Returning a long long value

Question

#include <stdio.h>

int main(void) 
{
    long long x = test();
    printf("%lld\n", x);

    return 1;
}

long long test()
{   
    return 1111111111111111111;
}

The output is 734294471 . If I replace the call to test() by a the number, the output is as I expect. I checked the value of x using a debugger and it wasn't set the to value returned by the function. What is going wrong?

I am using Visual Studio 2010 with the Visual C++ compiler.

Your IDE is not relevant for this sort of question, and `` is not a C++ header - if for some reason you needed to use the C-style IO library in C++ code, `#include ` instead. — Karl Knechtel, Jan 24 '11 at 23:02
@Karl I mention the IDE as it might affect the compiler settings. — Pulkit Sinha, Jan 24 '11 at 23:12
Excellent question, which pinpoints briefly and clearly exactly the problem I had. I'm surprised it didn't get more upvotes. — kotchwane, Mar 24 '21 at 08:58

score 4 · Accepted Answer · answered Jan 24 '11 at 22:26

4

You need to declare test before you call it, otherwise C assumes it returns int.

answered Jan 24 '11 at 22:26

Ned Batchelder

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1

Isn't this behaviour deprecated even in C nowadays? I would have expected a compilation error, actually. At least in C++ this -- using a function without having seen a declaration before -- is illegal. – sellibitze Jan 24 '11 at 22:30
1

this solved it! Enabling all compiler warnings showed the warning. – Pulkit Sinha Jan 24 '11 at 23:11
1

@tobyodavies: It is allowed in C89, which is the latest C standard that the Microsoft compiler supports. Presumably the OP is compiling this code as C, rather than C++. – caf Jan 25 '11 at 00:01
@Pulkit, you posted on SO _before_ reading the warnings from your compiler... seriously? – tobyodavies Jan 25 '11 at 00:21
@tobydavies I haven't worked with VC++ compiler a lot and didn't know that some warnings are disabled by default. The original compilation didn't show any warnings. – Pulkit Sinha Jan 25 '11 at 07:10

score 4 · Answer 2 · answered Jan 24 '11 at 22:28

4

IIRC, a long long constant in C/C++ is suffixed by 'LL'.

long long test() {
    return 1111111111111111111LL;
}

Your compiler is treating your constant as a 32-bit long (if you take your constant modulo 2^32, you get 734294471.)

answered Jan 24 '11 at 22:28

BobG

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I thought one L made it a `long` and a second made it a `long long` – tobyodavies Jan 24 '11 at 22:30
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The type of an unsuffixed decimal constant is the first out of `int`, `long int` and `long long int` that can represent the number. The `LL` suffix is only required to force a *smaller* number that would otherwise be of type `int` or `long` to be of type `long long` instead. – caf Jan 25 '11 at 00:04
the suffix is not necessary because if a constant doesn't fit in `int` it'll have type `long` or `long long` depending on its value http://stackoverflow.com/q/8108642/995714 – phuclv Feb 21 '16 at 10:02

score 1 · Answer 3 · answered Jan 24 '11 at 22:29

1

Try adding LL to your return value:

long long test()
{   
    return 1111111111111111111LL;
}

answered Jan 24 '11 at 22:29

mikhon

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score 0 · Answer 4 · answered Jan 24 '11 at 22:30

0

Add the suffix LL to your literal and see what happens. Presumably the compiler conberts the literal to an int. Are you getting any warnings from the compiler?

answered Jan 24 '11 at 22:30

Jörgen Sigvardsson

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Don't presume, *know*. For a decimal integer constant without a suffix, the compiler picks the first out of `int`, `long` and `long long` that is long enough to represent the number. – caf Jan 25 '11 at 00:07

Returning a long long value

4 Answers4