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lesantik [10]
3 years ago
9

Discuss about the main frame computer​

Computers and Technology
1 answer:
Nimfa-mama [501]3 years ago
7 0

Answer:

Mainframes (also called "big iron")[1] are powerful computers used for large information processingjobs. They are mainly used by governmentinstitutions and large companies for tasks such as census, industry and consumer statistics, enterprise resource planning, and financial transaction processing. Mainframe computers are specially used as servers on the World Wide Web (WWW). They are distinct from supercomputers.

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Your task is to write a C program that measures the latencies of various system calls. In particular, you want to know 1) the co
tensa zangetsu [6.8K]

Answer and Explanation:

#include <stdio.h>

#include<fcntl.h>

#include <sys/time.h>

#include<time.h>

#define MAX 1000

int main()

{

int pid;

int i,fd ;

char c[12];

FILE *fp;

struct timeval start,end;

double time1,time2,time3;

//open file for writing

fp=fopen("output.txt","w");

 

if(!fp)

{

printf("Not able to open the file output.txt\n");

return -1;

}

for(i = 0; i < MAX ; i++)

{

gettimeofday(&start,NULL);

//invoke getpid call

system(pid = getpid());

//printf("%d\n",start.tv_usec);

}

gettimeofday(&end,NULL);

//printf("%d\n",(end.tv_usec - start.tv_usec));

time1 = ((end.tv_sec - start.tv_sec) + (end.tv_usec - start.tv_usec) / 1000000.0)/MAX;

 

//wtite the time taken to execute getpid to

//to get micro second , divide multiply time by 1000000 , to get nano multiply time by 1000000000

printf("getpid(): %.10f %.10f\n",time1*1000000,time1*1000000000);

fprintf(fp,"getpid():%.10f %.10f\n",time1*1000000,time1*1000000000);

//in similar way execute other two commands ,open and read

 

for(i = 0; i < MAX ; i++)

{

gettimeofday(&start,NULL);

//invoke getpid call

system(open("/dev/null", O_RDONLY ));

//printf("%d\n",start.tv_usec);

}

gettimeofday(&end,NULL);

//printf("%d\n",(end.tv_usec - start.tv_usec));

time2 = ((end.tv_sec - start.tv_sec) + (end.tv_usec - start.tv_usec) / 1000000.0)/MAX;

 

//wtite the time taken to execute getpid to

printf("open(): %.10f %.10f\n",time2*1000000,time2*1000000000);

fprintf(fp,"open():%.10f %.10f\n",time2*1000000,time2*1000000000);

//in similar way execute other two commands ,open and read

fd = open("/dev/dev",O_RDONLY );

//printf("fd = %d\n",fd);

for(i = 0; i < MAX ; i++)

{

gettimeofday(&start,NULL);

//invoke getpid call

system( read(fd,c,10));

//printf("%d\n",start.tv_usec);

}

gettimeofday(&end,NULL);

//printf("%d\n",(end.tv_usec - start.tv_usec));

time3 = ((end.tv_sec - start.tv_sec) + (end.tv_usec - start.tv_usec) / 1000000.0)/MAX;

 

//wtite the time taken to execute getpid to

printf("read(): %.10f %.10f\n",time3*1000000,time3*1000000000);

fprintf(fp,"read(): %.10f %.10f\n",time3*1000000,time3*1000000000);

}

----------------------------------------------------------

//output

//I have written output to standard output also , you can remove that

getpid(): 0.1690000000 169.0000000000    

open(): 0.1890000000 189.0000000000    

read(): 3.1300000000 3130.0000000000

------------------------------------------------------

//Makefile content

prob2.o : prob2.c    

         gcc -c  prob2.c                                                                                                                                      

prob2 : prob2.o                                                                                                                                                

       gcc -o prob2 prob2.o                                                                                                                                    

all   :                                                                                                                                                        

       gcc -o prob2 prob2.c                                                                                                                                    

clean:                                                                                                                                                          

       rm -rf prob2.o  

---------------------------------------

use

$make all

then execute as below

$./prob2

3 0
3 years ago
Interpretations of​ Moore's law assert​ that:
jonny [76]

Answer:

Option (D) is the correct answer of this question.

Explanation:

Moore's Law relates to Moore's theory that the number of transistors on a microchip doubles every two years while device costs are halved.

Interpretations  of Moore's law assert the computing power doubles every 18 months.The Moore's Law theory states that development is exponential.Moore's Law states that every couple of years we should expect our computers to increase their speed and capacity and we'll pay less for them.h

Option(A),Option(B) ,Option(c) and option(E) do not belongs to Moore's law  so these options are incorrect options.

5 0
3 years ago
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