long for 12 digit number?

Question

I'm trying to use long for 12 digit number but it's saying "integer constant is too large for "long" type", and I tried it with C++ and Processing (similar to Java). What's happening and what should I use for it?

Can you provide a code sample so that we can see specific details? — Josh Peterson, Jun 21 '11 at 00:03
@Blindy, Sorry, I thought it might be useful. Obviously it is unnecessary though. — Josh Peterson, Jun 21 '11 at 00:14
will you be using this number for calculations? It's strange that you asking for a "12 digit number". Normally people are interested in a range and they specify if it is signed or not. If you are just dealing with a number like a credit card number or some phone number then it would be better to store as a string. — glowworms, Jun 21 '11 at 00:30
I didn't think it was strange. But then, I've had to write code to interoperate with COBOL. — dan04, Jun 21 '11 at 01:37
well the number is 600851475143 and I have to find out the largest prime factor. — Hikari Iwasaki, Jun 24 '11 at 01:27

score 5 · Answer 1 · answered Jun 21 '11 at 00:06

5

In C and C++ (unlike in Java), the size of long is implementation-defined. Sometimes it's 64 bits, sometimes it's 32. In the latter case, you only have enough room for 9 decimal digits.

To guarantee 64 bits, you can use either the long long type, or a fixed-width type like int64_t.

answered Jun 21 '11 at 00:06

dan04

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+1 for specific-width types, like those achievable through `cstdint` if your platform has it, or [`boost/cstdint.hpp`](http://www.boost.org/doc/libs/1_36_0/libs/integer/cstdint.htm) if it doesn't. – Wyatt Anderson Jun 21 '11 at 00:13
Just FYI - on some compilers (e.g. older GCC), `cstdint` is missing but the platform may have C99's `stdint.h`.... – Tony Delroy Jun 21 '11 at 01:15

score 3 · Answer 2 · answered Jun 21 '11 at 00:37

3

If you are specifying a literal constant, you must use the appropriate type specifier:

int i = 5;
unsigned i = 6U;

long int i = 12L;
unsigned long int i = 13UL;

long long int i = 143LL;
unsigned long long int i = 144ULL;

long double q = 0.33L;

wchar_t a = L'a';

answered Jun 21 '11 at 00:37

Kerrek SB

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I really think that in his case was the problem of a long int being 32bits – Vinicius Kamakura Jun 21 '11 at 01:26
@hexa: You're probably right. OP didn't say which platform/compiler/settings, I'm not sure which combinations of types and literal constants would cause which sort of warnings. – Kerrek SB Jun 21 '11 at 01:30

score 1 · Answer 3 · answered Jun 21 '11 at 00:04

1

Try using a long long in gcc or __int64 in msvc.

answered Jun 21 '11 at 00:04

Blindy

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Vinicius Kamakura · Accepted Answer · 2011-06-21T01:27:20.567

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I don't know in C++, but in C, there is a header file called <stdint.h> that will portably have the integer types with the number of bits you desire.

int8_t 
int16_t
int32_t
int64_t

and their unsigned counterpart (uint8_t and etc).

Update: the header is called <cstdint> in C++

edited Jun 21 '11 at 01:27

answered Jun 21 '11 at 00:09

Vinicius Kamakura

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can you please the above comment, its quite causing a problem when i use it – Oct 08 '13 at 18:16

long for 12 digit number?

4 Answers4