问题描述
我不明白为什么数组会衰减为模板函数中的指针.
I don't understand why the array decays to a pointer in a template function.
如果您查看以下代码:当参数被强制为引用(函数 f1)时，它不会衰减.在另一个函数 f 中，它衰减.为什么函数 f 中的 T 类型不是 const char (buff&)[3] 而是 const char*(如果我理解正确的话)?
If you look at the following code: When the parameter is forced to be a reference (function f1) it does not decay. In the other function f it decays. Why is the type of T in function f not const char (buff&)[3] but rather const char* (if I understand it correctly)?
#include <iostream>
template <class T>
void f(T buff) {
std::cout << "f:buff size:" << sizeof(buff) << std::endl; //prints 4
}
template <class T>
void f1(T& buff) {
std::cout << "f:buff size:" << sizeof(buff) << std::endl; //prints 3
}
int main(int argc, char *argv[]) {
const char buff[3] = {0,0,0};
std::cout << "buff size:" << sizeof(buff) << std::endl; //prints 3
f(buff);
f1(buff);
return 0;
}
推荐答案
因为数组不能作为函数参数按值传递.
当您按值传递它们时，它们会衰减为指针.
Because arrays can not be passed by value as a function parameter.
When you pass them by value they decay into a pointer.
在这个函数中:
template <class T>
void f(T buff) {
T 不能是 char (&buff)[3] 因为这是一个参考.编译器会尝试 char (buff)[3] 按值传递，但这是不允许的.因此，为了使其工作，数组衰减为指针.
T can not be char (&buff)[3] as this is a reference. The compiler would have tried char (buff)[3] to pass by value but that is not allowed. So to make it work arrays decay to pointers.
您的第二个函数有效，因为这里的数组是通过引用传递的:
Your second function works because here the array is passed by reference:
template <class T>
void f1(T& buff) {
// Here T& => char (&buff)[3]
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