INFORMIX-ESQL/C Programmer's Manual ESQL/C Function Library

The ifx_int8sub() function subtracts two int8 type values.

mint ifx_int8sub(n1,  n2,  difference)

	ifx_int8_t *n1;

	ifx_int8_t *n2;

	ifx_int8_t *difference;

 



n1

is a pointer to the int8 structure that contains the first operand.


n2

is a pointer to the int8 structure that contains the second operand.


difference  

is a pointer to the int8 structure that contains the difference of n1 and n2 (n1 - n2).

The difference can be the same as either n1 or n2.

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);

}

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.