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

The ifx_int8mul() function multiplies two int8 type values.

mint ifx_int8mul(n1,  n2,  product)

	ifx_int8_t *n1;

	ifx_int8_t *n2;

	ifx_int8_t *product;

 



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.


product  

is a pointer to the int8 structure that contains the product of n1  *  n2.

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

The file int8mul.ec in the demo directory contains the following sample program.

/*
* ifx_int8mul.ec *

The following program multiplies two INT8 numbers and
displays the result.
*/

#include <stdio.h>

EXEC SQL include "int8.h";

char string1[] = "480,999,777,666,345";
char string2[] = "80";
char result[41];

main()
{
mint x;
ifx_int8_t num1, num2, prd;

printf("IFX_INT8MUL 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);
}
if (x = ifx_int8mul(&num1, &num2, &prd))
{
printf("Error %d in multiplying num1 by num2\n", x);
exit(1);
}
if (x = ifx_int8toasc(&prd, result, sizeof(result)))
{
printf("Error %d in converting product to string\n", x);
exit(1);
}
result[40] = '\0';
printf("\t%s * %s = %s\n", string1, string2, result);

printf("\nIFX_INT8MUL Sample Program over.\n\n");
exit(0);
}

IFX_INT8MUL Sample ESQL Program running.

480,999,777,666,345 * 80 = 38479982213307600

IFX_INT8MUL Sample Program over.