Assigning unknown values in an array to variables

Question

Okay so I have an array of 9 integers. Some of values are given some and some are unknown. How would I assign a integer variable such as 'a' - 'z' to these unknown values? For example:

index [0] is unknown 
index [1] is 27 
index [2] is 6
index [3] is 9 
index [4] is unknown 
index [5] is 21 
index [6] is 24
index [7] is 3 
index [8] is unknown

I want

index [0] is a
index [1] is 27 
index [2] is 6
index [3] is 9 
index [4] is b
index [5] is 21 
index [6] is 24
index [7] is 3 
index [8] is c






  for (ii=0; ii<MAXLINE/2; ii++)
  {
    if (uniqueNumbers[ii] == UNKNOWN_INPUT)
    {

      printf("UNKOWN_INPUT at [%d]\n", ii);
    }
  }

Array is an indexed data structure of holding elements of same type. Why your array has a combination of int and char? — Mahesh
– Mahesh, Commented Feb 9, 2011 at 21:36
What would happen if you need to put negative values on the array though? — Argote
– Argote, Commented Feb 9, 2011 at 21:43

templatetypedef · Accepted Answer · 2011-02-09 21:40:50Z

3

This is a great question. There actually is no straightforward way to do this in C, since the type system, while weak, does enforce that you pick a type for each variable, meaning that without some extra structure you won't be able to have the array elements be either an int or a reference to a variable.

One way to do this is to have your array elements each be a tagged union, such as this one:

typedef struct {
    enum { Constant, Variable } type;
    union {
        int value; // If this is a constant
        int* ref;  // If this is a reference to a variable
    } value;
} Expression;

Now, your array elements can be either a Constant (in which the value field is set) or a Variable, in which case the ref field would be a pointer to the actual variable holding the value.

If this isn't quite what you want, you can easily make modifications. If you want to store symbolic references instead of hard references (e.g. To store that an entry is "the variable X" instead of "a pointer to some other value"), you could add another enumerated constant and a field holding the name of the variable.

answered Feb 9, 2011 at 21:40

templatetypedef

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1 Comment

Fred Foo Over a year ago

Unfortunately, anonymous unions aren't in C99. stackoverflow.com/questions/3228104/…

Argote · Accepted Answer · 2011-02-09 21:40:50Z

1

This is not possible with an array of type int. You should probably define a small struct with an int and a char and make an array of that.

struct values {
    int value;
    char variable;
};

Then you'd assign an arbitrary character (say ' ') to variable if you know the value of value or a letter if you don't.

When using this, if variable is set to ' ' (or whatever you defined) then go read the value of value, otherwise handle as an unknown.

edited Feb 9, 2011 at 21:40

answered Feb 9, 2011 at 21:34

Argote

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2 Comments

templatetypedef Over a year ago

I agree with the logic, but wouldn't a union be more appropriate here?

Argote Over a year ago

Well yes, I was trying to keep it relatively simple though.

Fred Foo · Accepted Answer · 2011-02-09 21:42:25Z

1

Expanding on @Argote's answer, the clean solution is to use an array of

struct Value {
    union {
        char var;
        int  i;
    } v;
    bool value_known;
};

The quick hack is to use the values -'a' through -'z' as variables. That only works if your values are always non-negative.

answered Feb 9, 2011 at 21:42

Fred Foo

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Comments

RedX · Accepted Answer · 2011-02-09 21:46:49Z

0

Since you do not seem to use negative values (you mark unknown with -1) you could just use the negative ASCII value of the chars to store this information.

index [0] is a (-97)
index [1] is 27 
index [2] is 6
index [3] is 9 
index [4] is b (-98)
index [5] is 21 
index [6] is 24
index [7] is 3 
index [8] is c (-99)

When printing use (char)(-1 * index[i]).

answered Feb 9, 2011 at 21:46

RedX

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Comments

user3458 · Accepted Answer · 2011-02-09 21:45:40Z

0

If what you want is to print a letter for any value that is -1, then you need to put support for this code into your output routine.

answered Feb 9, 2011 at 21:45

user3458

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