How do I print either string or integer in C? - c

Question

I have the sturct token_t down below. It has two things in it.
type (enum)
value (union)
The reason why value is union, cz a token can be either a number or string.
The main problem is that I cannot design a proper function that can print this token out for me. Check out my void print_token(token_t* token) function. I know it's terrible. Just becase the printf() prints only either %d (decimal) or %s (string) I cannot print my token out. I want my function to print the token no matter what the value is.
// ===========================================================================
typedef enum token_type_t
{
ID, // Identifier
STR, // String
WHS, // Whitespace
LPR, // Left Parenthesis
RPR, // Right Parenthesis
LCB, // Left Curly Bracket
RCB, // Right Curly Bracket
LSB, // Left Square Bracket
RSB, // Left Square Bracket
EOF, // End Of File
EQL, // Equal Sign
SEM, // Semicolon
} token_type_t;
typedef union token_value_t
{
char* str_value;
int int_value;
} token_value_t;
typedef struct token_t
{
token_type_t type;
token_value_t value;
} token_t;
// ===========================================================================
// Here's the problem. I want this function to print the token no matter what type
// the value is, I want it to be dynamic. Is there any way to do that?
void print_token(token_t* token)
{
printf("Token { type: %d, value: %s }\n", token->type, token->value);
}

Answer 1

How do I print either string or integer in C?
Use a different printf() depending on type.
void print_token(const token_t* token) {
printf("Token { type: %d, value: ", token->type);
switch (token->type) {
case ID: printf("%d", token->value.int_value); break;
case STR: printf("\"%s\"", token->value.str_value); break;
// TBD code for other cases
}
printf(" }\n");
}

Answer 2

One solution is to print only a string, but convert the integer to a string conditionally:
For example:
void print_token( const token_t* token )
{
char buf[21] ;
printf( "Token { type: %d, value: %s }\n",
token->type,
token->type == ID ? itoa( token->value.int_value, buf, 10 ) :
token->value.str_value );
}
Noting that itoa() is not a standard function - you might consider that prohibitive if you have to implement your own just for this.
You have not specified which token types are integers and which are strings so if perhaps ID is not the only integer, the condition may be more complex such as:
void print_token( const token_t* token )
{
char buf[21] ;
printf( "Token { type: %d, value: %s }\n",
token->type,
token->type == ID ||
token->type == EOF ? itoa( token->value.int_value, buf, 10 ) :
token->value.str_value );
}
If you have more more integer types than string types you would do better to swap the test.
If the are just one or two of one token type and the rest are of the other, and you have a suitable integer-to-string function, the above solution is succinct, but might get cumbersome if that are more than a few of both types. In that case consider the solution below.
A switch/case using of case "fall-through" cases and a default may be easier to maintain an does not need an integer-to-string fucntion:
void print_token( const token_t* token )
{
printf( "Token { type: %d, value: ", token->type ) ;
switch (token->type)
{
// Stack all the integer token cases here
case ID :
case EOF :
printf("%d", token->value.int_value) ;
break ;
// Everything is a string
default:
printf( "\"%s\"", token->value.str_value ) ;
break;
}
printf(" }\n");
}
Again you might swap the explicit cases and default if there are more string tokens than integer.