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3 years ago

Why was it important for the date format to be standardized by the

Computers and Technology

1 answer:

ipn [44]3 years ago

8 0

Answer:

This standard notation helps to avoid confusion in international communication caused by the many different national notations and increases the portability of computer user interfaces. In addition, these formats have several important advantages for computer usage compared to other traditional date and time notations.

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Numeric data is stored in what for direct processing

Ede4ka [16]

A string variables??? I'm not sure does this have any multiple choice

3 0

3 years ago

Use python

Degger [83]

Answer:

def rec_dig_sum( num ):

num_list = [ digit for digit in str(num)]

total = 0

for x in num_list:

total += x

return total

def dict_of_rec_dig_sums(low, high):

mydict = dict()

for number in the range(low, high+1):

mydict[rec_dig_sum(number)] = number

return mydict

Explanation:

The python program defines two functions, "rec_dig_sum" and "dict_of_rec_dig_sums". The former accepts a number and returns the sum of the digits of the number while the latter accepts a low and high number range.

The program returns a dictionary with the recursive sum as the keys and the number count as the values.

3 0

3 years ago

Given positive integer numinsects, write a while loop that prints that number doubled without reaching 100. follow each number w

hichkok12 [17]

Hi,

the program is as follows
___________________________________________________________

import java.io.*;
class doubleval

{
public static void main()throws IOException
{

DataInputStream dt=new DataInputStream(System.in);

System.out.println("Enter NUMBER WHOSE DOUBLE U WANT TO PRINT");
int n=Integer.parseInt(dt.readLine());

for(int i=n;i<=100;i=2*i)
{
System.out.println(i);
}
}
}

3 0

3 years ago

What is Key benefit of using ram in a computer

deff fn [24]

There is not really a key benifit, but it does help with little things.

7 0

3 years ago

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