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Delvig [45]
2 years ago
10

Write a loop that inputs words until the user enters STOP. After each input, the program should number each entry and print in t

his format:
#1: You entered _____
When STOP is entered, the total number of words entered should be printed in this format:

All done. __ words entered.
Computers and Technology
2 answers:
scoray [572]2 years ago
7 0

Answer:

oh wowwwwwwwww

Explanation:

krek1111 [17]2 years ago
6 0
Oh wowwwwwwwwwwwwwwwwwwwwwwwwwwwww
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I am not a living being, I am a cylindrical shape that has three to eight sides. I never die. I can build anything again. What i
Setler79 [48]

Answer:

A shape

Explanation:

4 0
3 years ago
What are the two components that make up the chipset?
svet-max [94.6K]
Northbridge and Southbridge components.
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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
2 years ago
Dates of birth were entered into a computer program . The data was stored in the format DAY/MONTH/YEAR.The program rejected this
Marina CMI [18]

Answer:

i. This value was not accepted because we have 12 months and not 13.

ii. This is because the user entered the value for the day as value for the month.

Explanation:

i. Why was this not accepted?

This value was not accepted because we have 12 months and not 13. The user entered a value that was above the maximum value registered for the number of months in a year- which is 12.

ii. Why do you think this error occurred

This is because the user entered the value for the day as value for the month.

This could be because the user uses a date system different from that of the program.

3 0
2 years ago
Which kind of file would be hurt most by lossy compression algorithm
Eduardwww [97]

Answer: an audio file containing speech

Explanation:

8 0
3 years ago
Read 2 more answers
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