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3 years ago

Which statement prints "hi" on the screen?

Computers and Technology

2 answers:

Nikitich [7]3 years ago

8 0

Answer:

Explanation:

Eva8 [605]3 years ago

5 0

Answer: (A) Puts("hi");

Explanation:

Puts() is the type function that uses the file handling in the programming language. It basically use to compose or write the function or line that displaying the output screen of the computer system.

The puts() function is the type of function which basically allow the user to read the characters or line include all the space until entering into the new character or line.

The declaration of the puts(0 function is as follows:

int puts (char *STRING) ;

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Microsoft ____________________ is a complex, full-featured firewall that includes stateful packet filtering as well as proxy ser

Lesechka [4]

Answer:

Internet Security & Acceleration Server (ISA)

Explanation:

<h2><u>Fill in the blanks </u></h2>

Microsoft <u>Internet Security & Acceleration Server (ISA)</u> is a complex, full-featured firewall that includes stateful packet filtering as well as proxy services, NAT, and intrusion detection.

4 0

3 years ago

Which section in a slide is not visible to the viewers of a presentation, but is visible to the presenter in the normal Slide vi

lyudmila [28]

Answer:

the speakers note

Explanation:

when putting it in presenter view it'll have a separate window so you'll have access and see speaker notes etc

3 0

3 years ago

Which of these is a benefit of using the Sort option?

Reika [66]

There are many benefits of using the shot options in Excel. Some benefits are allowing you to short by number,date,color, letter, columns,or text. This benefits the user because it allows the user to bring data up more easily.

I hope this answers is helpful

3 0

3 years ago

Using the C language, write a function that accepts two parameters: a string of characters and a single character. The function

Sav [38]

Answer:

#include <stdio.h>

void interchangeCase(char phrase[],char c){

for(int i=0;phrase[i]!='\0';i++){

if(phrase[i]==c){

if(phrase[i]>='A' && phrase[i]<='Z')

phrase[i]+=32;

else

phrase[i]-=32;

}

int main(){

char c1[]="Eevee";

interchangeCase(c1,'e');

printf("%s\n",c1);

char c2[]="Eevee";

interchangeCase(c2,'E');

printf("%s\n",c2);

}

Explanation:

Create a function called interchangeCase that takes the phrase and c as parameters.
Run a for loop that runs until the end of phrase and check whether the selected character is found or not using an if statement.
If the character is upper-case alphabet, change it to lower-case alphabet and otherwise do the vice versa.
Inside the main function, test the program and display the results.

8 0

3 years ago

Data Structure in C++

agasfer [191]

The code .cpp is available bellow

#include<iostream>

using namespace std;

//declaring variables

void merge(int* ip, int sz, int* opt, bool opt_asc); //merging

int* mergesort(int* ip, int sz);

void mergesort(int *ip, int sz, int* opt, bool opt_asc);

void merge(int* ip, int sz, int* opt, bool opt_asc)

{

int s1 = 0;

int mid_sz = sz / 2;

int s2 = mid_sz;

int e2 = sz;

int s3 = 0;

int end3 = sz;

int i, j;

if (opt_asc==true)

{

i = s1;

j = e2 - 1;

while (i < mid_sz && j >= s2)

{

if (*(ip + i) > *(ip + j))

{

*(opt + s3) = *(ip + j);

s3++;

j--;

}

else if (*(ip + i) <= *(ip + j))

{

*(opt + s3) = *(ip + i);

s3++;

i++;

}

if (i != mid_sz)

{

while (i < mid_sz)

{

*(opt + s3) = *(ip + i);

s3++;

i++;

}

if (j >= s2)

{

while (j >= s2)

{

*(opt + s3) = *(ip + j);

s3++;

j--;

}

else

{

i = mid_sz - 1;

j = s2;

while (i >= s1 && j <e2)

{

if (*(ip + i) > *(ip + j))

{

*(opt + s3) = *(ip + i);

s3++;

i--;

}

else if (*(ip + i) <= *(ip + j))

{

*(opt + s3) = *(ip + j);

s3++;

j++;

}

if (i >= s1)

{

while (i >= s1)

{

*(opt + s3) = *(ip + i);

s3++;

i--;

}

if (j != e2)

{

while (j < e2)

{

*(opt + s3) = *(ip + j);

s3++;

j++;

}

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

*(ip + i) = *(opt + i);

}

int* mergesort(int* ip, int sz)

{

int* opt = new int[sz];

mergesort(ip, sz, opt, true);

return opt;

}

void mergesort(int *ip, int sz, int* opt, bool opt_asc)

{

if (sz > 1)

{

int q = sz / 2;

mergesort(ip, sz / 2, opt, true);

mergesort(ip + sz / 2, sz - sz / 2, opt + sz / 2, false);

merge(ip, sz, opt, opt_asc);

}

int main()

{

int arr1[12] = { 5, 6, 9, 8,25,36, 3, 2, 5, 16, 87, 12 };

int arr2[14] = { 2, 3, 4, 5, 1, 20,15,30, 2, 3, 4, 6, 9,12 };

int arr3[10] = { 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 };

int *opt;

cout << "Arays after sorting:\n";

cout << "Array 1 : ";

opt = mergesort(arr1, 12);

for (int i = 0; i < 12; i++)

cout << opt[i] << " ";

cout << endl;

cout << "Array 2 : ";

opt = mergesort(arr2, 14);

for (int i = 0; i < 14; i++)

cout << opt[i] << " ";

cout << endl;

cout << "Array 3 : ";

opt = mergesort(arr3, 10);

for (int i = 0; i < 10; i++)

cout << opt[i] << " ";

cout << endl;

return 0;

}

4 0

4 years ago