C++ :: Calculating Equation Using Expression Builder
Jan 22, 2014How do to calculate the following equation"5+((56+8)/2)" using expression builder
Note: the equation may change "50+((26+8)/12)"
How do to calculate the following equation"5+((56+8)/2)" using expression builder
Note: the equation may change "50+((26+8)/12)"
It's been awhile since I've coded in C. I'm picking it back up to get some practice. I'm going through my old programs from a few years ago, and I ran into one where I'm calculating the entropy of a file. The problem I'm having is that everytime the program runs, no matter what file I choose for it to calculate the entropy on, it gives 3.00000 as a result. I've gone through it and can't seem to figure out why it's possible doing that.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define SIZE 256
int entropy_calc(long byte_count[], int length) {
float entropy;
float count;
[Code] .....
I am having a world of problems of what i thought to be rather simple.
Edit1->Text = "123401278034987120478102873401982734";
Now Edit1->Text is an AnsiString
And I would like to format this to a number value, so that i can say.
CSpinEdit1->Value = "123401278034987120478102873401982734"
In other words transferring the value from Edit1->Text to CpinEdit1->Value;
Integers won't work because the number is to big. So i tried going from String to Number like in this article Converting numbers to strings and string - C++ Articles
But that won't work because its an Ansistring, how would i go about this conversion?
I would need to know how to make the c++ builder application to be started with administrator rights (it's common behaviour for installers - when it is started, it asks for administrator rights).
View 4 Replies View Relatedhow to take two variables with values such as a 1; and b 2;. Now you take cout << a + b; and you will get 3, but I want to somehow have a 1; b 2; and c num; then have a + b = c. So what ever I have the user input is for a and be added together will become the value of the c variable.
View 1 Replies View RelatedThis is my code:
Code:
#include <math.h>
#include <stdio.h>
int main(void) {
float a,b,c,root_1,root_2;
printf("Please enter value a from the quadratic equation
[Code] ......
And I keep getting this error:
Code:
/tmp/ccgtUIun.o: In function `main':
assign345.c:(.text+0xc7): undefined reference to `sqrt'
assign345.c:(.text+0xef): undefined reference to `sqrt'
collect2: ld returned 1 exit status
I have a small project for school due today but i cant get my head around programming
Ive to create a program where ive to get the values from the user and add them to an equation. I have also to add a selection statement where it will ask for the material being tested which will have its value set in the program.
Write a program that creates the output shown in the Output Layout section below. The program should create 2 points with x and y coordinates as integers, prompt the user to input the x and y values for one of the points and randomly set the other (-99 to 99 range) and output the length of the radius line segment and the area of the circle that radius defines. The program should then end. Include an SDM for the program and any other appropriate documentation.
Special Calculations:
Distance between 2 points equation:
√((p0x – p1x)2 + (p0y – p1y)2) (This requires use of the math library)
Output Layout: (bold text represents user input)
Please enter the location of your first point.
Enter a value from -99 to 99 for your x coordinate: -2
Enter a value from -99 to 99 for your y coordinate: 17
The location of your second randomly set point.
Your x coordinate: 45
Your y coordinate: -89
The length of the radius line segment from point one to point two is 115.
The area of the circle with a radius of 115 is 41546.33.
Other:
int pAx;
int pAy;
int pBx;
int pBy;
soo quadratic equation solver. output is "x=nan x=nan".
#include <iostream>
#include <string>
#include <cmath>
using namespace std;
//prototyping
double ans_1(double,double,double)
[code].....
I'm writing an algorithm on a microcontroller in C.
One of the iterations that takes a long time involves the following equation, summing the product of pairs of variables:
signed char VarA;
signed char VarA;
signed long VarA;
Total+=VarA*VarB;
The vast majority of cases, at least one of the variables VarA and VarB will be 0, so nothing will be added to Total.
Is there a way to check for this condition which might give me a performance increase over doing a lot of multiplying-by-zeros?
I am creating a program that solves the quadratic equation ax^2 + bx +c.
I have this program almost complete except the output of the equation in the function called display_quadratic. I need the program to display the variables a,b,c in the equation ax^2 + bx + c but I am having 2 problems. My first problem is that I cannot get the right addition and subtraction signs for the equation.
For instance, if the program had the values for a,b,c as 2,2,3
it will display 2x^2 2x 3
How can I get it to display 2x^2 + 2x + 3? or if it was negative like 2x^2 - 2x - 3?
My next question is how do I get to not display the coefficients that are 1?
I had an if-else statement but no matter what I created it would overlap with another statement and print out twice. Here is the code:
Code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
void display_quadratic (float a, float b, float c);
float root1(float a, float b, float c);
[Code] ......
I am having trouble understanding how this loops would work. Give the function (x*x*x*x) - (10*x*x*x) + (35*x*x) - (50*x) + 24 Write a program that will use bisection method to find the roots of this function. Define lower limit and upper limit (e.g. -1.05 and 6.05) Starting at the lower limit step along the X axis at intervals of 0.1 for H calculating the function values f(x) and f(x +h), then f(x+h) and f(x+2h) until the upper limit is exceeded. If sign of the function value changes this indicates a root between ranges. Apply bi sectional method to this range until root has been found with an epsilon of 0.000001, then continue on until upper limit has been exceeded and all 4 roots found. If the function is within 0.000001 of zero then root has been found therefore no need for bi sectional method.
Code:
#include "stdafx.h"
#include "math.h"
#define H 0.1
#define epsilon 0.000001
double F(double x);
int main(void)
[code]....
So we have a function and a range. Program requires to work along the X axis at 0.1 increments until it reaches a point where the value changes from positive to negative or negative to positive. Then apply bisection method within that range to a given accuracy then print that root. Then continue on X axis until the next change of sign is found.
I am working on a program to find the value of the current in a coil. This value satisfies the following equation:
y'=sin(2t)-[(ey-1)/(ey+1)]
which is of the form y'=f(t,y)
I know that in order to solve this I need to use the trapezoidal method to solve a differential equation, the formula is:
yn+1=yn+.5*h(f(tn,yn)+f(tn+1,yn+1) where h=tn+1-tn
I have found examples of the standard trapezoidal method but I do not think they will work because of the difference in the formulas.
This is my code for the quadratic equation. It keeps telling me that my else is illegal since no matching if statement and my else statement is missing a statement
#include <iostream>
#include <cmath>
#include <string>
#include <iomanip>
using namespace std;
int main() {
string Name;
[Code] .....
My program is returning NaN even though the value in square root function is not negative -1*b + ((sqrt(pow(b,2) * -4 *(a*c))) / 2).
x = 10;
y = -1;
z = 1;
Heres the program:
main.cpp:
#include <iostream>
#include "Quad.h"
double x,y,z; // variable x = 'a', variable y = 'b' and variable z = 'c'.
[Code] ....
I would like it return the answer and not "NaN".
I've written a program that finds the maximum of a function. I now want to change it a little so that instead of evaluating the function y=x^2-7x-18, the program will ask the user to input an equation and then evaluate that equation. I'm really not sure of how to do so.
#include <iostream>
#include <cmath>
#include <cstdlib>
using namespace std;
int main() {
int a, b, delta, x, y;
double max= -1.8 * pow(10, 308);
[Code] .....
How do I enter the equation all at once instead of prompting for each piece separately?
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
[Code].....
I want to find the root of an equation f(x) by using the secant method [URL] .....
#include <math.h>
#include <iostream>
using namespace std;
//Define function f(x)
double f(double x, double a, double b) {
return sin((a*x)/(1+x*x))*atan(b*x)+atan(x);
[Code] ....
1) Do I have to keep the function f and RootFinderSMNew in different cpp files?
2) When compiling I get the following errors:
1>------ Build started: Project: Assignment2.1, Configuration: Debug Win32 ------
1> Find Roots Secant method.cpp
[Code] .....
What am I doing wrong ?
So, I successfully made a program that will perform the quadratic equation on three numbers, imaginary or real. however, i am now trying to simplify the result, as to get rid of the "/2a" on the bottom. Hence the simplify() function. I just started to create the simplification function, and am attempting to divide the imaginary part of the solution as well as the real part of the solution by 2a. Somehow, it gives the error, "error:invalid operands of types 'int' and 'double *' to binary 'operator*'" on lines 105 and 106. I suspect it has to do with the pointers and references that i am passing as parameters. Also, just an aside, I have never actually seen "/=" be used. It can be, right? I know "+=" can be.
#include <iostream>
#include <string>
#include <cmath>
#include <cstdlib>//simplify the answer
using namespace std;
int count=0;
//prototyping
double ans_1(double,double,double);
[Code] ....
I am using matlab R2o11 in which the programming medium is c++. I am having these variables=
T=373 t0 376
V=0.09 to 1
P=T/V
I just need a result showing ... First it should take T=373 and solve the equation using all volume values. Then it takes another temperature and do the same.
I need an equation that determines the lowest possible cost for laying power line.
The variables are as follows:
1. The width of a river
2. The distance a factory is downstream from a station on the other side of the river.
3. The cost for laying line underwater.
4. The cost for laying line on land.
All these variables will be user input each use of program.
Finally got a programme partially working, why the values are not showing? Is this down to incorrect formula or i havent declared the values etc?
Code:
#include <stdio.h>
#include <math.h>
float timeconstant(float R, float C);
float R;
float C;
float time_c;
[Code] ....
I've been trying to calculate the Second standard deviation but the average in the second loop isn't calculating correctly which is causing the standard deviation (method 2) to not calculate correctly. I can't find anything wrong.
Code:
#include <iostream>#include <iomanip>
#include <string>
#include <fstream>
#include <cmath>
usingnamespacestd;
int main ()
[Code]...
Currently I am working on a program where you enter in a date 14 03 2013 (Day, Month, Year) and you get the next day. I seem to be coming stuck with months with less than 31 days, and the whole leap year thing. Here is my code so far.
Code:
#include <stdio.h>
int day, month, year, next_day, next_month, next_year, calculation;
int main() {
printf("Enter a date in the form day/month/year: ");
scanf("%d %d %d", &day, &month, &year);
[Code] .....
As part of a project i want to display the speed of a small wheel on a 7 segment display. I am using a hall effect sensor to pick up the pulses of the rotation. I am not sure how to write a programe on C to calculate the RPM from this....
View 4 Replies View RelatedI am having trouble with my C++ program. My program asks the user to enter any number greater than 0 and less than 20, and if the user enters any number(s) outside the acceptable range, the output should display:
input outside of exceptable range > 0 and < 20
length: 1 width: 1 area: 1 perimeter: 1
But in my program, this output displays no matter what the user types (even if the numbers are in the acceptable range, when it should be displaying the length and width the user enters and also displaying the calculated perimeter and area whenever the user enters number(s) in the acceptable range. Here is my code:
#include <iostream>
using namespace std;
float class_rectangle_area();
float class_rectangle_perimeter();
float class_rectangle_length();
float class_rectangle_width();
[Code] .....