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VladimirAG
10 days ago
11

In order to write a successful algorithm, you must first be able 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
1 month ago
Andre is finding working in an online group challenging because there are people in the group from many different cultural backg
Harlamova29_29 [1022]

Answer:

Varied Communication Styles

Diverse Attitudes Toward Conflict

Different Methods for Completing Tasks

Various Decision-Making Approaches

Different Perspectives on Disclosure

Diverse Ways of Knowing

Explanation:

6 0
2 months ago
Read 2 more answers
A company operates on two types of servers: 2 large servers (L) and 4 smaller servers (S), with a combined total of 64GB RAM. Th
oksian1 [950]

Reasoning:

Let's denote the size of a large server as L and the size of a smaller server as S.

According to the data provided, we have two equations:

2L + 4S = 64.............(1)

and

L + 3S = 40...............(2)

To solve the equations, we proceed as follows:

2(2)-(1)

2L - 2L + 6S - 4S = 2*40 - 64

2S = 16

thus, S = 8..................(3), which indicates the size of the small server

Using (3) in (2) yields

L + 3(8) = 40

L = 40 - 24 = 16..............indicating the size of the large server

8 0
1 month ago
Identify the four basic categories of hardware in a personal computer system
Harlamova29_29 [1022]
The four primary categories include the motherboard, CPU, graphics card, and RAM.
Please choose Brainliest! Thanks!:)
5 0
1 month ago
Determine the number of bytes necessary to store an uncompressed binary image of size 4000 × 3000 pixels
maria [1035]

Response: 1,500,000 bytes.

Clarification:

If we assume the image dimensions are 4000 pixels in width and 3000 pixels in height, the total uncompressed image will consist of 4000*3000= 12,000,000 pixels.

In the case of a binary image, each pixel can have only two values, which necessitates one bit for each pixel.

This indicates that we need to accommodate 12,000,000 bits.

Given that 1 byte equals 8 bits.

So, to store an uncompressed binary image sized 4000 x 3000 pixels, 12,000,000/8 bytes is required ⇒ 1,500,000 bytes.

3 0
2 months ago
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