Array inside a Struct in C++

Question

I've been trying to figure out how to add an array into a struct... a struct of ints for example would look like this:

struct test{
    int a;
    int b;
    int c;
} test = {0,1,2};

but if I want to have an array for example:

struct test{
    int a;
    int b;
    int c;
    int deck[52];
} test;

is this doable? the initialization of the deck (of cards) happens in a different function. when I do it like this, I don't get an error in the struct but I get it when I try to use it... for example if I do this test.deck[i] = 1; it gives me this error:

Error C2143 Syntax Error missing ';' before '.'

if I were to use a , I could write test.a = 1;

Could anyone write a simple struct where a variable in it is an array and then just use it for a simple command?

Answer 1

If this is C++, no C, drop the test after the struct definition.

The following code works perfectly.

#include <iostream>

using namespace std;

struct Test {
  int a;
  int b;
  int c;
  int deck[52];
};

int main (int argc, char* argv[])
{
    Test t;
    t.deck[1] = 1;
    cout << "t.deck[1]: "<< t.deck[1] << endl;
    exit(0);
}

The problem: In C, you put the test after the definition to create a variable named test. So in C, test is not a type, it is a global variable , the way you wrote that.

This compiles:

#include <iostream>

using namespace std;

struct Test {
  int a;
  int b;
  int c;
  int deck[52];
} test;

int main (int argc, char* argv[])
{
    test.deck[1] = 1;
    cout << "test.deck[1]: "<< test.deck[1] << endl;
    exit(0);
}

Answer 2

The error:

Error C2143 Syntax Error missing ';' before '.'

is due to test being a type name. You need to define an instance:

int main() {
   test mytest;
   mytest.deck[1] = 1;
   return 0;
}