c++++ 模板编程中,类型推断失败时,可通过以下方法解决:显式指定模板参数。如:func
C++ 模板编程中的疑难解答:类型推断失败
问题:
使用 C++ 模板时,在类型推断过程中可能会遇到失败,导致编译错误。例如:
template<typename T>
void func(T t) {
// ...
}
int main() {
func<int>(); // 类型推断失败
}
解决方法:
为了解决类型推断失败,可以使用显式模板参数化,手动指定类型参数:
template<typename T>
void func(T t) {
// ...
}
int main() {
func<int>(10); // 显式指定类型参数
}
实战案例:
Consider the following program that uses an Array template to create an array of any type:
template <typename T>
struct Array {
T* data;
size_t size;
Array(size_t size) : data(new T[size]), size(size) {}
~Array() { delete[] data; }
T& operator[](size_t index) { return data[index]; }
};
int main() {
Array<int> arr(10);
for (size_t i = 0; i < arr.size; ++i) {
arr[i] = i * i;
}
for (size_t i = 0; i < arr.size; ++i) {
std::cout << arr[i] << " ";
}
std::cout << std::endl;
return 0;
}
This program demonstrates the type-safe behavior of C++ templates. The Array template is instantiated with the int type, creating an array of integers. The elements of the arrays can be accessed and modified using the operator[] method. The program prints the contents of the array, which are the squares of the integers from 0 to 9.
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