Any help is appreciated.

The following C program asks the user for two input null-terminated strings, each stored in uninitialized 100-byte buffer, and c

marissa [1.9K]

Answer:

Code is given below:

Explanation:

.data

str1: .space 20

str2: .space 20

msg1:.asciiz "Please enter string (max 20 characters): "

msg2: .asciiz "\n Please enter string (max 20 chars): "

msg3:.asciiz "\nSAME"

msg4:.asciiz "\nNOT SAME"

.text

.globl main

main:

li $v0,4 #loads msg1

la $a0,msg1

syscall

li $v0,8

la $a0,str1

addi $a1,$zero,20

syscall #got string to manipulate

li $v0,4 #loads msg2

la $a0,msg2

syscall

li $v0,8

la $a0,str2

addi $a1,$zero,20

syscall #got string

la $a0,str1 #pass address of str1

la $a1,str2 #pass address of str2

jal methodComp #call methodComp

beq $v0,$zero,ok #check result

li $v0,4

la $a0,msg4

syscall

j exit

ok:

li $v0,4

la $a0,msg3

syscall

exit:

li $v0,10

syscall

methodComp:

add $t0,$zero,$zero

add $t1,$zero,$a0

add $t2,$zero,$a1

loop:

lb $t3($t1) #load a byte from each string

lb $t4($t2)

beqz $t3,checkt2 #str1 end

beqz $t4,missmatch

slt $t5,$t3,$t4 #compare two bytes

bnez $t5,missmatch

addi $t1,$t1,1 #t1 points to the next byte of str1

addi $t2,$t2,1

j loop

missmatch:

addi $v0,$zero,1

j endfunction

checkt2:

bnez $t4,missmatch

add $v0,$zero,$zero

endfunction:

jr $ra

3 0

3 years ago

Modify any of the previous labs which would have crashed when non-numeric data was entered by adding exception handling so that

Mashutka [201]

Answer:

see explaination

Explanation:

import java.util.InputMismatchException;

import java.util.Scanner;

public class calculate {

static float a=0,b=0;

double cal()

{

if(a==0||b==0)

{

System.out.println("no values found in a or b");

start();

}

double x=(a*a)+(b*b);

double h=Math.sqrt(x);

a=0;

b=0;

return h;

}

float enter()

{

float val=0;

try

{

System.out.println("Enter side");

Scanner sc1 = new Scanner(System.in);

val = sc1.nextFloat();

return val;

}

catch(InputMismatchException e)

{

System.out.println("Enter correct value");

}

return val;

}

void start()

{

calculate c=new calculate();

while(true)

{

System.out.println("Enter Command");

Scanner sc = new Scanner(System.in);

String input = sc.nextLine();

switch(input)

{

case "A":

a=c.enter();

break;

case "B":

b=c.enter();

break;

case "C":

double res=c.cal();

System.out.println("Hypotenuse is : "+res);

break;

case "Q":

System.exit(0);

default:System.out.println("wrong command");

}

public static void main(String[] args) {

calculate c=new calculate();

c.start();

}

7 0

3 years ago

A motorist enters a freeway at 25 mi/h and accelerates uniformly to 65 mi/h. From the odometer in the car, the motorist knows th

Helga [31]

Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at

Replacing by the values already known for vo, vf and a, and solving for t, we get:

t = vf-vo /a = (29 m/s – 11.2 m/s) / 2.2 m/s = 8 sec

5 0

2 years ago

The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in

Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0

3 years ago

what do we call a landslide in which fine-grained soil moves cohesively but with extensive internal shearing?

pogonyaev

It is called an earth flow

5 0

2 years ago