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2 years ago

What is meant by encapsulating semaphores? Bring out the need for it

Computers and Technology

1 answer:

artcher [175]2 years ago

8 0

Answer:

Encapsulating semaphores basically allowing various codes to implemented in different modules.

Semaphores are used to implement in the synchronization process and give access to different types of variables. Encapsulate semaphore route so that the tasks performed and executed in a single routine.

Semaphore is the variable whose values basically indicates the status in the multitasking system. It is basically used to indicate the software and hardware flag in the system hiding and encapsulating data.

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The first computer introduced was the IBM PC. A. True B. False

Lerok [7]

The answer to your question is true

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2 years ago

Read 2 more answers

What are the differences between a cursor, insertion point and mouse pointer?

Anna35 [415]

Answer:

Explanation:

A cursor is a pointer which indicates the position of the mouse on a computer's display monitor.

An insertion point is a location in a document where additional information is inserted when the user begins to type.

A mouse pointer is an image used to activate/control certain elements in a GUI (graphical user interface).

I know you asked for the differences, though i thought I'd add a similarity. All of these 3 things are similar, as they show where the position of the info/mouse is. :)

Hope this helps!

-Biscuit08

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1 year ago

The following are reasons why many organizations are trying to reduce and right-size their information foot-print by using data

blagie [28]

Answer:

C. improve data quality and reduce redundancies, reduce increased and staggering storage management costs

Explanation:

Excess data retention can lead to retention of redundant and irrelevant data, which can reduce data handling and processing efficiency, while at the same time increasing the cost of data storage and management. This can be curbed be reducing data to the right size by using data governance techniques like data cleansing and de-duplication

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2 years ago

Explica la importancia que tiene el plan de trabajo (cronograma) en los proyectos y porque se realiza antes de ejecutar la soluc

zysi [14]

Answer:

El plan de trabajo permite delimitar la necesidad a la cual se debe responder con una solución (descripción del problema), los hitos que se deben alcanzar (objetivos general y específico), los recursos requeridos para lograr desarrollar y aplicar la solución (marco teórico), los posibles beneficios derivados del proceso exitoso (justificación del proyecto) y una administración racional y realistas de los recursos a disposición en cuanto a disponibilidad, tiempo y personal mediante una secuenciación de tareas (i.e. diagramas de Gantt), todo reduce el componente iterativo inherente a diseño y producción de un producto, permite prever posibles imprevistos y mantener una perspectiva realista y una eficiencia aceptable en cuanto a presupuesto, metodologías y tecnologías a usar.

El plan de trabajo debe incluir principios de ingeniería concurrente para minimizar para manera mejor los posibles imprevistos, usualmente difíciles de manera con principios de ingeniería secuencial.

Explanation:

El plan de trabajo permite delimitar la necesidad a la cual se debe responder con una solución (descripción del problema), los hitos que se deben alcanzar (objetivos general y específico), los recursos requeridos para lograr desarrollar y aplicar la solución (marco teórico), los posibles beneficios derivados del proceso exitoso (justificación) y una administración racional y realistas de los recursos a disposición en cuanto a disponibilidad, tiempo y personal mediante una secuenciación de tareas (i.e. diagramas de Gantt), todo reduce el componente iterativo inherente a diseño y producción de un producto, permite prever posibles imprevistos y mantener una perspectiva realista y una eficiencia aceptable en cuanto a presupuesto, metodologías y tecnologías a usar.

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2 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