8.7 复习题

1.经常被调用，逻辑简单，代码少，一般要求没有递归，没有循环。

2.

void song(const char * name, int times);

a.void song(const char * name, int times = 1);

b.加默认值只改原型就行了

c.void song( char * name = "oh, my god", int times = 1);

3.

void iquote(int x)
{cout << "\"" << x << "\""<<endl;
}

void iquote(double x)
{cout << "\"" << x << "\"" << endl;
}

void iquote(string x)
{cout << "\"" << x.c_str() << "\"" << endl;
}

int main()
{ iquote(1);iquote(2.123456);iquote("FableGame,http://blog.csdn.net/u012175089");cin.get();return 0;
}

4.

结构：

struct box
{char maker[40];float height;float width;float length;float volume;
};

//显示
void show(box& b)
{cout << "maker:" << b.maker << " height:" << b.height << " width:"<<b.width<< " length:" << b.length << " volume:"<< b.volume;
}

//计算体积
void computeVolume(box& b)
{b.volume = b.height * b.width * b.length;
}

//main函数
int main()
{box b{ "FableGame", 1, 2, 3, 0 };computeVolume(b);show(b);cin.get();return 0;
}

5.

//原来的代码

#include <array>
#include <string>
#include <set>
using namespace std;const int Seasons = 4;
const array<string, Seasons> Snames = { "Spring", "Summer", "Fall", "Winter" };
void fill(array<double, Seasons> * pa);
void show(array<double, Seasons> da);

int main()
{array<double, Seasons> expenses;fill(&expenses);show(expenses);cin.get();cin.get();return 0;
}

void fill(array<double, Seasons> * pa)
{for (int i = 0; i < Seasons; i++){cout << "Enter" << Snames[i] << " expenses: ";cin >> (*pa)[i];}
}

void show(array<double, Seasons> da)
{double total = 0.0;cout << "\nEXPENSES\n";for (int i = 0; i < Seasons; i++){cout << Snames[i] << ":$" << da[i] << endl;total += da[i];}cout << "Total Expenses: $" << total << endl;
}

//修改后的代码

#include <iostream>
#include <array>
#include <string>
#include <set>
using namespace std;const int Seasons = 4;
const array<string, Seasons> Snames = { "Spring", "Summer", "Fall", "Winter" };
void fill(array<double, Seasons>&pa);
void show(array<double, Seasons>& da);

int main()
{array<double, Seasons> expenses;fill(expenses);show(expenses);cin.get();cin.get();return 0;
}

void fill(array<double, Seasons>& pa)
{for (int i = 0; i < Seasons; i++){cout << "Enter" << Snames[i] << " expenses: ";cin >> pa[i];}
}

void show(array<double, Seasons>& da)
{double total = 0.0;cout << "\nEXPENSES\n";for (int i = 0; i < Seasons; i++){cout << Snames[i] << ":$" << da[i] << endl;total += da[i];}cout << "Total Expenses: $" << total << endl;
}

区别不是很大，不过原理是不同的。

6.

//a

double mass(double density, double volume = 1.0);

//或者重载

double mass(double density, double volume);
double mass(double density);

//b.默认函数只能放在后边,所以只能重载

void repeat(int times, const char* str);
void repeat(const char* str);

//c.可以使用模板函数

template< typename T> T average(T a, T b);

//也可以使用重载函数

int average(int a, int b);
double average(double a, double b);

//d.不能根据返回类型来决定函数的选择,两个版本的特征标相同。

7.

template<typename T> T getMax(T a, T b)
{return a > b ? a : b;
}

int main()
{cout << getMax(1, 2) << endl;cout << getMax(12.323, 3.333) << endl;cin.get();cin.get();return 0;
}

8.书上印错了，是复习题7的模板

#include <iostream>
#include <array>
#include <string>
#include <set>
using namespace std;struct box
{char maker[40];float height;float width;float length;float volume;
};

//显示
void show(box& b)
{cout << "maker:" << b.maker << " height:" << b.height << " width:" << b.width<< " length:" << b.length << " volume:" << b.volume<< endl;
}

//计算体积
void computeVolume(box& b)
{b.volume = b.height * b.width * b.length;
}

//函数模板
template<typename T> T getMax(T a, T b)
{return a > b ? a : b;
}

//模板的具体化
template<> box getMax(box a, box b)
{return a.volume > b.volume ? a : b;
}

//main
int main()
{cout << getMax(1, 2) << endl;cout << getMax(12.323, 3.333) << endl;box a{ "FableGame A", 12, 23, 33, 0 };computeVolume(a);show(a);box b{ "FableGame B", 41, 12, 13, 0 };computeVolume(b);show(b);box c = getMax(a, b);show(c);cin.get();cin.get();return 0;
}

9.对于变量加（），就是引用

int g(int x)
{return x;
}

int main()
{float m = 5.5f;float& rm = m;decltype(m) v1 = m;//floatdecltype(rm) v2 = m;//float&decltype((m)) v3 = m;//float&decltype(g(100)) v4 = g(12345);//intdecltype(2.0 * m) v5 = 2.0 * m;//doublecout << v1 << "\t" << v2 << "\t" << v3 << "\t" << v4 << "\t" << v5 << "\t" << endl;m = 123.123f;cout << v1 << "\t" << v2 << "\t" << v3 << "\t" << v4 << "\t" << v5 << "\t" << endl;cin.get();cin.get();return 0;
}