ifx_int8mul() 函数将两个 int8 类型值相乘。
语法
mint ifx_int8mul(n1, n2, product)
ifx_int8_t *n1;
ifx_int8_t *n2;
ifx_int8_t *product;
n1
指向包含第一个操作对象的 int8 结构的指针。
n2
指向包含第二个操作对象的 int8 结构的指针。
product
指向包含 n1 * n2 的乘积的 int8 结构的指针。
用法
product 可与 n1 或 n2 相同。
返回代码
0
运算成功。
-1284
运算导致溢出或下溢。
示例
demo 目录中的文件 int8mul.ec 包含下列样例程序。
/*
* ifx_int8mul.ec *
The following program multiplies two INT8 numbers and
GBase 8s ESQL/C 编程指南
南大通用数据技术股份有限公司
- 742 -
displays the result.
*/
#include
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);
}
GBase 8s ESQL/C 编程指南
南大通用数据技术股份有限公司
- 743 -
if (x = ifx_int8soasc(&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);
}
输出
IFX_INT8MUL Sample ESQL Program running.
480,999,777,666,345 * 80 = 38479982213307600
IFX_INT8MUL Sample Program over.