c++ - 无法解析符号 vector

Question

#include <iostream>
#include <vector>

using namespace std;


class block{
public:
    long nx,ny;
    vector<long> s;
    block(long &x, long &y):nx(x),ny(y),vector<long> s((x+1)*(y+1),0) {}
};

int main() {
    block B(2,2);
    for(int i=1;i<=9;i++) {
        cout<<B.s(i);
    }

    cout << "!!!Hello World!!!" << endl; // prints !!!Hello World!!!
    return 0;
}

已编译，错误消息显示无法解析符号“vector”。错误是什么？我想定义一个包含要初始化的可变维度 vector 的类。

---

#include <iostream>
#include <vector>

using namespace std;


class block{
public:
    long nx,ny;
    vector<long> s;
    block(long &x, long &y):nx(x),ny(y),s((x+1)*(y+1),0) {}
};

int main() {
    block B(2,2);
    for(int i=0;i<=9;i++) {
        cout<<B.s[i];
    }

    cout << "!!!Hello World!!!" << endl; // prints !!!Hello World!!!
    return 0;
}

在 block B(2,2) 仍然有问题；错误信息:构造函数“block::block”的实例与参数列表不匹配

为什么？谢谢!

Answer 1

首先:

  vector<long> s((x+1)*(y+1),0)

应该是:

  s((x+1)*(y+1),0)

不应为 重复类型。同时构造函数应该是:

  block(const long &x, const long &y):
    nx(x), ny(y), s((x + 1) * (y + 1), 0)
  {
  }

如果你真的需要引用。因为否则，当你这样做的时候

 block B(2,2);

在 main 中，它会给你错误，因为构造函数采用 long&，你传递的是 int 常量。深层原因与左值和右值有关:整型常量是右值，而 long& 是对非常量 long 的引用，是左值引用。根据这个博客:lvalues, rvalues and references

An lvalue reference (to non-const type) is a reference that can be initialized with an lvalue. Well, only with those lvalues that do not render const or volatile types. An rvalue reference (to non-const type) is a reference that can be initialized with an rvalue (again, only with those rvalues that do not designate const or volatile types). An lvalue reference to const type is a reference that can be initialized with rvalues and lvalues alike (rendering constant and non-constant types).

此外，根据 C++11 标准:第 4.1 节标准转换:

Standard conversions are implicit conversions defined for built-in types. A standard conversion sequence is a sequence of standard conversions in the following order:

— Zero or one conversion from the following set: lvalue-to-rvalue conversion, array-to-pointer conversion, and function-to-pointer conversion.

— Zero or one conversion from the following set: integral promotions, floating point promotion, integral conversions, floating point conversions, floating-integral conversions, pointer conversions, pointer to member conversions, and boolean conversions.

— Zero or one qualification conversion.

没有右值到左值的转换。这就是您看到编译错误的原因。在 long& 之前添加 const 使其能够使用右值进行初始化，这就是更改后错误消失的原因。

其次，

  cout<<B.s(i);

应该是:

cout<<B.s[i];

您应该使用 [] 来访问 vector 元素。

第三， vector 索引从0开始，所以

for(int i=1;i<=9;i++)

应该是

for(int i=0;i<9;i++)

否则索引越界。请在此处查看工作示例:http://ideone.com/YLT3mG