我的教授希望我从calculateArea输出“区域”作为字符/字符串。我不确定他的意思,但也许你们中的一些人可能会理解。
#include <iostream>
#include "math.h"
#include <cmath>
#include <sstream>
#include <string>
using namespace std;
const char& calculateArea(double diameter, double chord)
{
double length_1, length_2, angle; //This creates variables used by the formula.
angle = acos( (0.5 * chord) / (0.5 * diameter) ); //This calculates the angle, theta, in radians.
cout << "Angle: " << (angle * 180) / 3.14159 << "\n"; //This code displays the angle, currently in radians, in degrees.
length_1 = (sin(angle)) * 6; //This finds the side of the triangle, x.
cout << "X: " << length_1 << " inches "<< "\n"; //This code displays the length of 'x'.
length_2 = (0.5 * diameter) - length_1; /*This code finds the length of 'h', by subtracting 'x' from the radius (which is half the diameter).*/
cout << "h: " << length_2 << " inches" << "\n"; //This code displays the length of 'h'.
double area = ((2.0/3.0) * (chord * length_2)) + ( (pow(length_2, 3) / (2 * chord) ) ); /*This code calculates the area of the slice.*/
ostringstream oss;
oss << "The area is: "<< area << " inches";
string aStr = oss.str();
cout << "Debug: "<< aStr.c_str() << "\n";
const char *tStr = aStr.c_str();
cout << "Debug: " << tStr << "\n";
return *tStr;
//This returns the area as a double.
}
int main(int argc, char *argv[]) {
double dia, cho; //Variables to store the user's input.
cout << "What is your diameter? "; //
cin >> dia; // This code asks the user to input the diameter & chord. The function will calculate
cout << "What is your chord? "; // the area of the slice.
cin >> cho; //
const char AreaPrint = calculateArea(dia, cho); //Sends the input to the function.
cout << AreaPrint; //Prints out the area.
return 0;
}
虽然我得到了输出:
你的直径是多少? 12
你的和弦是什么? 10
角度:33.5573
X:3.31662英寸
h:2.68338英寸
调试:面积为:18.8553英寸
调试:面积为:18.8553英寸
Ť
我需要弄清楚如何返回字符串tStr指向。如果你不明白我说的话,抱歉,不确定教授的要求是什么。
答案 0 :(得分:1)
您正在查找char引用而不是字符串。 (* tStr说'给我指针的内容')
您正在尝试的更正确的版本是:
const char* calculateArea(double diameter, double chord)
和
return tStr; // no 'content of'
但这仍然很糟糕:当aStr超出范围时,返回的字符串是“超出范围”,所以你真的需要返回字符的副本或者只返回字符串本身(让std lib担心你的副本)
const string calculateArea(double diameter, double chord)
...
return aStr;