So I tried creating a test code for reading a negative number and positive number but whenever I enter a negative number it read it as being positive.
#include <stdio.h>
#include <iostream>
#include <iomanip>
[Code].....
PS: I am using char over int because the program that I am testing for requires me to use 8 bit variable.
My program uses a while loop to eventually get to an error of zero and a root of sqrt(3). I'm not understand why after the third iteration the program fails to compute a new x value. I'm using Visual Studio 2013. The code tag instructions were dubious.
Code:
#include <stdio.h>
#include <math.h>
main() {
/*This program uses the Newton-Raphson method to solve y = (x^3)-3 for it's roots.*/
printf("This program uses the Newton-Raphson method to solve y = (x^3)-3 for it's roots. Enter your estimate of the root.
");
float x,y,z;
int num;
num = 0;
[Code]...
What are positive and negative infinity for different data types in c++, are they represent maximum and minimum limit of a type? or positive infinity is not a finite value.can some explain this positive and negative infinity paradigm
View 3 Replies View Relatedint Fib1 = 1;
int Fib2 = 2;
int Fib3 = 0;
int randomynumber;
int Loop;
[code].....
this returns negative numbers sometimes.what did i do wrong side note this is not the complete program it is only the part with the problem because the complete code is sort of longish and very confusing
#include <iostream>
using namespace std;
int initialization (int []);
int identifying (int [],int);
[Code].....
Write a program to enter the number till 100 till the user want and at the end it should display the count of positive and negative and zero entered.
View 8 Replies View RelatedConsider a new data type, the mikesint, which can hold 9 bits.
(a) What is the largest integer that an unsigned mikesint can hold?
(b) What is the largest positive integer that a signed mikesint can hold?
(c) What is the largest negative integer that a signed mikesint can hold?
Not sure how to determine this. I'm stuck.
I have a problem with my assignment. I would like to ask how to transfer positive and negative values from array temperature to arrays positive and negative?
#include <iostream>
#include <iomanip>
using namespace std;
int main(){
int n=0, d=0, temperature[20], sum;
int positive[], negative[];
float avg;
[code]....
I have an assigment and I don't know yet how to write this program :" I introduce the average monthly temperatures of a year(12 values for 12 months). When I compile the program it needs to show the highest negative temparature and the lowest positive temperature of that year.
Example: Entry data: -4 -6 0 5 10 20 24 25 17 8 -1 -7 exit data : max negative= -1 and min positive= 5 "
So I have to make a program that allows the user to enter both positive and negative numbers and the program is suppose to calculate the sum of only the positive values while ignoring the negative values. Also it is to be a sentinel-controlled loop with a number ending the set of values.
View 4 Replies View RelatedHaving error . I multiplied 0 by -4 and my result is -0 instead of 0. I tried to change the data type put It won't work. This is my code:
#include <iostream>
int main () {
double b, c;
std::cout<<"b: ";
std::cin>>b;
std::cout<<"c: ";
std::cin>>c;
std::cout<<b*c<<std::endl;
return 0;
}
I have a program where the user inputs a line of numbers, and the two highest ones are displayed. It works fine, until negative values are entered at which point it shows 0 as the result.
Code: #include <iostream>
using namespace std;
int main( ) {
int num = 0;
int highest = 0;
[Code].....
im supposed to create a program that reads in a list of integers from the terminal and writes the negative numbers to one file and the positive numbers to another file.
i got most of it doen but for some reason its not writting the negative numbers. on what im doing wrong?
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int main(){
int pos_num = 0;
int neg_num = 0;
int positive_numbers = pos_num % 5;
[Code]...
I am trying to average the negative numbers and positive number and of course the total average.
This will read in a list of n values, where n is not known ahead of time. The number of values read into the array will be saved in n.
vector<int> readList() {
std::vector<int> result;
ifstream inFile;
inFile.open("setA.txt");
for (int x; inFile >> x; ) {
result.push_back(x);
[code]....
array is a one-dimensional array of integers and n is the number of elements in that array that contain valid data values. Both of these are input parameters to the function. The function must calculate 1) the average of the n integers in array, storing the result in ave; 2) the average of the positive numbers (> 0), storing the result in avePos, and 3) the average of the negative numbers (< 0), storing the result in aveNeg.
void avgs (std::vector &array, int &ave, int &avePos, int &aveNeg) {
int sum = 0, pos_sum = 0, neg_sum = 0, pos_count = 0, neg_count = 0;
for (auto i : array) {
sum += i;
if (i > 0) { pos_sum += i; ++pos_count; }
[code]....
ignoring negative numbers when I am trying to add up only positive numbers.
SAMPLE:
if (num>=0) {
sum= sum + num;
}
else
how would the else in this case being a negative number not be included in the sum
Well for some reason this doesn't work.
Code:
#include <stdio.h>#include <ctype.h>
int main(void)
{
int lottoNums[150];
int num;
[Code]....
I have this case statement that is not performing as expected. For some reason ASE_OK and ASE_NotPresent both print out. Why is this? ASE_OK is a return value of 0 while ASE_NotPresent is -1.
Code: // Try to create the buffers and store it
switch (theAsioDriver->createBuffers(info, numChannels, bufferSize, callbacksInfo)) {
case ASE_OK:
std::cout << "ASE_OK";
input->bufferSize = output->bufferSize = bufferSize; // Buffer size is stored in both input and output
case ASE_NotPresent:
std::cout << "ASE_NotPresent";
errorMessage = "Could not find input or output devices.";
return NO_DEVICES_FOUND;
[code]....
okay so like case 2 and 3 is not working, also the program is still not finish but why isn't case 2 and 3 not functioning well on the switch part? case 1 works just fine
Code:
#include<stdio.h>
main ()
{
//passcode part (passcode is 1234)
[Code].....
I'm trying to get an if/else statement to work using a character condition, but it is a no-go. After I input either a 'y' or 'n', the program just sits there until I press 'Enter' again, at which point it ends. It never actually goes through either of the 'if' statements. At first I only had a single '=' in the condition, but I found that that was wrong. When I corrected it to '==' it still didn't work. I also tried clearing the buffer by adding a 'getchar()' at the beginning of the program, but that didn't work either.
#include <stdio.h>
#define std_rate .8855
int main(void) {
char ans;
int counter = 1, euros = 5;
double dollars = 0, ex_rate;
[Code] .....
The compiler (Visual Studio 2013) requires me to use 'scanf_s' instead of just 'scanf' for some reason, and if I don't have two 'getchar()' commands at the end, then it won't stay open long enough for me to see the results.
The switch statement is giving me trouble. Worked fine in the main, but when I had to put it in separate functions per my professor, I had several issues. I guessing I am not calling the function correctly. Our group worked on it but could not find a solution to fix the switch statement.
#include <iostream>
#include <iomanip>
#include <fstream>
#include <conio.h>
using namespace std;
//function prototypes
int menu(); //function to show main menu
void Seating_Chart(); //function to show seating chart
[Code] ....
was making a somewhat of a Binary to Hex convertor but only 10/15 cases work and the non working are in the middle; 0010, 0011, 0100, 0101, 0110, 0111;;
// Test Code.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include <iostream>
#include <cmath>
#include <conio.h>
using namespace std;
int main(void)
{int A;
cout << "Enter the binary starting with the MSB
[code]....
Why is my program outputting zero for the variable pop?
Code:
#include <iostream>
int main(void){
using namespace std;
//capture data
cout << "Enter the world's population: ";
[Code] ....
It's obviously not std::cout, but I'm thinking its the way I'm operating on long?
so basically my project goes like this:
Write a C++ program that reads in two positive integers that are 20 or fewer digits in length and outputs the sum of the two numbers.
Your program will read the digits as values of type char so that the number 1234 is read as four characters '1', '2', '3' and '4'. After they are read into the program, the characters are changed to values of type int. The digits will be read into a partially filled array and you might find it useful to reverse the order of the elements in the array after array is filled with data from the keyboard.
Your program will perform the addition by implementing the usual pencil and paper addition algorithm. The result of the addition is stored in an array of size 20 and the result is written to screen. if the result of the addition is an integer with more than maximum number of digits(that is more than 20 digits) then your program should issue a message saying that it has encountered "integer overflow".
You should be able to change the maximum length of the integers by changing only one globally defined constant. Include the loop that allows the user to continue to do more additions until the user says the program should end. What I have so far is
#include <iostream>
#include <cstdlib>
using namespace std;
void reverseArr(int a[], int liu);
void addLargeInt(int a1[], int liu1, int a2[], int liu2, int sum[], int& liu_sum);
int main() {
cin.get(next);
[Code]...
Create a program that keeps on reading positive numbers until user enters a zero value, and then outputs the last positive number entered.
I do not know how to output the 'last positive number' and I do not even know am I doing things correctly so far....
Code:
#include <stdio.h>
int main() {
double num;
for(;;) {
printf("Enter a positive number: ");
scanf("%lf",&num);
[Code] ....
I was going through the following statement in a c book:
"interpreting -1 as unsigned int leads to a big +ve number"
-1 is already unsigned .... then what does the above statement mean ??
Assume you want to make sure that the user enters a positive number that is divisible by 10 with no remainder. Write the condition you would use in the following do-while loop.
do {
cout << “Enter a positive number that is divisible by 10 with no remainder” << endl;
cin >> number;
}
while ( ____________________________________________________________);