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Sergio039 [100]
3 years ago
9

Please tell fast plzzzzzzzzzzz.​

Computers and Technology
1 answer:
kumpel [21]3 years ago
6 0

Answer:

9

Explanation:

1×2^3+0+0+1×2^0

8+1=9

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Write a MIPS assembly language program that accomplishes the following tasks:The program will prompt the user to enter an intege
Svetlanka [38]

Answer:

Here's the code below

Explanation:

# All the comment will start with #

# MIPS is assembly language program.

# we store the all the data inside .data

.data:

k: .asciiz "Enter the kth value:\n"

.text

.globl main

li $s0, 0 # $s0 = 0

li $t0, 0 # $t0 = 0

la $n0, k # $n0 is kth value address

syscall # call print_string()

li $v0, 0 # $v0

syscall

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addi $sp, $sp, -4 # move the stack pointer down

sw $ra, 0 ($sp) # save the return address for main

jal fncheck # call fncheck(n);

fncheck:

beq $n, 0, fnfirst #if n==0 then fnfirst

beq $n, 0, fnfirst #if n==0 then fnfirst

ble $n0, 5, fnrec # if (n <=5) then 5

bge $n0, 5, fnjoke

#first function for n 0 and 1

fnfirst:

li $t0, 0 # initialize $t0 = 0

addi $t0, $t0, 20 # $t0 = 20

move $v0, $t0 # return 20

jr $ra

#function to execute 5*function1(n-2) + n;

fnrec:

sub $sp, $sp, 12 # store 3 registers

sw $ra, 0($sp) # $ra is the first register

sw $n0, 4($sp) # $n0 is the second register

addi $n0, $n0, -2 # $a0 = n - 2

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sw $v0, 8($sp) # store $v0, the 3rd register to be stored

lw $n0, 4($sp) # retrieve original value of n

lw $t0, 8($sp) # retrieve first function result (function1 (n-2))

mul $t0, $t0, 5 # $t0 = 5 * function1(n-2)

add $v0, $v0, $t0

lw $ra, 0($sp) # retrieve return address

addi $sp, $sp, 12

jr $ra

#function to print joke

fnjoke:

la "Joke section"

4 0
4 years ago
How do I write this code in java? input "Enter Your Age to Order a Beer"
Klio2033 [76]

Answer:

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Explanation:

5 0
3 years ago
5)What are the differences in the function calls between the four member functions of the Shape class below?void Shape::member(S
Stells [14]

Answer:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

Explanation:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

The s1 and s2 objects are passed by value as there is no * or & sign with them. If any change is made to s1 or s2 object, there will not be any change to the original object.

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

The s1 and s2 objects are passed by pointer as there is a * sign and not & sign with them. If any change is made to s1 or s2 object, there will be a change to the original object.

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. If any change is made to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. The major change is the usage of const keyword here. Const keyword restricts us so we cannot make any change to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. const keyword restricts us so we cannot make any change to s1 or s2 object as well as the Shape function itself.

5 0
3 years ago
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