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NemiM [27]
3 years ago
5

How do you pick the brainliest answer on brainly.com?

Computers and Technology
2 answers:
Svet_ta [14]3 years ago
7 0
After two people answer, there is a button over each answer that lets you pick the brainliest
xeze [42]3 years ago
6 0
I 2 people have answered ur question you have an option to make one the "brainliest answer"
You might be interested in
Which tools is used to bundle cables neatly inside and outside of a computer?​
Ugo [173]

Answer:

Cable Ties

Explanation:

6 0
3 years ago
What are the three general methods for delivering content from a server to a client across a network
Tju [1.3M]

Answer:

Answered below.

Explanation:

The three general methods consist of unicasting, broadcasting and multicasting.

Casting implies the transfer of data from one computer (sender) to another (recipient).

Unicasting is the transfer of data from a single sender to a single recipient.

Broadcasting deals with the transfer of data from one sender to many recipients.

Multicasting defines the transfer of data from more than one sender to more than one recipients.

8 0
3 years ago
Assume that we would like to expand the MIPS register file to 128 and keep the size of all other fields as in the original MPIS
garri49 [273]

Answer:

These changes has the ability to decrease the size of a program, it

can do so by going ahead to reduce the instances in which the register contents “spill” into other registers or memory, it also does so by allowing more complex operations to be implemented in one instruction as agianst requiring multiple instructions.

On the other hand they are likely able to increase the size of a program by requiring adding bits to the opcode and register fields, thereby increasing memory width

Explanation:

See answer

7 0
3 years ago
Read 2 more answers
A company wishes to increase security at the employee entrance. Which of the following technologies would be the most useful?
AVprozaik [17]
I believe a metal detector would be a fine choice for an employee entrance. 
5 0
4 years ago
Read 2 more answers
Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3
Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

3 0
3 years ago
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