C++ 11 从C++ primer第五版的学习笔记

1. auto （page107）

auto 推断会忽略const

 

const int ci = i, & cr = ci; auto b = ci; // b is an int (top-level const in ci is dropped) auto c = cr; // c is an int (cr is an alias for ci whose const is top-level) auto d = & i; // d is an int*(& of an int object is int*) auto e = & ci; // e is const int*(& of a const object is low-level const)

 

如果需要推断const，需要明确加上const

const auto f = ci; // deduced type of ci is int; f has type const int

 

We can also specify that we want a reference to the auto-deduced type. Normal

initialization rules still apply:

 

; ; // ok: we can bind a const reference to a literal

 

When we ask for a reference to an auto-deduced type, top-level consts in the

initializer are not ignored（当我们需要引用来推断类型，top-level const在初始化时没有被忽略）. As usual, consts are not top-level when we bind a reference to an initializer.

(ps:We use the term top-level const to indicate that the pointer itself is a const. When a pointer can point to a const object, we refer to that const as a low-level const.)

 

When we define several variables in the same statement, it is important to

remember that a reference or pointer is part of a particular declarator and not part of

the base type for the declaration. As usual, the initializers must provide consistent

auto-deduced types:

 

auto k = ci, &l = i; // k is int; l is int& auto &m = ci, *p = &ci; // m is a const int&;p is a pointer to const int // error: type deduced from i is int; type deduced from &ci is const int auto &n = i, *p2 = &ci;

 

********************************************************************************************

需要注意的是，auto不能用来声明函数的返回值。但如果函数有一个尾随的返回类型时，auto是可以出现在函数声明中返回值位置。这种情况下，auto并不是告诉编译器去推断返回类型，而是指引编译器去函数的末端寻找返回值类型。在下面这个例子中，函数的返回值类型就是operator+操作符作用在T1、T2类型变量上的返回值类型。

1 2 3 4 5 6

template <typename T1, typename T2> auto compose(T1 t1, T2 t2) -> decltype(t1 + t2) {    return t1+t2; } auto v = compose(2, 3.14); // v's type is double

来源： <http://blog.jobbole.com/44015/>

 

********************************************************************************************

2.decltype(page109)

返回值 decltype(表达式)，表达式是不会运行的

[返回值的类型是表达式参数的类型]

 

decltype(f()) sum = x; // sum has whatever type f returns

 

 

decltype不会忽略 top-level const

, ; // x has type const int decltype(cj) y = x; // y has type const int& and is bound to x decltype(cj) z; // error: z is a reference and must be initialized

 

Some expressions willcause decltype to yield a reference type. Generally speaking, decltype returns a reference type for expressions that yield objects that can stand on the left-hand side of the assignment:(有时候表达式会导致decltype返回引用类型。一般而言，decltype返回表达式左值的引用类型)

 

, ) b; // ok: addition yields an int; b is an (uninitialized) int decltype(*p) c; // error: c is int& and must be initialized

 

r是引用类型，decltype(r)也是引用类型，但是(r+0)这个表达式返回的左值类型不是引用类型。岁b是 int类型

 

On the other hand, the dereference operator is an example of an expression for

which decltype returns a reference. As we’ve seen, when we dereference a pointer,

we get the object to which the pointer points. Moreover, we can assign to that object.

Thus, the type deduced by decltype(*p) is int&, not plain int. (大概意思，*操作符修饰表达式，decltype返回表达式（*p）的lvalue（参考(4)lvalue rvalue的解释），所以最后decltype(*p)是int&，而不是int)

 

另一个auto和decltype推断类型不同的地方是decltype还依赖给予的表达式。将变量放入小括号中将会影响decltype的返回类型。将变量放val入小括号，decltype把(val)当做表达式，这种表达式的lvalue。

 

//decltype of a parenthesized variable is always a reference decltype((i)) d; // error: d is int& and must be initialized,because (i) is an lvalue decltype(i) e; // ok: e is an (uninitialized) int

 

 

3.Range-Based  for Loops(page)

语法格式：

for (declaration : expression)

    statemen

expression是一个包含顺序元素的类型

 

遍历一个string

string str("some string"); // print the characters in str one character to a line for (auto c : str) // for every char in str cout << c << endl; // print the current character followed by a newline

 

 

string s(; // count the number of punctuation characters in s for (auto c : s) // for every char in s if (ispunct(c)) // if the character is punctuation ++punct_cnt; // increment the punctuation counter cout << punct_cnt<< " punctuation characters in " << s << endl;

这个地方似乎就感觉到了decltype的方便性了。

 

如果需要修改元素，需要用引用，auto& c,输出结果：HELLO WORLD!!!

string s( "Hello World!!!" ); // convert s to uppercase for ( auto & c : s) // for every char in s (note: c is a reference) c = toupper(c); // c is a reference, so the assignment changes the char in s; cout << s << endl;

 

 

Warning

The body of a range for must not change the size of the sequence over

which it is iterating.

 

4.模板右边括号（page142）

在C++ 98中，vector<vector<int>> vctTemp是一个非法的表达式，编译器会认为右边的>>是一个移位操作符，因此必须修改为vector<vector<int> > vctTemp，即在右边的两个>中间添加一个空格。在C++ 11中，这将不再是一个问题，编译器将能够识别出右边的双括号是两个模板参数列表的结尾。

 

Warning

Some compilers may require the old-style declarations for a vector of

vectors, for example, vector<vector<int> >

 

5.初始化列表（page143）

在引入C++ 11之前，只有数组能使用初始化列表。在C++ 11中，vector、list等各种容器以及string都可以使用初始化列表了。初始化列表对应的类为initializer_list，vector、list等各种容器以及string之所以可以使用初始化列表，是因为它们重载了参数类型为initializer_list的构造函数（称为初始化列表构造函数）和赋值函数（称为初始化列表赋值函数）。

 

vector , , , , }; map , }); Print({ , , , , });
getchar(); ; }

 

vector); }; , ); , }; // v4 has two elements with values 10 and 1

 

 

vector}; , "hi"}; // v8 has ten elements with value "hi"

虽然v5,v7,v8都有括号，但只有v5是初始化列表。我们不能用int初始化string，编译器会根据给予的值选择另一种方式初始化。

 

 

 

constexptr,begin(),end(),cbegin,cend,using 的typedef用法

参考

(1)C++ primer 第五版

(2)http://www.cnblogs.com/hujian/archive/2012/02/20/2358853.html C++11 语法甜点1

(3)http://www.cnblogs.com/pzhfei/archive/2013/03/02/CPP_new_feature.html C++11 新特性

(4)http://blog.chinaunix.net/uid-7471615-id-83794.html rvalue lvalue的解释