Say I wanted to overload the modulus operator to return the remainder of a division between two floating point numbers. Why isn't a custom double operator%(double, double) allowed even though that function isn't available in the standard anyway?
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where are operators defined in C/C++? in headers or compiled code?
looking for the definition of how % modulus is calulated
I heard that the speed of floating point multiplication is much faster than division. Is it still the case today?
View 14 Replies View RelatedI have to build a program that calculates the remainder of the expression
(2^10)!/((2^10-500)! * 500!)
when divided by 10^9+7
where x^y = x*x*x*x...*x (y times)
and x! = x*(x-1)...*1
How can I do that? I know how to calculate the remainder of x! and the remainder of y!, but I do not know how t calculate the remainder of x!/y!. I can´t even store this in a variable because x! is very large.
Function is not returning a decimal point value. Here is my function
int meanValueFunction(vector<int> arrayValues){
int sum = 0;
sum = sumFunction(arrayValues);
float meanValue = sum/arrayValues.size();
cout<< meanValue << endl;
return meanValue;
I want result in decimal point i.e 27.2 for the values (2 4 20 10 100) but it returns 27 instead.
If I have the following code:
long lSecondsSum = 8039;
double dNumDays = lSecondsSum / (24 * 3600);
I expect to get 0.093044 but for some reason I am getting dNumDays = 0.0000000000.
However, if I write the code as follows:
long lSecondsSum = 8039;
double dNumDays = lSecondsSum/24;
dNumDays = dNumDays/3600;
then I get correct dNumDays = 0.092777777777777778.
Also, how do I avoid all these floating point errors.
I want to find the remainder of the division between a and b, but without using the reminder operator a%b.I thought to subtract b from a as long as a>b, that will give the remainder, but I don't know how to write it in code.
View 11 Replies View Relatedif we have a decimal number like c=3.46
And i want to set two number, a and b
now a= static_type<int>(c); so a=3;
and i want b= 46
which is the two numbers after the decimal how can I do that ? how can I set b = 46 ?
P.S: i do not know what c equals to. now it's two number after the floating point but it might be more or less
I need to implement a C++ program that asks the user for four floating-point numbers. The program should then calculate the average using two different functions, one value returning and one void. The program should output the average of the four numbers. For this program I need to use float instead of int for the types of variables. Below is a proto-type code that I am able to use to do this program.
#include <iostream>
using namespace std;
int sum(int,int);
[Code].....
I'm displaying a table of floating point numbers with setprecision(5). If the number is "1.25" it will display "1.2500" which is what I want. However, if the number is "0.25" it will display "0.25000"
How can I make numbers with a base number of zero display properly?
I have the problem of trying to find the smallest natural number that makes two consecutive terms in single precision floating point notation in the riemann zeta function equal. So basically the riemann function of 2 is given by:
sum of 1/(k^2) from k=1 until infinity, so : 1/(1^2) + 1/(2^2) + 1/(3^2) + ...... until infinity.
Now the question asks to stop at the smallest natural number n, at which the sum 1/1^2 + 1/2^2 + ......+ 1/(n^2) is equal to the sum 1/1^2 + 1/2^2 + ..... + 1/((n+1)^2) in single precision floating point notation.
Now well the obvious way to look for n would be on this way:
float i = 1.0;
float n = 1/(i);
float n1 = 1/(i+1.0);
while ( n != n1){
i += 1.0;
n = 1/i;
n1 = 1/(i+1.0);}
But first of all this is obviously completely inefficient and I dont think it will yield a valid answer for any float variable, i.e. I dont think the sum until 1/n^2 and 1/(n+1)^2 will ever differ. I tried it out with the largest possible value of a variable of type float and the values were still seen as unequal in C++. How to do this? Will C++ find a value for n for which the condition holds? Is the compiler or my hardware important for this, i.e. would I get different results on a different pc?
So I am using Visual Studio 2012 Professional, this is C++ code. I am just trying to get the remainder from a very simple division. Nothing difficult, heres the code:
double getProbability(){
int rd = random();
int max = numeric_limits<int>::max();
double result = rd % max;
cout << "Probability: " << result << "
";
return result;
}
When I look at the values in debug I get:
max 2147483647
rd 1804289383
result 1804289383.0000000
That is completely wrong. The answer should be 0.840188. What is going on here?
random() just returns a number from a vector that was prepopulated with "random" integers. Not really random, but that isn't all that important. What is important is why on earth is a % operation returning such a huge number. I assigned the values to variables so I could look at them in the debugger. I know I am going to probably get a thousand different ways that I could do this "better" but again, that isn't what I am looking for. I would just like to know why the % operation is doing what it is doing?
Try to implement overloading << operator. If I done it void then everything work fine (see comment out) if I make it class of ostream& then the operator return to me some memory address.
Code:
#ifndef Point_HPP // anti multiply including gates
#define Point_HPP
#include <sstream>
class Point {
private:// declaration of private data members
double x;// X coordinate
double y;// Y coordinate
[Code] .....
I have a program which generates lots of data points in 2D (x and y co-ordinates). Perhaps 1,000,000+ of these points. These are floating point numbers. The values all fall within a specific range -R/2 and +R/2.
I want to impose an 'imaginary' grid. Such that I can collect the number of each of these points falling within a specific grid location. The actual 'grid' size is variable.
Should I try and collate this data from my C++ program and then post-process it in some other s/w like matlab?
The question was to evaluate postfix expression (floating point numbers). I had been able to implement stack data structure using one way singly linked list linked list but I am not been able to extract the original input by the user expressions like
ex:
1. 252.124 3453.7 * 46.3 346.2 23.6 ^/$
2.45.23 87.045 * 6.5 ^$
etc,($ELIMETER)
How to take such inputs from the user for proper evaluation . Previously I tried to extract separate digits from integer and decimal fields and computed numbers. The method is very lengthy. Any optimised way for taking such input!
I'm stuck on the last part of my program. The directions are the following~
Expand the program to add an overloaded function to handle floating point numbers (i.e., doubles). Include output for one list of integers and one list of doubles. Use this function prototype: double avgx(double&, double&, int, ...);
Compile and run. You should have one function named avg, one named davg, and two functions named avgx
My code does not compile and I think I'm not declaring my function prototype correctly?
#include <iostream>
using std::cout;
using std::endl;
#include <cstdarg>
// function prototype(s)
int avg(int, ...);
[Code] ....
For class I need to write a program that inputs a file (the dividend), performs binary division on the file (using 0x12 as the divisor), and outputs the remainder(checksum).
I have researched binary division algorithms and I get the general gist, but I'm still unsure where to start. How would I store the dividend and divisor? As two arrays of bits?
Then, I need to figure out how to perform shifts and XORs on the the binary numbers. Maybe I should use bitwise operations?
I have a text file like below read.txt:
1.0 2.0 3.0 4.0
2.0 3.0 4.0 6
5.0 7 1.0 5.0
calc.cpp:
void main() {
FILE *fp;
fp=fopen("read.txt","r");
double *read_feature = new double*[3];
[Code] ....
I want to count all the numbers in my text file (read.txt). Read text file consist of floating and integer number. Answer for the above file would be integer=2 and float =10.
I have a program written to add 2 complex numbers. Everything is working, except when the sum gets written, it gives me a number that is way off.
#include <iostream>
#include <complex>
#include <fstream>
#include <cstdlib>
#include <math.h>
class complex {
public:
complex();
complex(double r, double i){
[Code] .....
And my output ends up being Enter a complex number (a+bi) :
1+2i
Enter a complex number (a+bi) :
2+3i
x 1+2i
y 2+3i
4.8784e-270+4.85593e-270i
I have tried to understand the concept of linked lists and I have read the assigned chapter 2 times. My teacher is a little laid back when it comes to teaching! This is only a portion of my program. This function is supposed to add 2 binary numbers 11101+1101 and store the result in the temp list. The answer I get is 10000.I don't think that it is adding the carry.
binary operator+( binary num1, binary num2) {
binary temp;
int carry, sum;
[Code]..
Code:
while ((y % 12) != 0) {
y++;
}
I liked that the above code does not put the result into a variable and then test the variable which would use more memory, and more lines of code. Is this thinking bad?
how can we use modulus operator with double type variables
View 2 Replies View RelatedQuestion about instead of using the division operator to display the output of user"s input....
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#include<stdio.h>
#include<conio.h>
void main()
{
float i;
i=0.7;
[Code] ....
If i do run the above program in turbo C/C++ complier, it outputs "h". But,if i change the code as i=0.6 and if (i<0.6), it outputs "w". Even if i change it to i=0.8 and if(i<0.8), then also it outputs "w".
Code:
#include<stdio.h>
#include<conio.h>
float square(float);
void main() { clrscr();
float a,b;
printf("ENter a Number");
scanf("%f",&a);
[Code] ....
In the above program, I am calculating the square of float number. But sometimes the number is entered as NAN and sometimes Output is NAN. What is NAN? I am entering floating point number, then y NAN is entered?
SEE the Image attached for the OUTPUT.
I am having trouble understanding the mantissa of a floating point number. I have divided up the floating point number into the sign bit, the exponent and the mantissa, I have found the exponent, but I am not sure what to do with the mantissa? From what I have gathered so far i divide the mantissa by ten until I get a number between 1 and 10. after that i convert the number into a decimal with everything after the decimal point (or radix) being a fractional number. But when I do that on paper I dont get my intended number. How do i put the exponent and mantissa together to make a decimal from my floating point?
ex. input is 00111010000111111111011000001000
sign is 0
exponent is 01110100 which is 64+32+16+4-127=-11
mantissa is 00111111111011000001000 which would be 1.11111111011000001
When i convert that i get 1.99756622314 i dont know what to do with the -11 exponent and the answer i want is 6.1e-4
