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galben [10]
3 years ago
13

Jennifer frequently uses a few programs on her laptop. Where will she find all the frequently used program icons in her computer

?
A. the left pane of the start menu

B. the right pane of the start menu

C. the bottom left corner of the the start menu

D. the bottom right corner of the start menu
Computers and Technology
1 answer:
Mkey [24]3 years ago
7 0
Jennifer can find the list of all the frequently used program icons on the left pane of the start menu. Below this list, still on the left pane, all programs installed on her computer will be listed in alphabetical order. This is specific for windows OS.
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8 0
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Given two character strings s1 and s2. Write a Pthread program to find out the number of substrings, in string s1, that is exact
anyanavicka [17]

Answer:

Explanation:

#include <stdlib.h>

#include <stdio.h>

#include <string.h>

#define MAX 1024

int total = 0 ;

int n1, n2;

char s1, s2;

FILE fp;

int readf(FILE fp)

{

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

   printf("ERROR: can’t open string.txt!\n");

   return 0;

 }

 s1=(char)malloc(sizeof(char)MAX);

 if(s1==NULL) {

   printf("ERROR: Out of memory!\n");

   return 1;

 }

 s2=(char)malloc(sizeof(char)MAX);

 if(s1==NULL) {

   printf("ERROR: Out of memory!\n");

   return 1;

 }

 /* read s1 s2 from the file */

 s1=fgets(s1, MAX, fp);

 s2=fgets(s2, MAX, fp);

 n1=strlen(s1); /* length of s1 */

 n2=strlen(s2)-1; /* length of s2 */

 if(s1==NULL || s2==NULL || n1<n2) /* when error exit */

   return 1;

}

int num_substring(void)

{

 int i, j, k;

 int count;

 for(i=0; i<=(n1-n2); i++) {

   count=0;

   for(j=i, k=0; k<n2; j++, k++){ /* search for the next string of size of n2 */

   if((s1+j)!=(s2+k)) {

     break;

   }

   else

     count++;

   if(count==n2)

     total++; /* find a substring in this step */

   }

 }

 return total;

}

int main(int argc, char argv[])

{

 int count;

 readf(fp);

 count=num_substring();

 printf("The number of substrings is: %d\n", count);

 return 1;

}

3 0
3 years ago
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