[英]How do I store strings (or any type) into an array of type T?
I'm trying to store a string (or any type) inside an array of type T, but I get two errors:我试图在 T 类型的数组中存储一个字符串(或任何类型),但出现两个错误:
a reference of type "std::string &" (not const-qualified) cannot be initialized with a value of type "const char [3]" “std::string &”类型的引用(非const限定)不能用“const char [3]”类型的值初始化
'bool container::insertBack(T &)': cannot convert argument 1 from 'const char [3]' to 'T &' “bool container::insertBack(T &)”:无法将参数 1 从“const char [3]”转换为“T &”
I've tried changing the type to int instead of string: I received similar error messages.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");
}
g++
gives a slightly different output for the same error:对于相同的错误, g++
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++
: clang++
:
non-const lvalue reference to type 'std::__cxx11::basic_string<char>' cannot bind to a value of unrelated type 'const char [3]'
The solution is to take the argument as a const&
.解决方案是将参数作为const&
。
What you'll find next is that T arr[CAPACITY] = { 0 };
接下来你会发现T arr[CAPACITY] = { 0 };
gives a runtime exception like: basic_string::_M_construct null not valid
.给出一个运行时异常,如: basic_string::_M_construct null not valid
。
You are not zero initializing arr
like that.你不是零初始化arr
那样。 In fact, you are creating a new arr
and trying to construct it with nullptr
, which will not work for std::string[]
.事实上,您正在创建一个新的arr
并尝试使用nullptr
构造它,这不适用于std::string[]
。
You might as well use an unsigned integer as size_t
for counting elements as the standard containers do to make future interactions with standard functions/algorithms easier.您也可以像标准容器一样使用无符号整数作为size_t
来计算元素,以便将来与标准函数/算法的交互更容易。
With that fixed::有了这个固定::
#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");
}
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