oracle 是什么导致了这个 PL/SQL 函数中的数值溢出？

What Causes the Numeric Overflow in this PL/SQL Function?

提问by theory

I can run this command on Oracle 10.2 without problem:

我可以在 Oracle 10.2 上毫无问题地运行此命令：

SQL> select instr(unistr('foo'), chr(4050596145)) as hazit from dual;

     HAZIT
----------
     0

So I tried to encapsulate it into a function:

于是我试着把它封装成一个函数：

CREATE OR REPLACE FUNCTION hazit(string IN VARCHAR2) RETURN INTEGER
AS
BEGIN
    RETURN instr(unistr(string), chr(4050596145));
END;
/

Function created.

But I get a numeric overflow error when I try to use it:

但是当我尝试使用它时出现数字溢出错误：

SQL> select hazit('foo') FROM DUAL;
select hazit('foo') FROM DUAL
       *
ERROR at line 1:
ORA-01426: numeric overflow
ORA-06512: at "DWHEELER.HAZIT", line 4

What gives?

是什么赋予了？

采纳答案by A.B.Cade

I don't have an explanation but this seems to work:

我没有解释，但这似乎有效：

CREATE OR REPLACE FUNCTION hazit(string IN VARCHAR2) RETURN NUMBER IS
i number;
BEGIN
    select instr(unistr(string), chr(4050596145))
      into i from dual;
    return i;
END;
/

Hereis a fiddle

这是一个小提琴