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QveST
18 days ago
14

Subroutines in MIPS Determines the minimum of two integers Functions within the MIPS slides describe how one can use subroutines

(also called procedures, functions, and methods) in MIPS. Because of the importance of subroutines in modern programming, most hardware designers include mechanisms to help programmers. In a high-level language like C or Java, most of the details of subroutine calling are hidden from the programmer.
Engineering
1 answer:
pantera1 [306]18 days ago
4 0

Answer:

This question is not complete, here’s the full question:

Subroutines in MIPS Determine the minimum of two integers Functions within the MIPS slides explain how subroutines (also known as procedures, functions, and methods) can be utilized in MIPS. Given the significance of subroutines in contemporary programming, most hardware developers integrate mechanisms to support programmers. In high-level languages like C or Java, many aspects of subroutine calls are obscured from the saw. MIPS features special registers for passing information to and from a subroutine. The registers $a0, $a1, $a2, $a3 serve the purpose of conveying arguments into the relevant subroutine. The registers $v0 and $v1 transmit parameters back from the subroutine. The stack (and stack pointer register $sp) has various roles during the use of subroutines. It aids in transferring extra parameters to and from subroutines. Moreover, it holds temporary memory values that are necessary for a subroutine. Most crucially, it preserves the current state so that the subroutine can return to the caller once finished. This includes saving the frame pointer ($fp), the return address register ($ra), and the caller-saved registers ($s0-$s7). Suppose you need a program that accepts two integers from the user, determines the smaller of the two, and then prints the minimum value. One approach would involve creating a subroutine that takes two inputs and returns the lesser value. The following program illustrates one tactic to achieve this.

Explanation:

# minimum function to compute min($a0, $a1):

In the code below,

   $a0 and $a1 are the input parameters

   $v0 and $v1 represent the return value (holding the smaller one)

   $ra indicates the return address

min:

blt $a0, $a1, firstIsLessThanLabel

blt $a1, $a0, secondIsLessThanLabel

  firstIsLessThanLabel:

move $v0, $a0

  secondIsLessThanLabel:

move $v0, $a1

jr $ra

Please note, for outputting, the following code should be used instead.

firstIsLessThanLabel: // this procedure is defined

li $v0, 4

la $a0, labelP1IsLess

syscall

b exitLabel

secondIsLessThanLabel:   // this procedure is defined

li $v0, 4

la $a0, labelP2IsLess

syscall

b exitLabel

exitLabel: // this procedure is defined

li $vo, 10

syscall

The code below displays the entire program without the procedures:

.data

p1:.asciiz "Please enter the 1st integer: "

p2:.asciiz "Please enter the 2nd integer: "

labelP1IsLess:.asciiz "first number is less than second number"

labelP2IsLess:.asciiz "second number is less than first number"

.text

main:

li $v0, 4

la $a0, p1

syscall

li $v0, 5

syscall           //read the input

move $8, $v0

li $v0, 4

la $a0, p2

syscall

li $v0, 5

syscall

move $9, $v0

blt $8, $9, firstIsLessThanLabel

blt $9, $8, secondIsLessThanLabel

b exitLabel

firstIsLessThanLabel:

li $v0, 4

la $a0, labelP1IsLess

syscall

b exitLabel

secondIsLessThanLabel:

li $v0, 4

la $a0, labelP2IsLess

syscall

b exitLabel

exitLabel:

li $vo, 10

syscall

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1.015 ha.

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Answer:

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Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = -\frac{dp}{dV/V}................1

And -\frac{dV}{V} = \frac{d\rho}{\rho}............2

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Final volume v2 = \frac{\pi }{4} *d^2*l(f)................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 =  \frac{\pi }{4} *d^2*(l(f)-l(i))

dV = \frac{\pi }{4} *d^2*dl

Substituting all values yields

0.0130 = \frac{\pi }{4} *2^2*dl

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

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