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

The decadd() function adds two decimal type values.

mint decadd(n1,  n2,  sum)

	dec_t  *n1;

	dec_t  *n2;

	dec_t  *sum;

 



n1

is a pointer to the decimal structure of the first operand. 


n2

is a pointer to the decimal structure of the second operand. 


sum

is a pointer to the decimal structure that contains the sum (n1  +  n2). 

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

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

/*
* decadd.ec *

The following program obtains the sum of two DECIMAL numbers.
*/

#include <stdio.h>

EXEC SQL include decimal;

char string1[] = " 1000.6789"; /* leading spaces will be ignored */
char string2[] = "80";
char result[41];

main()
{
mint x;
dec_t num1, num2, sum;

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

if (x = deccvasc(string1, strlen(string1), &num1))
{
printf("Error %d in converting string1 to DECIMAL\n", x);
exit(1);
}
if (x = deccvasc(string2, strlen(string2), &num2))
{
printf("Error %d in converting string2 to DECIMAL\n", x);
exit(1);
}
if (x = decadd(&num1, &num2, &sum))
{
printf("Error %d in adding DECIMALs\n", x);
exit(1);
}
if (x = dectoasc(&sum, result, sizeof(result), -1))
{
printf("Error %d in converting DECIMAL result to string\n", x);
exit(1);
}
result[40] = '\0';
printf("\t%s + %s = %s\n", string1, string2, result); /* display result */

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

DECADD Sample ESQL Program running.

1000.6789 + 80 = 1080.6789

DECADD Sample Program over.