如何在不知道 C++ 中有多少可选 arguments 的情况下在循环中使用 va_arg？

Question

I want to write a function which takes at least two integer and returns the sum of all integer passed to the function:

int sumOfAtLeastTwoIntegers(int a, int b, ...){
   
    int sum = a+b;
    va_list ptr;
    va_start(ptr,b);
    for(){
        sum += va_arg(ptr, int)
    }

    va_end(ptr);
    return sum;
}

I want to know how the expression in the for loop has to look like such that the loop continues until all optional arguments were added to the sum. How would I achieve this without knowing how many optional arguments were passed to the function? The function call would look like this:

sumOfAtLeastTwoIntegers(2,3,4,5,1,0,200);

Answer 1

There are generally two ways to handle variable arguments, pre-knowledge and post-knowledge.

Pre-knowledge is like with printf("%d %c\n", anInt, aChar) , there's an argument up front which you can use to figure out how many remain. An example of that would be:

int sumOfInts(size_t count, int a, ...); // Needs "count" integers.
int eleven = sumOfInts(2, 4, 7);

Post-knowledge requires you to have a sentinel value that tells you when to stop, such as with:

int sumOfNonZeroInts(int a, ...); // Needs non-zero integers, stops at 0.
int eleven = sumOfInts(4, 7, 0);

---

One other thing you may want to consider is steering clear of variable argument lists, there are much more expressive ways in C++ for doing what you want with, for example, vectors. The following provides one way of doing this:

#include <iostream>
#include <vector>

template<typename T> T sumOf(const std::vector<T> &vec) {
    T acc = T();
    for (const T &item: vec)
        acc += item;
    return acc;
}

int main() {
    auto eleven = sumOf<int>({4, 7}); 
    std::cout << "Four plus seven is equal to " << eleven << '\n';
}

This isn't necessarily as fast as variable arguments, but my default position nowadays is to generally optimise for readability first:-)

Answer 2

I suggest to use a variadic template like the following. It works only if u give it two integers at least. And all of the args must be integers too.

#include<iostream>
#include <type_traits>

template< typename ... Args>
std::enable_if_t<std::is_same_v<std::common_type_t<Args...>, int>, int>
sum(int arg1, Args...args)
{
    return (args+...+arg1);
}
int main(){

    std::cout << sum(1,2);//working
    std::cout << sum(1,.2);//compile error
    std::cout << sum(1);//compile error
 
}

Answer 3

There are three ways to handle variable-argument functions:

1. Either have some kind of terminator , an argument whose sole meaning is to terminate the argument list.

2. Make the first argument be a count of the number of arguments passed.

3. Some kind of argument-matching format, like eg printf or scanf .

For your kind of function, the first two variants could both be good solutions (depending on the range of values allowed in the arguments).

Answer 4

Use the first argument as an "input size" just how you'd do with regular arrays especially in C.

int sumOfAtLeastTwoIntegers(int howMany, ...) {
   
    int sum = 0
    va_list ptr;
    va_start(ptr,b);
    while(howMany--){
        sum += va_arg(ptr, int)
    }

    va_end(ptr);
    return sum;
}