The ifx_int8sub() function
The ifx_int8sub() function subtracts two int8 type values.
Syntax
mint ifx_int8sub(n1, n2, difference)
ifx_int8_t *n1;
ifx_int8_t *n2;
ifx_int8_t *difference;
- n1
- A pointer to the int8 structure that contains the first operand.
- n2
- A pointer to the int8 structure that contains the second operand.
- difference
- A pointer to the int8 structure that contains the difference of n1 and n2 (n1 - n2).
Usage
The difference can be the same as either n1 or n2.
Return codes
- 0
- The subtraction was successful.
- -1284
- The subtraction resulted in overflow or underflow.
Example
The file int8sub.ec in
the demo directory contains the following sample
program.
/*
*int8sub.ec *
The following program obtains the difference of two INT8
type values.
*/
#include <stdio.h>
EXEC SQL include "int8.h";
char string1[] = "6";
char string2[] = "9,223,372,036,854,775";
char string3[] = "999,999,999,999,999.5";
char result[41];
main()
{
mint x;
ifx_int8_t num1, num2, num3, sum;
printf("IFX_INT8SUB Sample ESQL Program running.\n\n");
if (x = ifx_int8cvasc(string1, strlen(string1), &num1))
{
printf("Error %d in converting string1 to INT8\n", x);
exit(1);
}
if (x = ifx_int8cvasc(string2, strlen(string2), &num2))
{
printf("Error %d in converting string2 to INT8\n", x);
exit(1);
}
/* subtract num2 from num1 */
if (x = ifx_int8sub(&num1, &num2, &sum))
{
printf("Error %d in subtracting INT8s\n", x);
exit(1);
}
if (x = ifx_int8toasc(&sum, result, sizeof(result)))
{
printf("Error %d in converting INT8 result to string\n", x);
exit(1);
}
result[40] = '\0';
printf("\t%s - %s = %s\n", string1, string2, result); /* display result */
if (x = ifx_int8cvasc(string3, strlen(string3), &num3))
{
printf("Error %d in converting string3 to INT8\n", x);
exit(1);
}
/* notice that digits right of the decimal are truncated. */
if (x = ifx_int8sub(&num2, &num3, &sum))
{
printf("Error %d in subtracting INT8s\n", x);
exit (1);
}
if (x = ifx_int8toasc(&sum, result, sizeof(result)))
{
printf("Error %d in converting INT8 result to string\n", x);
exit(1);
}
result[40] = '\0';
printf("\t%s - %s = %s\n", string2, string3, result); /* display result */
printf("\nIFX_INT8SUB Sample Program over.\n\n");
exit(0);
}
Output
IFX_INT8SUB Sample ESQL Program running.
6 - 9,223,372,036,854,775 = -9223372036854769
9,223,372,036,854,775 - 999,999,999,999,999.5 = 8223372036854776
IFX_INT8SUB Sample Program over.