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Allisa [31]
3 years ago
8

Write a program which populates an array with integer values read from a file. We do not know how many integers are in the file,

so you must loop until the end of the file is reached. For this problem, you may NOT use the function feof(). Instead, use the result returned from fscanf() to determine the end-of-file. Recall, we can set the result of fscanf() to an integer variable, and check to see if the integer variable is equal to the EOF marker. The program must take the items in the array and reverse them. You may use one array only to solve this problem.
Computers and Technology
1 answer:
max2010maxim [7]3 years ago
5 0

Answer:

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

int main()

{

int arr[100];

int i = 0;

int j = 0;

char c[10];

char temp;

int sum = 0;

FILE* fp;

if ((fp = fopen("test.txt", "r")) == NULL) {

printf("cannot open the file");

return;

}

else {

do {

temp = fgetc(fp);

if (temp == ' ' || temp == '\n') {

c[j] = '\0';

arr[i++] = atoi(c);

j = 0;

continue;

}

c[j++] = temp;

} while (temp != EOF);

for (j = i - 1; j >= 0; j--) {

printf("%d\n", arr[j]);

}

}

getchar();

}

Explanation:

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Write a c program to count the total number of commented characters and words in a c file taking both types of c file comments (
Tanzania [10]

#include<stdio.h>

#include<stdlib.h>

int comment1(FILE *fp)

{

   char ch;

   int count=0;

   while(fscanf(fp,"%c",&ch)!=EOF)

   {

       if(ch=='\n')

       {

           return count;

       }

       count++;

   }

   return count;

}

int comment2(FILE *fp)

{

   char ch;

   int count=0;

   while(fscanf(fp,"%c",&ch)!=EOF)

   {

       if(ch=='*')

       {

           fscanf(fp,"%c",&ch);

           if(ch=='/')

           {

               return count;

           }

           count++;

       }

       count++;

   }

   return 0;

}

int main()

{

   printf("Enter the file name:");

   char s[1000],ch,ch1;

   scanf("%s",s);

   FILE*fp;

   fp = fopen(s,"r");

   int count=0;

   while(fscanf(fp,"%c",&ch)!=EOF)

   {

       if(ch=='\"')

       {

           while(fscanf(fp,"%c",&ch)!=EOF)

           {

               if(ch=='\"')

               {

                   break;

               }

               if(ch=='\\')

               {

                   fscanf(fp,"%c",&ch);

               }

           }

       }

       else if(ch=='/')

       {

           fscanf(fp,"%c",&ch);

           if(ch=='/')

           {

               count += comment1(fp);

           }

           else if(ch=='*')

           {

               count += comment2(fp);

           }

       }

   }

   printf("%d\n",count);

   return 0;    

}

3 0
3 years ago
Read 2 more answers
Complete the steps to evaluate the following
kirill115 [55]

Answer:

log base 3a= -0.631.log a/3 base 3

Now, -log m= log 1/m

hence,

log base 3a= 0.631.log 3/a base 3

log base 3a/log 3/a base 3 =0.631

log base 3 ( a.3/a) =.631 since, log m/logn =log n(m)

log base 3 3=0.631

Hence, answer is log base 3 3=0.631

Explanation:

Please check the answer section.

3 0
3 years ago
Read 2 more answers
A company uses DHCP servers to dynamically assign IPv4 addresses to workstations. The address lease duration is set as 5 days. A
nikitadnepr [17]

Answer:

FF-FF-FF-FF-FF-FF and 255.255.255.255

Explanation:

FF-FF-FF-FF-FF-FF can be defined as the layer 2 address broadcast which is often used on ethernet frames as well as help to broadcast all equipment due to the fact broadcast is made possible through Ethernet networks in which the Frames are addressed to reach every computer system on a given LAN segment as far as they are addressed to MAC address FF:FF:FF:FF:FF:FF.

255.255. 255.255 can be seen as the layer 3 address which help to address the exact same hosts because it enables the broadcast address of the zero network which is the local network due to the fact that the IP broadcasts are often used by BOOTP and DHCP clients to find and send requests to their respective servers in which a message sent to a broadcast address may be received by all network-attached hosts.

Therefore the Layer 2 and Layer 3 destination addresses which the message contain are FF-FF-FF-FF-FF-FF and 255.255.255.255

7 0
3 years ago
Which range of values would result in 10 elements stored in an array?
Paraphin [41]

Answer:

0-9

Explanation:

count 0 as 1

len(0,1,2,3,4,5,6,7,8,9)=10

3 0
3 years ago
A typical broadcast live events and use streaming technology in which audio and video files are continuously downloaded to your
Nikitich [7]

Answer:

Webcasts

Explanation:

The rest of the options don't need to be streamed as there isn't a continuous flow of information.

7 0
3 years ago
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