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Translate 0×abcdef12into decimal

Short Answer

Expert verified

The translation is:

2882400018

Step by step solution

01

Define Hexadecimal

Hexadecimal is the base 16 number system that represents the 4-bit binary numbers in a simplified way. In hexadecimal the values from 10-15 is represented by the alphabets A-F.

In machine programs, most values will be considered in a hexadecimal format which makes the execution faster.

02

Determine the decimal value

For the given value, the decimal value can be calculated by multiplying each term by the powers of 16 and summing all values.

The calculation is as follows:

0×abcdef12=2×160+1×161+15×162+14×163+13×164+12×165+11×166+10×167

The decimal value of 0×abcdef12is 2882400018

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Most popular questions from this chapter

Assume that registers \(s0 and \)s1 hold the values 0x80000000 and 0xD0000000, respectively.

1. What is the value of \(t0 for the following assembly code?

add \)t0,\(s0,\)s1

2. Is the result in \(t0 the desired result, or has there been an overflow?

3. For the contents of registers \)s0 and \(s1 as specified above, what is the value of \)t0 for the following assembly code?

sub \(t0,\)s0,\(s1

4. Is the result in \)t0 the desired result, or has there been an overflow?

5. For the contents of registers \(s0 and \)s1 as specified above, What is the value of \(t0 for the following assembly code?

add \)t0,\(s0,\)s1

add \(t0,\)t0,\(s0

6. Is the result in \)t0 the desired result, or has there been an overflow?

For the MIPS assembly instructions in Exercise 2.4, rewrite the assembly code to minimize the number if MIPS instructions (if possible) needed to carry out the same function.

Question: If the current value of the PC is 0x00000600, can you use a single jump instruction to get to the PC address as shown in Exercise 2.39?

Question: Write the MIPS assembly code to implement the following C code:

lock(lk);

shvar=max(shvar.x);

unlock(lk);

Assume that the address of the lk variable is in \(a0, the address of the shvar variable is in \)a1, and the value of variable x is in $a2. Your critical section should not contain any function calls. Use ll/sc instructions to implement the lock() operation and the unlock() operation is simply an ordinary store instruction.

Question: Assume \(t0 holds the value 0x00101000. What is the value of \)t2 after the following instructions?

slt \(t2, \)0, \(t0

bne \)t2, \(0, ELSE

j DONE

ELSE: addi \)t2,$t2, 2

DONE:

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