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Blizzard [7]
2 years ago
13

Write a new ARMv8 assembly file called "lab04b.S" which is called by your main function. It should have the following specificat

ions: Create a function that computes factorials: n! = n · (n − 1) · ... · 3 · 2 · 1. Use the my_mul function in question 1 for all multiplications Use the concept of recursion. Do NOT directly use the concept of iteration. Do not forget that using the recursion is the same as using the STACK. Make sure you comment on the code. Demonstrate your code running to the TA. Turn in your lab04b.S file here when complete.
Engineering
1 answer:
Len [333]2 years ago
6 0

Answer:

my_mul:

.globl my_mul

my_mul:

   //Multiply X0 and X1

   //   Does not handle negative X1!

   //   Note : This is an in efficient way to multipy!

   SUB SP, SP, 16       //make room for X19 on the stack

   STUR X19, [SP, 0]    //push X19

   ADD X19, X1, XZR     //set X19 equal to X1

   ADD X9 , XZR , XZR //set X9 to 0

mult_loop:

   CBZ X19, mult_eol

   ADD X9, X9, X0

   SUB X19, X19, 1

   B mult_loop

mult_eol:

   LDUR X19, [SP, 0]

   ADD X0, X9, XZR      // Move X9 to X0 to return

   ADD SP, SP, 16       // reset the stack

   BR X30

Explanation:

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wolverine [178]

Answer:

σ = 391.2 MPa

Explanation:

The relation between true stress and true strain is given as:

σ = k εⁿ

where,

σ = true stress = 365 MPa

k = constant

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ε = (61.8 - 54.8)/54.8 = 0.128

n = strain hardening exponent = 0.2

Therefore,

365 MPa = K (0.128)^0.2

K = 365 MPa/(0.128)^0.2

k = 550.62 MPa

Now, we have the following data:

σ = true stress = ?

k = constant = 550.62 MPa

ε = true strain = Change in Length/Original Length

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σ = (550.62 MPa)(0.181)^0.2

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2 years ago
For laminar flow over a flat plate, the local heat transfer coefficient hx is known to vary as x−1/2, where x is the distance fr
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Answer:

2

Explanation:

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h_x=cx^{-1/2}

We integrate between 0 to x for obtain the value of the coefficient, so\bar{h}_x =\frac{1}{x} \int\limit^x_0 h_x dx\\\bar{h}_x = \frac{c}{x} \int\limit^x_0 \frac{1}{\sqrt{x}}dx\\\bar{h}_x = \frac{c}{c} (2x^{1/2})\\\bar{h}_x = 2cx^{-1/2}

Substituing

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