Parts of a computer software

Answers

Answer 1

Answer:

I think application software

utility software

networking software


Related Questions

You will write code to manipulate strings using pointers but without using the string handling functions in string.h. Do not include string.h in your code. You will not get any points unless you use pointers throughout both parts.
You will read in two strings from a file cp4in_1.txt at a time (there will be 2n strings, and you will read them until EOF) and then do the following. You alternately take characters from the two strings and string them together and create a new string which you will store in a new string variable. You may assume that each string is no more than 20 characters long (not including the null terminator), but can be far less. You must use pointers. You may store each string in an array, but are not allowed to treat the string as a character array in the sense that you may not have statements like c[i] = a[i], but rather *c *a is allowed. You will not get any points unless you use pointers.
Example:
Input file output file
ABCDE APBQCRDSETFG
PQRSTFG arrow abcdefghijklmnopgrstuvwxyz
acegikmoqsuwyz
bdfhjlnprtvx

Answers

Solution :

#include [tex]$< \text{stdio.h} >$[/tex]

#include [tex]$< \text{stdlib.h} >$[/tex]

int [tex]$\text{main}()$[/tex]

{

[tex]$\text{FILE}$[/tex] *fp;

[tex]$\text{fp}$[/tex]=fopen("cp4in_1.txt","r");

char ch;

//while(1)

//{

while(!feof(fp))

{

char *[tex]$\text{s1}$[/tex],*[tex]$\text{s2}$[/tex],*[tex]$\text{s3}$[/tex];

[tex]$\text{s1}$[/tex] = (char*) [tex]$\text{malloc}$[/tex]([tex]$20$[/tex] * [tex]$\text{sizeof}$[/tex](char));

[tex]$\text{s2}$[/tex] = (char*) [tex]$\text{malloc}$[/tex]([tex]$20$[/tex] * [tex]$\text{sizeof}$[/tex](char));

[tex]$\text{s3}$[/tex] = (char*) [tex]$\text{malloc}$[/tex]([tex]$40$[/tex] * [tex]$\text{sizeof}$[/tex](char));

[tex]$\text{int i}$[/tex]=0,j=0,x,y;

while(1)

{

ch=getc(fp);

if(ch=='\n')

break;

*(s1+i)=ch;

i++;

}

while(1)

{

ch=getc(fp);

if(ch=='\n')

break;

*(s2+j)=ch;

j++;

}

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

{

*(s3+x)=*(s1+x);

}

for(y=0;y<j;x++,y++)

{

*(s3+x)=*(s2+y);

}

for(x=0;x<i+j;x++)

{

printf("%c",*(s3+x));

}

printf("\n");

getc(fp);

}

}

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