error: a function-definition is not allowed here before '{' token

[imstuux]

Write your question here.

#include <iostream>
#include <windows.h>
using namespace std;

int main()
{
    cout << "Test with variants.\nChoose a,b,c,or d as the answer by typing the coresponding letter.\n";
    Sleep(3000);
    cout <<"Get ready for the first question.You have 5 seconds to answer\n";
    Sleep(5000);
    cout<<"Question one:\n What's 48+52?\n";
        int timer()
        {
            
        
            int a=5;
                while(a>=0)
                    cout<<a<<","<<--a;
        }
    cout<<"a=96\n";
    cout<<"b=100\n";
    cout<<"c=104\n";
    cout<<"d=98\n";
    int ans1,a,b,c,d;
    cin>>ans1;
    Sleep(1000);
        if(ans1=a)
        {
            cout<<"Correct,the answer is 96.\n";
        }
        else
        {
            cout<<"Wrong.The right answer was 96\n";
        }

I would like to make a test where you insert one of the given variants,in 5 seconds.I fail when I try to add the timer.The error is <<a function-definition is not allowed here before '{' token>>

EDIT:Forgot to say that if the time runs out ,the message <<Question skipped>> to appear and to automatically skip to second question.

[AbstractionAnon]

Line 12: What is this line? Is it an int declaration? A function call? Or a function prototype?
The compiler thinks you're trying to define a nested function, which is not allowed.

[keskiverto]

Lines 12-19 seem to contain a definition of a function, which takes no arguments, has name "timer" and returns an int value.

The function definition is inside the body of an another function. That is not legal. (C++11 did add an exception though: lambda functions.)

[imstuux]

I wrote that function outside of main function,how do I use it in main after line 11?

[AbstractionAnon]

Simply make a function call to it.

  timer();

BTW, your function appears to be an int function, but does not return anything.

[imstuux]

What exactly should it return?I'm a starter here,it's my second year of study in c++....

[AbstractionAnon]

There doesn't appear to be any value that it is trying to return to the caller.
If that's the case, then make the type of the function void.