'foo'未在此范围内声明c ++

Question

我刚刚学习c ++(自从我多年前参加一个为期一周的夏令营以来第一天看它)

我正在将我正在使用Java编写的程序转换为C++:

#ifndef ADD_H
#define ADD_H
#define _USE_MATH_DEFINES
#include <iostream>
#include <math.h>

using namespace std;

class Evaluatable {
public:
  virtual double evaluate(double x);
};

class SkewNormalEvalutatable : Evaluatable{
public:
  SkewNormalEvalutatable();
  double evaluate(double x){
    return 1 / sqrt(2 * M_PI) * pow(2.71828182845904523536, -x * x / 2);
  }
};

SkewNormalEvalutatable::SkewNormalEvalutatable()
{
}

double getSkewNormal(double skewValue, double x)
{
  SkewNormalEvalutatable e ();
  return 2 / sqrt(2 * M_PI) * pow(2.71828182845904523536, -x * x / 2) * integrate(-1000, skewValue * x, 10000, e);
}

// double normalDist(double x){
//   return 1 / Math.sqrt(2 * Math.PI) * Math.pow(Math.E, -x * x / 2);
// }

double integrate (double start, double stop,
                                     int numSteps, 
                                     Evaluatable evalObj)
{
  double stepSize = (stop - start) / (double)numSteps;
  start = start + stepSize / 2.0;
  return (stepSize * sum(start, stop, stepSize, evalObj));
}

double sum (double start, double stop,
                               double stepSize,
                               Evaluatable evalObj)
{
  double sum = 0.0, current = start;
  while (current <= stop) {
    sum += evalObj.evaluate(current);
    current += stepSize;
  }
  return(sum);
}

// int main()
// {
//   cout << getSkewNormal(10.0, 0) << endl;
//   return 0;
// }
#endif

错误是:

SkewNormal.h: In function 'double getSkewNormal(double, double)' :
SkewNormal.h: 29: error: 'integrate' was not declared in this scope
SkewNormal.h: In function 'double integrate(double, double, int, Evaluatable)':
SkewNormal.h:41: error: 'sum' was not declared in this scope

积分和求和都应该是函数

这是Java代码,或多或少相同:

public static double negativelySkewed(double skew, int min, int max){
    return randomSkew(skew) * (max - min) + min;
}

public static double randomSkew(final double skew){
    final double xVal = Math.random();
    return 2 * normalDist(xVal) * Integral.integrate(-500, skew * xVal, 100000, new Evaluatable() {

        @Override
        public double evaluate(double value) {
            return normalDist(value);
        }
    });
}

public static double normalDist(double x){
    return 1 / Math.sqrt(2 * Math.PI) * Math.pow(Math.E, -x * x / 2);
}

/** A class to calculate summations and numeric integrals. The
 *  integral is calculated according to the midpoint rule.
 *
 *  Taken from Core Web Programming from 
 *  Prentice Hall and Sun Microsystems Press,
 *  http://www.corewebprogramming.com/.
 *  © 2001 Marty Hall and Larry Brown;
 *  may be freely used or adapted. 
 */

public static class Integral {
  /** Returns the sum of f(x) from x=start to x=stop, where the
   *  function f is defined by the evaluate method of the 
   *  Evaluatable object.
   */

  public static double sum(double start, double stop,
                           double stepSize,
                           Evaluatable evalObj) {
    double sum = 0.0, current = start;
    while (current <= stop) {
      sum += evalObj.evaluate(current);
      current += stepSize;
    }
    return(sum);
  }

  /** Returns an approximation of the integral of f(x) from 
   *  start to stop, using the midpoint rule. The function f is
   *  defined by the evaluate method of the Evaluatable object.
   */

  public static double integrate(double start, double stop,
                                 int numSteps, 
                                 Evaluatable evalObj) {
    double stepSize = (stop - start) / (double)numSteps;
    start = start + stepSize / 2.0;
    return(stepSize * sum(start, stop, stepSize, evalObj));
  }
}

/** An interface for evaluating functions y = f(x) at a specific
 *  value. Both x and y are double-precision floating-point 
 *  numbers.
 *
 *  Taken from Core Web Programming from 
 *  Prentice Hall and Sun Microsystems Press,
 *  http://www.corewebprogramming.com/.
 *  &copy; 2001 Marty Hall and Larry Brown;
 *  may be freely used or adapted. 
 */
public static interface Evaluatable {
      public double evaluate(double value);
}

我确定这很简单

另外,我该如何打电话

getSkewNormal(double skewValue, double x)

从SkewNormal.h以外的文件？

Answer 1

在C++中,您应该在使用它们之前声明函数.在您的代码integrate中,在第一次调用之前未声明integrate.这同样适用于sum.因此错误.重新排序定义,以便函数定义在第一次调用该函数之前,或者为每个函数引入[forward]非定义声明.

另外,在C++中禁止使用头文件中定义外部非内联函数.你的定义SkewNormalEvalutatable::SkewNormalEvalutatable,getSkewNormal,integrate等有没有企业在头文件之中.

SkewNormalEvalutatable e();C++中的声明也声明了一个函数e,而不是e你想象的对象.simple SkewNormalEvalutatable e;将声明一个由默认构造函数初始化的对象.

此外,收到的最后一个参数integrate(和sum)由值作为对象Evaluatable类型.这意味着尝试传递SkewNormalEvalutatable作为最后一个参数integrate将导致SkewNormalEvalutatable切片Evaluatable.因此,多态性不起作用.如果您需要多态行为,则必须通过引用或指针接收此参数,而不是按值接收.

Answer 2

在C++中,您的源文件通常在一次传递中从上到下进行解析,因此必须先声明任何变量或函数,然后才能使用它们.这有一些例外,比如在类定义中定义函数内联时,但是代码不是这种情况.

移动integrate上面的定义getSkewNormal,或者在上面添加前向声明getSkewNormal:

double integrate (double start, double stop, int numSteps, Evaluatable evalObj);

这同样适用sum.

Answer 3

通常,在C++函数中必须在调用之前声明它们.所以在定义之前的某个时候getSkewNormal(),编译器需要看到声明:

double integrate (double start, double stop, int numSteps, Evaluatable evalObj);

人们所做的大部分是将所有声明(仅)放在头文件中,并将实际代码(函数和方法的定义)放入单独的源(*.cc或*.cpp)文件中.这巧妙地解决了需要声明所有函数的问题.