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andrezito [222]
3 years ago
5

What is the output for the code below?

Computers and Technology
1 answer:
alisha [4.7K]3 years ago
5 0
What is output by the code below? int[] array = {33,14,37,11,27,4,6,2,6,7}; System .out.println(array.length); ... int[] array = {5,10,3,6,9,15}; ... int total = 0; ... output by the code below? int j=1, tally=0; while(j<9) { tally++; j++; } System.out.print(tally);.
From quizlet
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WINSTONCH [101]

i think its B) Theme

7 0
3 years ago
A computer reads a sequence from top to bottom and left to right? True or False
Nat2105 [25]
The answer is False
5 0
3 years ago
Diane, a developer, needs to program a logic component that will allow the user to enter a series of values
oee [108]
<h2>The Examples Of Users:</h2>
  • Form
  • Table
  • Query
<h2>The Answers Are:</h2>
  • Formatted summary of information from a database

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  • Stores raw data in a relational database

  • Retrieves specific information from a database. Can also be used to update, edit, and remove data
<h2>Hope it helps*^-^*All Correct!?</h2>
3 0
2 years ago
In the original UNIX operating system, a process executing in kernel mode may not be preempted. Explain why this makes (unmodifi
Elena L [17]

Answer:

the preemption is -> The ability of the operating

system to preempt or stop a currently

scheduled task in favour of a higher priority

task. The scheduling may be one of, but not

limited to, process or 1/0 scheduling etc.

Under Linux, user-space programs have always

been preemptible: the kernel interrupts user

space programs to switch to other threads,

using the regular clock tick. So, the kernel

doesn't wait for user-space programs to

explicitly release the processor (which is the

case in cooperative multitasking). This means

that an infinite loop in an user-space program

cannot block the system.

However, until 2.6 kernels, the kernel itself was

not preemtible: as soon as one thread has

entered the kernel, it could not be preempted to

execute an other thread. However, this absence

of preemption in the kernel caused several

oroblems with regard to latency and scalability.

So, kernel preemption has been introduced in

2.6 kernels, and one can enable or disable it

using the cONFIG_PREEMPT option. If

CONFIG PREEMPT is enabled, then kernel code

can be preempted everywhere, except when the

code has disabled local interrupts. An infinite

loop in the code can no longer block the entire

system. If CONFIG PREEMPT is disabled, then

the 2.4 behaviour is restored.

So it suitable for real time application. Only

difference is we don't see many coders using it

5 0
3 years ago
(1) Prompt the user for a string that contains two strings separated by a comma. (1 pt) Examples of strings that can be accepted
denis-greek [22]

Answer:

Check the explanation

Explanation:

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#define MAX_LIMIT 50

int checkComma(char *input)

{

int flag = 0;

for(int i = 0; i < strlen(input); i++)

{

if(input[i] == ',')

{

flag = 1;

break;

}

}

return flag;

}

int main(void)

{

char input[MAX_LIMIT];

char *words[2];

char delim[] = ", ";

printf("\n");

do

{

printf("Enter input string: ");

fgets(input, MAX_LIMIT, stdin);

size_t ln = strlen(input) - 1;

if (*input && input[ln] == '\n')

input[ln] = '\0';

if(strcmp(input, "q") == 0)

{

printf("Thank you...Exiting\n\n");

exit(1);

}

else

{

if(checkComma(input) == 0)

{

printf("No comma in string.\n\n");

}

else

{

char *ptr = strtok(input, delim);

int count = 0;

while(ptr != NULL)

{

words[count++] = ptr;

ptr = strtok(NULL, delim);

}

printf("First word: %s\n", words[0]);

printf("Second word: %s\n\n", words[1]);

}

}

}while(strcmp(input, "q") != 0);

return 0;

}

Kindly check the attached image below for the output.

4 0
3 years ago
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