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MAXImum [283]
2 years ago
11

Cn someone help me plz

Computers and Technology
1 answer:
3241004551 [841]2 years ago
8 0

Answer:

A

Explanation:

if we put x=7 then the the condition will be true because after that x  which is 7 will be greater than 5

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What is needed to create a good problem statement?
exis [7]
A good problem statement should be:
Concise. The essence of your problem needs to be condensed down to a single sentence. ...
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Measurable. ...
Specify what is Impacted.
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2 years ago
What is the main storage location of a computer
gogolik [260]
I'm pretty sure its the hard drive, that's the main storage
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For the following array x [10] = { 45, 20, 50, 30, 80, 10, 60, 70, 40, 90} show the contents of x after the function call split
Valentin [98]

Answer:

The array index by using split function.

45 20 50 30 80 10 60 70 40 90

45 20 40 30 80 10 60 70 50 90

45 20 40 30 10 80 60 70 50 90

10 20 40 30 45 80 60 70 50 90

Explanation:

In the following statement, there is an integer array type variable x and its index value is 10 that means it contains only 10 numeric values and then, they set split function which split the array and the split function is the built-in function which is used to separate the string or an array. So, the array is split into the following parts.

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2 years ago
Write a client program that writes a struct with a privateFIFO name (call it FIFO_XXXX, where XXXX is the pid that you got from
ivann1987 [24]

Answer:

client code:

#include <stdio.h>

#include <fcntl.h>

#include <sys/stat.h>

#include <sys/types.h>

#include <unistd.h>

#include <stdlib.h>

#include <string.h>

int main (void)

{

 struct values

 {

   char privateFIFO[14];

   int intbuff;

 }input;

 int fda;  // common FIFO to read to write to server

 int fdb;      // Private FIFO to read from server

 int clientID;

 int retbuff;

 char temp[14];

 clientID = getpid();

 strcpy(input.privateFIFO, "FIFO_");

 sprintf(temp, "%d", clientID);

 strcat(input.privateFIFO, temp);

 printf("\nFIFO name is %s", input.privateFIFO);

 // Open common FIFO to write to server

 if((fda=open("FIFO_to_server", O_WRONLY))<0)

    printf("cant open fifo to write");

 write(fda, &input, sizeof(input));    // write the struct to the server

 close(fda);

 // Open private FIFO to read

 if((fdb=open(input.privateFIFO, O_RDONLY))<0)

    read(fdb, &retbuff, sizeof(retbuff));

 printf("\nAll done\n");

 close(fdb);

}

server code:

#include <stdio.h>

#include <errno.h>

#include <fcntl.h>

#include <sys/stat.h>

#include <sys/types.h>

#include <unistd.h>

#include <stdlib.h>

#include <string.h>

struct values

 {

   char privateFIFO[14];

   int intbuff;

 }input;

int main (void)

{

 int fda;  //common FIFO to read from client

 int fdb;  //private FIFO to write to client

 int retbuff;

 int output;

// create the common FIFO  

 if ((mkfifo("FIFO_to_server",0666)<0 && errno != EEXIST))

       {

       perror("cant create FIFO_to_server");

       exit(-1);

}

// open the common FIFO

 if((fda=open("FIFO_to_server", O_RDONLY))<0)

    printf("cant open fifo to write");

output = read(fda, &input, sizeof(input));

// create the private FIFO

 if ((mkfifo(input.privateFIFO, 0666)<0 && errno != EEXIST))

 {

   perror("cant create privateFIFO_to_server");

   exit(-1);

 }

 printf("Private FIFO received from the client and sent back from server is: %d", output);

 //open private FIFO to write to client

if((fdb=open(input.privateFIFO, O_WRONLY))<0)

   printf("cant open fifo to read");

 write(fdb, &retbuff, sizeof(retbuff));

 close(fda);

 unlink("FIFO_to_server");

 close(fdb);

 unlink(input.privateFIFO);

}

Explanation:

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What is application software used for
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Applications software (also called end-user programs) include such things as database programs, word processors, Web browsers and spreadsheets. A word processor could be classed as general purpose software as it would allow a user to write a novel, create a restaurant menu or even make a poster.

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