C++ 中简单可变参数 function 的不匹配模板？

Question

I am writing a Scheme to C++ compiler (although Scheme is not so related to this question), and I have a set of functions that are invoked in the Scheme code that can be used equivalently in the resulting C++ code. I generated the C++ function "f" (seen below), and I cannot figure out for the life of me why it doesn't work (the error is below). I have three templated parameters, and I provide the function with three arguments as well. If you have any insight into this problem, please let me know - it's so incredibly frustrating!

// from stdarg_macros.h
#include <stdarg.h>
#define INIT va_list args; va_start(args, nargs);
#define LOOP for (int i = 0; i < nargs; i++)
#define DEINIT va_end(args); return r;

// from io.cpp
#include <iostream>
template <typename T>
void display(T var) {
        std::cout << var << std::endl;
}

// from operators.c
int add(int nargs, ...) {INIT int r = 0; LOOP r += va_arg(args, int); DEINIT}
double add_d(int nargs, ...) { INIT double r = 0; LOOP r += va_arg(args, double); DEINIT}
int sub(int nargs, ...) {INIT int r; LOOP {if (i == 0) r = va_arg(args, int); else r -= va_arg(args, int);} DEINIT}
double sub_d(int nargs, ...) {INIT int r; LOOP {if (i == 0) r = va_arg(args, double); else r *= va_arg(args, double);} DEINIT}
int mul(int nargs, ...) {INIT int r = 1; LOOP r *= va_arg(args, int); DEINIT}
double mul_d(int nargs, ...) {INIT double r = 1; LOOP r *= va_arg(args, double); DEINIT}
double div_d(int nargs, ...) {INIT double r; LOOP {if (i == 0) r = va_arg(args, double); else r /= va_arg(args, double);} DEINIT}

// my transpiler's output file
template <typename T, typename S, typename M, typename x>
T f(S a, M b, x c) {
return mul_d(3, a, b, div_d(2, 3.0, c));
};

int main() {
        auto m = mul_d(3, 8.0, 2.0, div_d(2, 5.0, 3.0));
        display(m);
        display(f(1, 2, 3));
        return 0;
}

$ g++ -std=c++14 Output/math.cpp && ./a.out
    Output/math.cpp:9:9: error: no matching function for call to 'f'
    display(f(1, 2, 3));
        ^
    Output/math.cpp:3:3: note: candidate template ignored: couldn't infer template argument 'T'
    T f(g a, i b, L c) {
      ^
    1 error generated.

Answer 1

The template arguments could be deduced only from function arguments. In this case the 1st template parameter T of f can't be deduced , it's only used to specify function's return type.

When possible, the compiler will deduce the missing template arguments from the function arguments.

You have to specify the argument explicitly like

display(f<int>(1, 2, 3));
//       ^^^^^

Or you can remove the template parameter from f (and let the return type to be deduced):

template <typename S, typename M, typename x>
auto f(S a, M b, x c) {
    return mul_d(3, a, b, div_d(2, 3.0, c));
};

then you can just call it like

display(f(1, 2, 3));