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liq
14 days ago
8

Personal computer hard disk platters typically have storage capacities ranging from 40 gb to ____.

Computers and Technology
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Write a program using the standard library I/O functions (fopen, fread, fwrite, fclose, ferror, feof, and fflush) to read a text
amid [951]

Answer:

Refer to the explanation

Explanation:

#include <stdio.h>

#include <stdlib.h>

#include<string.h>

int main(int argc, char *argv[])

{

FILE *fr,*fr1,*fr3; /* declare the file pointer */

char filename[256];

char search[256];

char line[256];

int count=0;

char replace[256];

printf("Enter FileName: "); // ask for filename

scanf("%s",filename);

fr3 = fopen (filename, "r");

if(ferror(fr3)) //check for any error assosicated to file

{

fprintf(stderr, "Error in file"); // print error in stderr

printf("Error");

}

printf("Input file data is");

while(!feof(fr3)) // checking end of file condition

{ //printf("inside");

fgets(line,256,fr3);

printf("%s",line);

}

printf("\nEnter String to search: "); //ask for string to search

scanf("%s",search);

printf("Enter string with whom you want to replace: "); //ask for string with whom you want to replace one time

scanf("%s",replace);

fr = fopen (filename, "r"); // open first file in read mode

fr1 = fopen ("output.txt", "w"); //open second file in write mode

if(ferror(fr)) //check for any error assosicated to file

{

fprintf(stderr, "Error in file"); // print error in stderr

printf("Error");

}

while(!feof(fr)) // checking end of file condition

{ //printf("inside");

fgets(line,256,fr);

int r=stringCompare(line,search); // comparing every string and search string

// printf("%d",r);

if(r==1 && count==0)

{

fwrite(replace, 1, sizeof(replace), fr1 ); // writing string to file.

printf("%s\n",replace);

count++;

}

else{

printf("%s",line);

fwrite(line, 1, sizeof(line), fr1 );

}}

printf("\n");

fflush(fr1); // it will flush any unwanted string in stream buffer

fflush(fr);

fflush(fr3);

fclose(fr); //closing file after processing. It is important step

fclose(fr1);

fclose(fr3);

system("PAUSE");

return 0;

}

// Compare method which is comparing charaacter by character

int stringCompare(char str1[],char str2[]){

int i=0,flag=0;

while(str1[i]!='\0' && str2[i]!='\0'){

if(str1[i]!=str2[i]){

flag=1;

break;

}

i++;

}

if (flag==0 && (str1[i]=='\0' || str2[i]=='\0'))

return 1;

else

return 0;

}

7 0
2 months ago
A video conferencing application isn't working due to a Domain Name System (DNS) port error. Which record requires modification
Rzqust [1037]

Answer:

Service record (SRV)

Explanation:

Service records, known as SRV records, contain information defining aspects of the DNS like port numbers, server details, hostnames, priority, weight, and the IP addresses of designated service servers.

The SRV record serves as a valuable reference for locating specific services, as applications needing those services will search for the corresponding SRV record.

When configured, the SRV provides the necessary ports and personal settings for a new email client; without this, the parameters within the email client will be incorrect.

8 0
4 months ago
Which decimal value (base 10) is equal to the binary number 1012?
Natasha_Volkova [1026]

Answer:

The decimal representation of 101₂² from base 2 equals 25 in base 10.

Explanation:

To derive the decimal equivalent of 101₂²;

101₂ × 101₂ results in 101₂ + 0₂ + 10100₂.

In this expression, we observe that the '2' in the hundred's place must be converted to '0' while carrying over '1' to the thousand's position, leading to;

101₂ + 0₂ + 10100₂ = 11001₂.

This shows that;

101₂² =  11001₂.

Next, we convert the outcome of squaring the base 2 number, 11001₂, into base 10 through the following method;

Converting 11001₂ to base 10 results in;

1 × 2⁴ + 1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰.

The calculation yields;

16 + 8 + 0 + 0 + 1 = 25₁₀.

7 0
3 months ago
Use the following data definitions data myBytes BYTE 10h,20h,30h,40h myWords WORD 3 DUP(?),2000h myString BYTE "ABCDE" What will
Natasha_Volkova [1026]

Answer:

Given Data:

myBytes BYTE 10h, 20h, 30h, 40h

myWords WORD 3 DUP(?), 2000h

myString BYTE "ABCDE"

From the supplied information, we can derive that:

(a).     a. EAX = 1

         b. EAX = 4

         c. EAX = 4

         d. EAX = 2

         e. EAX = 4

         f. EAX = 8

         g. EAX = 5

8 0
3 months ago
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