C macro : turn a number into a string

Question

I have a table that defines symbols appearance on a 5x7 dot display. Something like:

extern UINT8 symbols[][5] = {
    {0x0,0x0,0x0,0x0,0x0},
    {0x0,0x0,0x5F,0x0,0x0},
    {0x0,0x7,0x0,0x7,0x0},
    {0x14,0x7F,0x14,0x7F,0x14}, // etc.

The leading part of the table matches ASCII table, followed by a set of special symbols, e.g. an arrow, or a check-mark. To reference those I have a list of macros:

#define SYMBOL_LEFT_ARROW 120 // 120 is the entry in the table
#define SYMBOL_RIGHT_ARROW (SYMBOL_LEFT_ARROW+1)    
#define SYMBOL_UP_ARROW (SYMBOL_RIGHT_ARROW+1)

Now I need to say something like (won't compile):

const char * const message = "Next" + SYMBOL_RIGHT_ARROW;

Question: How do I turn SYMBOL_RIGHT_ARROW into "\x79", or whole string into "Next\x79" AT COMPILE TIME so I can have the string in R/O section?

Freescale HC08 C-compiler.

Answer 1

You can concatenate strings in C source:

printf("%s\n", "forty" "two"); /* prints "fortytwo" */
/* NOTE:             ^^^ no punctuation */

To do that with your symbols is a lot of work, but maybe you can live with that.

#define SYMBOL_LEFT_ARROW 120
#define SYMBOL_LEFT_ARROW_STR "\x79"
#define SYMBOL_RIGHT_ARROW (SYMBOL_LEFT_ARROW + 1)
#define SYMBOL_RIGHT_ARROW_STR "\x83"
const char * const message = "Next" SYMBOL_RIGHT_ARROW_STR;

UPDATE

If you can make the value of the symbol match its position in the symbol table (120 match "\x78"), try these macros

#include <stdio.h>

#define ADD_ZERO_X(y) 0x ## y
#define SYMBOL_NUM(x) ADD_ZERO_X(x)

#define STRINGIZE(z) #z
#define ADD_SLASH_X(y) STRINGIZE(\x ## y)
#define SYMBOL_STR(x) ADD_SLASH_X(x)

#define SYMBOL_LEFT_ARROW 78 /* must write in hexadecimal without any prefix */
#define SYMBOL_RIGHT_ARROW 79
#define SYMBOL_UP_ARROW 7a

int main(void) {
  printf("%d\n", SYMBOL_NUM(SYMBOL_LEFT_ARROW));
  printf("%s\n", SYMBOL_STR(SYMBOL_LEFT_ARROW));
  printf("%d\n", SYMBOL_NUM(SYMBOL_RIGHT_ARROW));
  printf("%s\n", SYMBOL_STR(SYMBOL_RIGHT_ARROW));
  printf("%d\n", SYMBOL_NUM(SYMBOL_UP_ARROW));
  printf("%s\n", SYMBOL_STR(SYMBOL_UP_ARROW));
  return 0;
}

---

Edit (SO doesn't like my browser)

After macro expansion SYMBOL_NUM(32) is transformed to a integer literal (0x78); and SYMBOL_STR(78) is transformed to a string literal ("\x78").

You can use the literals as if you had typed them in.

const char *test = "Next" SYMBOL_STR(78) " one";
/* same as
   const char *test = "Next\x78 one";
*/

Answer 2

I came up with this little program:

#include <stdio.h>

#define TEST_CHR '\x77'
#define VAL(x) #x
#define STRINGIFY(x) VAL(x)
int main()
{
   int x = TEST_CHR;
   char *yyy = "%d " STRINGIFY(TEST_CHR) "\n";

   printf(yyy,x);

   return 0;

}

the indirection in the macro is necessary so that your character gets expanded before the "#" turns it into a string. notice that the '\x77' value turns into a valid int when you use it that way...