add? C considers signed integer overflow to be undefined behavior, and programs which trigger undefined behavior semantically have no meaning. This means that a compiler writer is free to assume that signed integer overflow will never occur, according to the standard.

As such, compilers will pick addu over addas it is simpler. Assume the compiler performs a naive translation to assembly, and doesn't exploit any special features of C. Since a char is represented with one 8-bit byte, the initial value of x will be 1, and then x is incremented continually. Adding one more to that results in binarywhich is Why is it important that a C compiler not exploit any special features of C here?

In particular, what feature of C might cause this program to do something differently? Some questions about this code follow:.

When you add one to the largest unsigned 8-bit value, it wraps around back to 0 and continues incrementing, forever. Unlike with the previous review question, it was not stipulated whether or not we assume the compiler exploits any special features of C. Is this information relevant to this question? Interestingly enough, while signed integer overflow is undefined in C, this sort of wrapping done by unsigned values is defined. That is, this program does not trigger undefined behavior, so we can assume the compiler will translate this into something similar.

It's just that in this case, it's unlikely that **01111 binary system** behavior observed is what was intended. Consider the following C code, which is intended to extract the lowest 7 bits of the given input i:. Consider the following C code, which is intended to extract the next 7 bits of **01111 binary system** given input i:.

Consider the following C code, which is intended to extract the lowest 7 bits of the given input itreating the result as a signed value:.

Consider the following C code, which is intended to extract the next 7 bits of the given input itreating the result as a signed value:. How 01111 binary system this be multiplied by 4, without using an instruction intented for division or multiplication? How can this be divided by 4, without using an instruction intented for division or multiplication?

There is only one form of shift left, but there are two forms of shift right. Some questions follow about this fact:. One is for signed values in two's complement, and the other for unsigned values. For example, consider the following 4-bit number:. If the original value was unsigned, then we should always fill in with zeros.

However, if the value was signed, then we'd potentially have a problem if we're using shift right to divide: That is, we could end up getting a positive result back if we divide a negative number by a positive number, which wouldn't be correct.

To address this problem, there is a version of shift right 01111 binary system shift right which will fill in the XX 's above with whatever the leftmost bit remaining is in this case, 1. In summary, this means that the non-arithmetic version of shift right always fills in zeros on the left resulting in in the above exampleand the arithmetic version fills in whatever bit preserves the signedness of the original value resulting in in the above example.

Fill in higher order bits with 0s or 1s, according to the mechanism described in the previous answer. There are no leftmost bits to fill in. Substituting zeros on the right is the correct behavior regardless of the signedness of the value.

Division by a power of two can be achieved via the clever use of shift right. However, this won't always get the same result as actual division would. Some questions follow about these facts:. The difference in results is viewable in a difference between rounding. What is the difference in rounding here? Integer division rounds towards zero, whereas shift right rounds towards negative infinity.

When dealing with positive values, this distinction is 01111 binary system, as there is no situation under which the behavior is different. However, 01111 binary system negative values, this causes problems. Mathematically, this results in Using integer division gets -2but using shift right on the equivalent binary values results in It loads the immediate value 2nd operand into the target register first operand. All MIPS instructions are exactly 32 bits large. Therein lies a problem: It is for this reason that li is a psuedoinstruction which can be automatically translated to multiple instructions, whenever we need to specify a constant using the full 32 bits.

The variables used below should be placed in the register with the same name. You do not need to exit the program properly.

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