How to access any variable name according to loop index

Question

I have some integer variables, I named them n0 to n9. I want to access them using a loop. I tried this code to do that:

int n0 = 0, n1          


        

Answer 1

Simple answer: declare an array instead, as int n[10].

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Advanced answer: it doesn't seem to be the case here, but in the case where you do need to use individual variable names of array items, for whatever reason, you can use an union:

typedef union
{
  struct
  {
    int n0;
    int n1;
    int n2;
    ... // and so on
    int n9;
  };

  int array[10];

} my_array_t;

In case you have an old dinosaur compiler, then declare the struct with a variable name such as struct { ... } s;

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How to use the above type in a practical, real world program:

  my_array_t arr = {0};

  for(int i=0; i<10; i++)
  {
    arr.array[i] = i + 1;
  }

  // access array items by name:    
  printf("n0 %d\n", arr.n0); // prints n0 1
  printf("n1 %d\n", arr.n1); // prints n1 2

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Or you could initialize members by name:

  my_array_t arr = 
  { 
    .n0 = 1, 
    .n1 = 2,
    ...
  };

---

Silly, artificial example of how to use the above type to assign values to the variables without using array notation:

  my_array_t arr = {0};

  // BAD CODE, do not do things like this in the real world:

  // we can't use int* because that would violate the aliasing rule, therefore:
  char* dodge_strict_aliasing = (void*)&arr;

  // ensure no struct padding:
  static_assert(sizeof(my_array_t) == sizeof(int[10]), "bleh");

  for(int i=0; i<10; i++)
  {
    *((int*)dodge_strict_aliasing) = i + 1;
    dodge_strict_aliasing += sizeof(int);
  }

  printf("n0 %d\n", arr.n0); // prints n0 1
  printf("n1 %d\n", arr.n1); // prints n1 2

  for(int i=0; i<10; i++)
  {
    printf("%d ",arr.array[i]); // prints 1 2 3 4 5 6 7 8 9 10
  }