我有这段代码:
#include <iostream>
#include <cmath>
#include <stdlib.h>
using namespace std;
double f(double x);
double biseccion(double a, double b, double tolerancia, int maxiter);
int main()
{
double a, b, raiz;
double tolerancia=0.00000;
int maxiter=25;
cout << "Input begin of interval: ";
cin >> a;
cout << "Input end of interval: ";
cin >> b;
cout << "\n";
cout << " # de"<<"\n"<<"Iteration"<<"\t"<<" A"<<"\t"<<" B"<<"\t"<<" C"<<"\t"<<" f(c)"<<endl;
raiz=biseccion(a,b,tolerancia,maxiter);
cout << "\n";
cout << "The root is: "<< raiz <<endl;
return 0;
}
double f(double x)
{
return x*x*x-x-2;
}
double biseccion(double a, double b, double tolerancia, int maxiter)
{
double c;
int numiter=1;
do
{
c=(a+b)/2;
if(f(a)*f(c)<0)
{
b=c;
}
else
{
a=c;
}
cout<<" "<<numiter<<"\t"<<"\t"<<a<<"\t"<<b<<"\t"<<c<<"\t"<<f(c)<<endl;
numiter++;
}
while((abs(f(c))>tolerancia)&&(numiter<maxiter));
return c;
}
不是在我的代码中写“x * x * x-x-2”,而是希望用户在请求间隔开始之前输入它。我怎么能这样做?
我尝试使用变量存储“x * x * x-x-2”,但都没有效果。
答案 0 :(得分:6)
你需要解析输入,它可能不像你想的那么容易,但有一些库可以帮助你。
muparser.sourceforge.net/
code.google.com/p/expressionparser /
partow.net/programming/exprtk/index.html
这里也是c#的解决方案,也可能对你有所帮助。