将数组传递给函数时，C++ 错误变量声明为 void

Question

I'm trying to refresh myself in C++ before my college starts again and I ran into some problems. I'm using a bubblesort function given by my professor and I'm struggling to run it in my int main. The function parameters is bubblesort(int *a, int length) , so I used bubblesort(a, sizeof(a)/sizeof(*a) ) . The compiler shows an error ' a ' is declared void. I tried searching up for an answer if I made a mistake but I couldn't catch my error. If you understand why I am getting this error can you please explain in detail what I'm missing or doing wrong.

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
using namespace std;

void bubblesort(int *a, int length)
{  
    int i, temp, finished = 0; 
   while(!finished)
   {  
       finished = 1;
      for( i = 0; i< length-1; i++)
      {  
        if(a[i] > a[i+1]) 
         {  
            temp = a[i]; 
            a[i] = a[i+1]; 
            a[i+1] = temp;
            finished = 0; 
         } 
      }
   }
}

int main()
{
   int a[] = {5,1,7,9,4,3};
   for (int i = 0;i < sizeof(a)/sizeof(*a);i++){ cout << a[i]; }
   cout << endl;
   void bubblesort(a, sizeof(a)/sizeof(*a));
   for (int i = 0;i < sizeof(a)/sizeof(*a);i++){ cout << a[i]; }
}

Answer 1

You don't need return type in function call.

Remove void from the line

void bubblesort(a, sizeof(a)/sizeof(*a));

and make it

bubblesort(a, sizeof(a)/sizeof(*a));

Answer 2

You can't specify the return type of a function when making a function call, so you need:

bubblesort(a, sizeof(a)/sizeof(*a));  // no void at the beginning

However, you have a fixed-size array, so the call could also simply be:

bubblesort(a, std::size(a));

---

Also, this loop:

for (int i = 0;i < sizeof(a)/sizeof(*a);i++){ cout << a[i]; }

can be rewritten like this:

for (int elem : a)
  cout << elem;