编译器如何不使用size参数处理此模板化函数

Question

In the code below, I have a function where a length parameter must be passed, like this:

int recv(char* buf, int len);

But a convenience is to write a helper function like this:

    template <size_t N>
    int recv(char(&array)[N]) {
        return recv(array, N);
    }

Which means you can pass an array and the compiler somehow 'knows' the size so you don't have to pass it.

But when I have used templates before, I needed to pass the type, eg

std::vector<int> myvec;

But how does that work? Internally what is the compiler doing to resolve this? What is the name of this feature?

Please explain the syntax.

#include <string.h>
#include <iostream>

class test_recv
{
public:
    test_recv() {
        strcpy(arr, "ABCDEFGHIJKLMNOPQRSTUVWXYZ");
    }
    int recv(char* buf, int len) {
        int tmp = pos;
        while (len-- && pos < 26)
            *buf++ = arr[pos++];

        return pos - tmp;
    }
    template <size_t N>
    int recv(char(&array)[N]) {
        return recv(array, N);
    }

private:
    int pos = 0;
    char arr[26+1];
};


int main() {
    test_recv test;
    char buffer[10];
    int bytes;
    while ((bytes = test.recv(buffer, 10)) != 0) {
        for (int i = 0; i < bytes; ++i)
            std::cout << buffer[i];
    }

    std::cout << '\n';
    test_recv test2;
    char buffer2[10];
    while ((bytes = test2.recv(buffer2)) != 0) {
        for (int i = 0; i < bytes; ++i)
            std::cout << buffer2[i];
    }
}

Answer 1

This feature is called "template argument deduction" . See the link for more details.

Short version: If, in a call to a templated function, the template arguments (types, or in your case, array size) can be deduced from the arguments the function is called with (in your case, the actual array), you don't have to specify them explicitly.