如何将字符串(或任何类型)存储到T类型的数组中

Question

我试图将字符串(或任何类型(存储在T类型的数组中，但我遇到了两个错误：

类型为"std:：string&"(非const限定(的引用不能使用类型的值初始化

"bool container:：insertBack(T&("：无法转换从"const char[3]"到"T&"的参数1

我尝试将类型更改为int而不是string：我收到了类似的错误消息。

#include <iostream>
#include <string>
template<typename T>
class container
{
public:
container();
// Postcondition: data member n is initialized to -1 and all elements in the empty arr array are initialized to zero
bool isFull();
// Postcondition: returns true if the container object (i.e., the arr array) is full; returns false otherwise
bool insertBack(T& val);
//  Precondition: the container object is not full
// Postcondition: if arr array is not full, n is incremented by 1; returns true with val is inserted at the end of the arr array 
//                 Otherwise, returns false; the value is not inserted and program execution continues.
private:
static const int CAPACITY = 10;     // physical size of the arr array or the storage capacity of a container object
T arr[CAPACITY];            // arr array can store up to CAPACITY  (10 in our case) of any type 
int n;                      // n is used as the subscript for the arr array. n is initialized to -1 for an empty array
// Each time a new value is inserted into the arr array, n must first be incremented 
// by 1. Since n has been initialized to -1, the first inserted value is stored in arr[0],
// and the 2nd inserted value will be in arr[1], etc.  and the nth inserted value will be 
// stored in arr[n – 1]. Obviously, n + 1 represents the actual number of elements
// stored in the array after n rounds of insertion.         
};
template<typename T>
container<T>::container()
{
n = -1;
T arr[CAPACITY] = { 0 };
}
template<typename T>
bool container<T>::isFull()
{
return n == CAPACITY - 1;
}
template<typename T>
bool container<T>::insertBack(T& val)
{
if (!isFull())
{
n++;
arr[n - 1] = val;
return 1;
}
else
{
return 0;
}
}
int main()
{
container<std::string> s1;
s1.insertBack("aa");
}

Answer 1

g++对相同错误给出了略有不同的输出：

cannot bind non-const lvalue reference of type 'std::__cxx11::basic_string<char>&' to an rvalue of type 'std::__cxx11::basic_string<char>'

clang++:

non-const lvalue reference to type 'std::__cxx11::basic_string<char>' cannot bind to a value of unrelated type 'const char [3]'

解决方案是将参数作为const&。

接下来你会发现T arr[CAPACITY] = { 0 };
给出了一个运行时异常，如：basic_string::_M_construct null not valid。

你不是零初始化arr那样。事实上，您正在创建一个新的arr，并试图用nullptr构建它，这对std::string[]不起作用。

您还可以像标准容器那样使用无符号整数size_t来计算元素，以使未来与标准函数/算法的交互更容易。

固定：：

#include <iostream>
#include <string>
template<typename T, size_t Capacity = 10>       // made the capacity selectable
class container {
public:
container() : arr{}, n(0) {}                 // using member initializer list
bool isFull() const { return n == Capacity; } // correct test with a 0 based counter
bool insertBack(const T& val) {               // const
if (isFull()) return false;
// return true or false, not 1 or 0 
arr[n++] = val;
return true;
}
private:
T arr[Capacity];
size_t n;
};
int main() {
container<std::string> s1;
s1.insertBack("aa");
}