Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN: 9780133594140
Author: James Kurose, Keith Ross
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Related questions
Question
Transcribed Image Text:How many times
CppBuzz.com is
printed?
int main()
{
int i=0;
Ibl:
cout<<"CppBuzz.co
m";
i++;
if (i<5)
{
goto Ibl;
}
return 0;
}
(A) Error
(B) 5 times
(C) 4 times
(D) 6 times
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by stepSolved in 2 steps
Knowledge Booster
Similar questions
- Match the C-function on the left to the Intel assemble function on the right. W: cmpl $4 movl %edi , %edi jmp .L4(,%rdi,8) %edi .L3: movl $17, %eax ret .15: movl $3, %eax int A ( int x , int y) { int a ; if ( x == 0 ) else i f ( x == 1 ) a = 3 ; else i f ( x == 2 ) a = 2 0 ; else i f ( x == 3 ) a = 2 ; else i f ( x == 4 ) a = 1 ; ret .L6: a = 17; movl $20, %eax ret .L7: movl $2, %eax ret else a = 0; .L8: return a ; movl $1, %eax .L2: ret . section .rodata . L4: .quad .L3 .quad .L5 .quad .L6 .quad .L7 .quad .L8 X: testl %edi, %edi je cmpl je cmpl je стр1 je cmpl .L16 $1, %edi .L17 $2, %edi .L18 $3, %edi int B (int x, int y) { int a; switch (x) { .L19 $4, %edi %al movzbl %al, %eax case 0: a = 17; break; sete break; case 1: a = 3; case 2: a = 20; break; case 3: a = 2; break; case 4: a = 1; a = 0; } return a; ret .L16: break; movl $17, %eax ret .L17: movl $3, %eax } ret .L18: movl $20, %eax ret .L19: movl ret $2, %eaxarrow_forwardC programming #include <stdio.h>int main() {int i, j, n ;printf("height? ") ;scanf("%2d", &n) ;for (i = 1 ; i <= n ; i++) {// printf("%d: ", i) ;for (j = 1 ; j <= i ; j++) {// Pick *one* of the following// printf("%d", j % 10) ;printf("*") ;}printf("\n") ;}return 0 ;} How can I get output like this by editing the given programming -- height? 5************************* and then how can I get the following?- height? 7 1 123 12345 1234567 123456789 12345678901 1234567890123arrow_forward#include<stdio.h>#include<stdlib.h> int cent50=0;int cent20=0;int cent10=0;int cent05=0; void calculatechange(int* change){if(*change>0){if(*change>=50){*change-=50;cent50++;}else if(*change>=20){*change-=20;cent20++;}else if(*change>=10){*change-=10;cent10++;}else if(*change>=05){*change-=05;cent05++;}calculatechange(change);}}void printchange(){if(cent50)printf("\n50cents:%d coins",cent50);if(cent20)printf("\n20cents:%d coins",cent20);if(cent10)printf("\n10cents:%d coins",cent10);if(cent05)printf("\n05cents:%d coins",cent05);cent50=0;cent20=0;cent10=0;cent05=0;}void takechange(int* change){scanf("%d",change);getchar();}int main(){int change=0;int firstinput=0;while(1){if(!firstinput){printf("\nEnter the amount:");firstinput++;}else{printf("\n\nEnter the amount to continue or Enter -1 to…arrow_forward
- For C programming languagearrow_forwardQuestion 37 public static void main(String[] args) { Dog[] dogs = { new Dog(), new Dog()}; for(int i = 0; i >>"+decision()); } class Counter { private static int count; public static void inc() { count++;} public static int getCount() {return count;} } class Dog extends Counter{ public Dog(){} public void wo(){inc();} } class Cat extends Counter{ public Cat(){} public void me(){inc();} } The Correct answer: Nothing is output O 2 woofs and 5 mews O 2 woofs and 3 mews O 5 woofs and 5 mews Oarrow_forwardDetermine all the output from the following program as it would appear on the screen. void func1(int); void func2(int = 4, int = 5, int = 2); int func3(int &, int, int); %3D int main() { int x = 0, z = 0, y = 2; func1(y); cout << y << endl; func2(x, y, z); func2(); func3(x, y, z); = Z func1(x); cout << x << " " << y << " " << z << endl; return 0; } void func1(int b) { static int a;arrow_forward
- void fun(int i) { do { if (i % 2 != 0) cout =1); cout << endl; } int main() { int i = 1; while (i <= 8) { fun(i); it; } cout <arrow_forwardComputer Science Part C: Interactive Driver Program Write an interactive driver program that creates a Course object (you can decide the name and roster/waitlist sizes). Then, use a loop to interactively allow the user to add students, drop students, or view the course. Display the result (success/failure) of each add/drop.arrow_forwardint main(){ long long int total; long long int init; scanf("%lld %lld", &total, &init); getchar(); long long int max = init; long long int min = init; int i; for (i = 0; i < total; i++) { char op1 = '0'; char op2 = '0'; long long int num1 = 0; long long int num2 = 0; scanf("%c %lld %c %lld", &op1, &num1, &op2, &num2); getchar(); long long int maxr = max; long long int minr = min; if (op1 == '+') { long long int sum = max + num1; maxr = sum; minr = sum; long long int res = min + num1; if (res > maxr) { max = res; } if (res < minr) { minr = res; } } else { long long int sum = max * num1; maxr = sum; minr = sum; long long int res = min * num1;…arrow_forwardusing System; class HelloWorld { static void Main() { int rows=21; int columns=76; for(int i=0;i<rows;i++) //for loop for printing the pattern for 21 rows { for(int j=1;j<=columns;j++) //for loop for printing the pattern in 76 columns { if((i+j)%7!=1 ) //if the addition of i and j does not give remainder of 1 when divided by 7 then Console.Write("-"); //print dash - on console else //otherwise peint a space on console Console.Write(" "); } Console.WriteLine(); //print a newline for printing on new row... } } }arrow_forward#include<stdio.h>#include<stdlib.h> int cent50=0;int cent20=0;int cent10=0;int cent05=0; void calculatechange(int* change){if(*change>0){if(*change>=50){*change-=50;cent50++;}else if(*change>=20){*change-=20;cent20++;}else if(*change>=10){*change-=10;cent10++;}else if(*change>=05){*change-=05;cent05++;}calculatechange(change);}}void printchange(){if(cent50)printf("\n50cents:%d coins",cent50);if(cent20)printf("\n20cents:%d coins",cent20);if(cent10)printf("\n10cents:%d coins",cent10);if(cent05)printf("\n05cents:%d coins",cent05);cent50=0;cent20=0;cent10=0;cent05=0;}void takechange(int* change){scanf("%d",change);getchar();}int main(){int change=0;int firstinput=0;while(1){if(!firstinput){printf("\nEnter the amount:");firstinput++;}else{printf("\n\nEnter the amount to continue or Enter -1 to…arrow_forwardarrow_back_iosarrow_forward_ios
Recommended textbooks for you
- Computer Networking: A Top-Down Approach (7th Edi...Computer EngineeringISBN:9780133594140Author:James Kurose, Keith RossPublisher:PEARSON
Computer Organization and Design MIPS Edition, Fi...Computer EngineeringISBN:9780124077263Author:David A. Patterson, John L. HennessyPublisher:Elsevier Science
Network+ Guide to Networks (MindTap Course List)Computer EngineeringISBN:9781337569330Author:Jill West, Tamara Dean, Jean AndrewsPublisher:Cengage Learning 
Concepts of Database ManagementComputer EngineeringISBN:9781337093422Author:Joy L. Starks, Philip J. Pratt, Mary Z. LastPublisher:Cengage Learning
Prelude to ProgrammingComputer EngineeringISBN:9780133750423Author:VENIT, StewartPublisher:Pearson Education
Sc Business Data Communications and Networking, T...Computer EngineeringISBN:9781119368830Author:FITZGERALDPublisher:WILEY
Computer Networking: A Top-Down Approach (7th Edi...
Computer Engineering
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:PEARSON
Computer Organization and Design MIPS Edition, Fi...
Computer Engineering
ISBN:9780124077263
Author:David A. Patterson, John L. Hennessy
Publisher:Elsevier Science
Network+ Guide to Networks (MindTap Course List)
Computer Engineering
ISBN:9781337569330
Author:Jill West, Tamara Dean, Jean Andrews
Publisher:Cengage Learning
Concepts of Database Management
Computer Engineering
ISBN:9781337093422
Author:Joy L. Starks, Philip J. Pratt, Mary Z. Last
Publisher:Cengage Learning
Prelude to Programming
Computer Engineering
ISBN:9780133750423
Author:VENIT, Stewart
Publisher:Pearson Education
Sc Business Data Communications and Networking, T...
Computer Engineering
ISBN:9781119368830
Author:FITZGERALD
Publisher:WILEY