How can I write very big number?

I have an Oracle table with the column "col_name number(20)".
I use the following code for binding:
OCIBindByName(stmt, &number,
errhp, (text *) ":col_name", -1, (dvoid *) &number,
(sword) sizeof(number), SQLT_INT, (dvoid *) 0, (ub2 *) 0,
(ub2 *) 0, (ub4) 0, (ub4 *) 0, OCI_DEFAULT)But the data, written in the table, are incomplete.
I tried SQLT_UIN but results are the same.
Variable number is declared as unsigned long

If your value is truncated, it means it cannot be held into an 32bits integer value.
Your column is declared as NUMBER(20) so it can hold values that exceed the capacity of an 32bits integer value and thus you need to use 64bits integers.
OCI does not support yet (one day maybe, lol) binding 64bits C integers. You need to use OCINumber type for that.
Here is a extract of an answer i made in a previous post about the same subject :
Vincent Rogier wrote:
SQLT_INT or SQLT_UIN are limited to 32bits integers.
To use 64bits integers, the C type is long long and unsigned long long.
Using binary_double is not working because with really big values, there is loss of data.
The only way i found in order to manipualte 64bits integer properly is using OCINumber.
By example in order to define an output placeholder to get data from an number(20) column, you have to :
* define using SQLT_VNU with size = sizeof(OCINumber)
* fetch
* then the buffer is an OCINumber
* use OCINumberToInt() with the sizeof(long long) for rsl_lenght parameter
* then you've got a correct value
You can use as well OCINumberSign() before calling OCINumberToInt() to find out if the value is signed or not, then you pass an unsigned long long or signed long long to OCINumberToInt()

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