如何在标准C ++ 17中将临时数组传递给函数 [英] How to pass a temporary array to a function in standard C++17

问题描述

我具有以下签名的功能：

I have a function of the following signature:

void foo(int argc, char const * const argv[]);

我想用一种简洁的方式来称呼它，类似但不必如下： / p>

I would like to call it in a concise way, similar, but not necessary identical as below:

foo(3, {"one", "two", "three"});

我知道C仅为此目的支持复合文字（参考）。

I know that C supports compound literals just for this purpose (reference).

我也知道如何使用模板（参考）来解决问题。

I also know how to solve the problem using templates (reference).

但是，我的问题略有不同。函数签名是固定的-通常接受传递给 main 的参数，并且数组的大小不是预定义的。我正在为此功能编写测试，就我而言，所传递的数组在编译时是已知的。

However, my problem is slightly different. The function signature is fixed - it normally takes arguments passed to main and the size of the array is not predefined. I am writing tests for this function and in my case the passed arrays are known at compile-time.

是否可以调用 foo 而不使用下面的临时变量或包装函数？

Is there a way to call foo without using temporary variables or wrapper functions like below?

char const * array[3] = {"one", "two", "three"};
foo(3, array);

template<int N>
void wrapper(char const * const (&array)[N])
{
    foo(N, array);
}

推荐答案

尽管我不知道如何

template<size_t N>
void foo(int argc, const char *(&&argv)[N]){
    int i=0;
    for( auto o: argv) {
        ++i;
        cout<<"array"<<i<<" = " << o<<"\n";
    }

    cout<< "\narray size "<<i<<"\n"<<argc<<"\n";
    return;
}

int main(){

    foo(3, {"one", "two", "three"});

}

array1 = one
array2 = two
array3 = three

array size 3
3

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