'foo' 未在此范围内声明 c++

#39;foo#39; was not declared in this scope c++(foo 未在此范围内声明 c++)
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问题描述

我只是在学习 C++(自从我多年前参加了 1 周的夏令营以来的第一天)

I'm just learning c++ (first day looking at it since I took a 1 week summer camp years ago)

我正在将我正在用 Java 编写的程序转换为 C++:

I was converting a program I'm working on in Java to 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

Integrate 和 sum 都应该是函数

Integrate and sum are both supposed to be functions

这是Java代码,大致相同:

Here is the Java code, more or less the same:

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/.
 *  &copy; 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);
}

我确定这是非常简单的事情

I'm certain it's something very simple

还有,我怎么打电话

getSkewNormal(double skewValue, double x)

来自 SkewNormal.h 之外的文件?

From a file outside SkewNormal.h?

推荐答案

在 C++ 中,您应该先声明函数,然后才能使用它们.在您的代码中,integrate 在第一次调用 integrate 之前没有声明.这同样适用于 sum.因此错误.要么重新排序您的定义,使函数定义先于对该函数的第一次调用,要么为每个函数引入一个 [forward] 非定义声明.

In C++ you are supposed to declare functions before you can use them. In your code integrate is not declared before the point of the first call to integrate. The same applies to sum. Hence the error. Either reorder your definitions so that function definition precedes the first call to that function, or introduce a [forward] non-defining declaration for each function.

此外,在 C++ 中,在头文件中定义外部非内联函数.您对 SkewNormalEvalutatable::SkewNormalEvalutatablegetSkewNormalintegrate 等的定义在头文件中没有作用.

Additionally, defining external non-inline functions in header files in a no-no in C++. Your definitions of SkewNormalEvalutatable::SkewNormalEvalutatable, getSkewNormal, integrate etc. have no business being in header file.

Also SkewNormalEvalutatable e(); C++ 中的声明声明了一个函数 e,而不是您似乎假设的对象 e.简单的 SkewNormalEvalutatable e; 将声明一个由默认构造函数初始化的对象.

Also SkewNormalEvalutatable e(); declaration in C++ declares a function e, not an object e as you seem to assume. The simple SkewNormalEvalutatable e; will declare an object initialized by default constructor.

此外,您将integrate(和sum)的最后一个参数按值作为Evaluatable的对象接收代码>类型.这意味着尝试将 SkewNormalEvalutatable 作为 integrate 的最后一个参数传递将导致 SkewNormalEvalutatable 被分割为 Evaluatable.因此,多态性将不起作用.如果你想要多态行为,你必须通过引用或指针接收这个参数,而不是通过值.

Also, you receive the last parameter of integrate (and of sum) by value as an object of Evaluatable type. That means that attempting to pass SkewNormalEvalutatable as last argument of integrate will result in SkewNormalEvalutatable getting sliced to Evaluatable. Polymorphism won't work because of that. If you want polymorphic behavior, you have to receive this parameter by reference or by pointer, but not by value.

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