i was suppose to write a program that gives a divide error but output is not showing anything like it.I am using orwell devc++ 5.4.2 and all the default built tools(gcc compiler etc.) with default settings.so, what should i do to see this error?
Code:
#include<stdio.h>
#include<conio.h>
int main()
{
int a,b,c;
float x;
}
I am trying to get a remainder of a number with a = 0.9144, rm = ry % a; however, I keep getting a divide by zero error (I believe due to the program rounding 0.9144 down to the integer 0).
I need to write a program that accepts a number from the user, divides that number by two until it reaches one or cannot be divided evenly anymore, then adds all of the quotients from each division and displays them.
So something like this should be displayed:
Please enter a number: 8
8/2=4 4/2=2 2/2=1 4+2+1= 7
I thought about using an array to possibly store the quotients but I just can't see how that would work.
I am new to C programming and I am just wondering how to multiply / divide two different variables which the user type in as the promt is asking like this:
Code: void inmatning3 (double *a, double *b) { printf("Mata in tv217 stycken flyttal: "); /* asks you to type in 2 numbers */ scanf("%lf %lf", a, b); }
When you've enterd the two numbers I need to eather multiply or divide the two variables "a" & "b" .....
I need to make a program that takes in a user’s number and keep dividing it by 2 until it is 1.I need to display the division steps ...
I do not know how to keep making it divide by 2 until 1. I tired this but it obviously did not work..
Code: #include <iostream> using namespace std; int main() { //User Input int input,total; cout <<"Enter A Number To Be Divided By 2 Until 1" << endl << endl;
I've been stuck on a divide a conquer algorithm problem for about an hour now, and I'm not sure how to solve it. I need to use divide-and-conquer to implement an algorithm that finds the dominant element of an array of positive integers and returns -1 if the array does not have a dominant element (a dominant element is one that occurs in more than half the elements of a given array).
No sorting may be used, and the only comparison that may be used is a test for equality.
I understand the general process I need to follow to solve this problem, but I'm not sure exactly how to convert my thoughts to code. I know that if a number x is the dominant element of an array A, the x must be the dominant element in either the first half of A, the second half of A, or both.
Here is what I have so far.
#include <iostream> #include <string> #include <vector> using namespace std; int find_dominant(int *A, int p, int r) {
I'm practicing so I wrote this simple program that suppose to add and divide two numbers. It does that but the result comes out with a 0 at the front and don't know why.
#include<iostream> using namespace std; int main() { int a ,b; int result; a = 0; b = 0;
The first line indicates how many paths exist in my design. The second and third lines are those lines(the first number indicates 2 pairs, so 10,1 and 2,7 are 2 pairs) The first element in the third line indicates 3 pairs and the pairs are 8,3 and 7,7 and 10,7.
I have been able to read the text file and store these into ONE array. However, I need to divide them up and have each line in a seperate array.
So for example
array alpha to contain 10,1 and 2,7 array beta to contain 8,3 and 7,7, and 10,7
For example, to calculate GPA, we add all the grade point of a course and divide by the number of courses we are taking. How do I create a program in such a way that when the user enters grade point for two courses, it will add the grade points for the two course and divide by 2. And if another user enters grade points for six courses, it adds the grade points and divides by 6.
The problem is that you have a set of numbers and you need to divide that set into two subsets where the difference between the sums of the subset is minimal.
Example: a set of numbers {1,5,9,3,8}, now the solution is two subsets, one subset with elements {9,3} and the other {8,5,1} the sum of the first one is 13 and the sum of the second is 13 so the difference between the sums is 0. The result shows the difference between the sums.
Another example: a set of numbers where the difference between the subsets cannot be zero, {9 51 308 107 27 91 62 176 28 6}, the minimal difference between the two subsets is 2.
I want to know how the function finds the two subsets, it works great because I've tested it for up to 300 inputs which sum adds up to 100,000.
Code: #include <iostream> #include <stdio.h> #include <math.h> #include <stdlib.h> #include <limits.h> using namespace std; int BalancedPartition ( int a[] , int n ) {
Create a program that adds, subtracts, multiplies, or divides two integers. The program will need to get a letter (A for addition, S for subtractions, M for multiplication, or D for division) and two integers from the user. If the user enters an invalid letter, the program should display an appropriate error message before the program ends. If the letter is A (or a), the program should calculate and display the sum of both integers. If the letter is S (or s), the program should display the difference between both integers. When calculating the difference, always subtract the smaller number from the larger one. If the letter is M (or m), the program should display the product of both integers. If the letter is D (or d), the program should divide both integers, always dividing the larger number by the smaller one."
And here is the test data. I am posting the results from my desk-check table.
operation first integer second integer answer A 10 20 30 a 45 15 60 S 65 50 15 s 7 13 6 G -1 M 10 20 200 d 45 15 3 d 50 100 2
Then, I transferred my program into a source file. Here it is:
//Exercise16.cpp - display answer of two integers
#include <iostream> using namespace std; int main() { //declare variables int firstInteger = 0;
[Code] ....
After putting in the data, everything worked fine, except the last two operations, which are M (multiplication) and D (division). All the answers for the last two operations essentially give me a 0.
Cannot manage to find the error source when i try running the program, the first part of the program runs just fine its when i try to get the temperature one that i get the error
#include <iostream> #define pi 3.141592 using namespace std; int main() { double r, h; //declare variables for radious and height double Surfacearea;
I've just recently started to learn C++, and I'm encountering some errors I can't seem to figure out.
InventoryItem.h:
Code: #pragma once class InventoryItem { public: InventoryItem(string name, int amount); ~InventoryItem(void); string getName(void); int getAmount(void);
I am trying to make a program like a virtual machine, and therefore I need to have a virtual hard drive. I have it split across four files, each being exactly 67,108,864 bytes (at this point, the files consist of 0x00 throughout the entire file). When I try to write to the beginning of one of the files, I get a "EXC_BAD_ACCESS (code=1, address=0xff)" from Xcode.
I have no error compiling, but running my project it stops before entering a function and debugging I have an error about Segmentation fault. The function:
I've a code and it works on my linux laptop; however it doesnt work on a matrix server. Im getting error Code: file.c: In function CondCheck:file.c:40:3: warning: this decimal constant is unsigned only in ISO C90 [enabled by default] The code of the programme
Code:
#include <stdio.h> #include <string.h> #include <math.h> /*DECIMAL TO BINARY CONVERTER*/ int BinaryConverter (int bina) }
While trying to redo everything in c i have an error on a do while loop and i don't understand i've corrected everything else but i don't understand why the error occurs even though it says how to fix it it says error expected ) before token -line 18 part of code
Code:
int main(){ Beep (523,1000); // sound at 523 hertz for 1 000 milliseconds char cPresent; do }
[code]....
this is just extra bits i've added on to the assignment but I've worked on the c++ code for a week now i have less than a day to redo it