The ifx_int8cvflt() function

The file int8cvflt.ec in the demo directory contains the following sample program.

/*
   * ifx_int8cvflt.ec *

   The following program converts two floats to INT8 types and displays
   the results.
*/

#include <stdio.h>

EXEC SQL include "int8.h";

char result[41];

main()
{
    mint x;
    ifx_int8_t num;

    printf("IFX_INT8CVFLT Sample ESQL Program running.\n\n");

    printf("Float 1 = 12944.321\n");

/* Note that in the following conversion, the digits to the
   right of the decimal are ignored. */

    if (x = ifx_int8cvflt(12944.321, &num))
        {
        printf("Error %d in converting float1 to INT8\n", x);
        exit(1);
        }

/* Convert int8 to ascii to display value. */

    if (x = ifx_int8toasc(&num, result, sizeof(result)))
        {
        printf("Error %d in converting INT8 to string\n", x);
        exit(1);
        }
    result[40] = '\0';
    printf("  The INT8 type value is = %s\n", result);
    printf("Float 2 = -33.43\n");

/* Note that in the following conversion, the digits to the
   right of the decimal are ignored. */

    if (x = ifx_int8cvflt(-33.43, &num))
        {
        printf("Error %d in converting float2 to INT8\n", x);
        exit(1);
        }
    if (x = ifx_int8toasc(&num, result, sizeof(result)))
        {
        printf("Error %d in converting INT8 to string\n", x);
        exit(1);
        }
    result[40] = '\0';
    printf("  The second INT8 type value is = %s\n", result);

    printf("\nIFX_INT8CVFLT Sample Program over.\n\n");
    exit(0);
}