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

The invdivinv() function divides an interval value by another interval value.

mint invdivinv(i1, i2, num)

	intrvl_t *i1, *i2;

	double *num;

 



i1 

is a pointer to an interval variable that is the dividend.


i2 

is a pointer to an interval variable that is the divisor.


num  

is a pointer to the double value that is the quotient.

The invdivinv() function divides the interval value in i1 by i2, and stores the result in num. The result can be either positive or negative.

Both the input and output qualifiers must belong to the same interval class: either the year to month class or the day to fraction(5) class. If necessary, the invdivinv() function extends the interval value in i2 to match the qualifier for i1 before the division.

The demo directory contains this sample program in the file invdivinv.ec.

/*
* invdivinv.ec *

The following program divides one interval value by another and
displays the resulting numeric value.
*/

#include <stdio.h>

EXEC SQL include datetime;

main()
{
mint x;
char out_str[16];

EXEC SQL BEGIN DECLARE SECTION;
interval hour to minute hrtomin1, hrtomin2;
double res;
EXEC SQL END DECLARE SECTION;

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

printf("Interval #1 (hour to minute) = 75:27\n");
incvasc("75:27", &hrtomin1);
printf("Interval #2 (hour to minute) = 19:10\n");
incvasc("19:10", &hrtomin2);

printf("---------------------------------------------\n");
invdivinv(&hrtomin1, &hrtomin2, &res);
printf("Quotient (double) = %.1f\n", res);

printf("\nINVDIVINV Sample Program over.\n\n");
}

INVDIVINV Sample ESQL Program running.

Interval #1 (hour to minute) = 75.27
Interval #2 (hour to minute) = 19:10
------------------------------------
Quotient (double) = 3.9

INVDIVINV Sample Program over.