Why is long long int not as long as promised??

[Oliver Graeser]

I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte =
64 bit right? but when I try to assign a value with more than 32 bit,
it fails. To illustrate:

for (i=0; i<64; i++){
long long int k = 1<<i;

cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}
results in

1 2 8
2 4 8
3 8 8
4 16 8
5 32 8
6 64 8
7 128 8
8 256 8
9 512 8
10 1024 8
11 2048 8
12 4096 8
13 8192 8
14 16384 8
15 32768 8
16 65536 8
17 131072 8
18 262144 8
19 524288 8
20 1048576 8
21 2097152 8
22 4194304 8
23 8388608 8
24 16777216 8
25 33554432 8
26 67108864 8
27 134217728 8
28 268435456 8
29 536870912 8
30 1073741824 8
31 -2147483648 8
32 1 8
33 2 8
34 4 8
35 8 8
36 16 8
......

and so on.

I'm beyond confused - can anyone here help me out? Machine is a Core 2
Duo Macbook Pro with OSX10.5, gcc, Xcode.

Thanks

Oliver

[Captain Trips]

Oliver Graeser wrote:

I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte =
64 bit right? but when I try to assign a value with more than 32 bit,
it fails. To illustrate:

for (i=0; i<64; i++){
long long int k = 1<<i;

unsinged long int k = 1<<i; // use this

>
cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

That happens because you have only half of the numbers for positive, the
other half is used for negatives.

Look here: http://www.cplusplus.com/doc/tutorial/variables.html

[A. W. Dunstan]

Oliver Graeser wrote:

I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte =
64 bit right? but when I try to assign a value with more than 32 bit,
it fails. To illustrate:

for (i=0; i<64; i++){
long long int k = 1<<i;

cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}
results in

1 2 8
2 4 8
3 8 8
4 16 8
5 32 8
6 64 8
7 128 8
8 256 8
9 512 8
10 1024 8
11 2048 8
12 4096 8
13 8192 8
14 16384 8
15 32768 8
16 65536 8
17 131072 8
18 262144 8
19 524288 8
20 1048576 8
21 2097152 8
22 4194304 8
23 8388608 8
24 16777216 8
25 33554432 8
26 67108864 8
27 134217728 8
28 268435456 8
29 536870912 8
30 1073741824 8
31 -2147483648 8
32 1 8
33 2 8
34 4 8
35 8 8
36 16 8
.....

and so on.

I'm beyond confused - can anyone here help me out? Machine is a Core 2
Duo Macbook Pro with OSX10.5, gcc, Xcode.

Thanks

Oliver

1 is an int, so when you shift it left by more than 32 you wrap around. The
(int) result fits nicely into a 'long long', so the compiler has no reason
to complain. Try this:

const long long int ONE=1L;
for (i=0; i<64; i++){
long long int k = ONE << i;
cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

--
Al Dunstan, Software Engineer
OptiMetrics, Inc.
3115 Professional Drive
Ann Arbor, MI 48104-5131

[jellybean stonerfish]

On Thu, 25 Sep 2008 21:45:13 +0800, Oliver Graeser wrote:

I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte =
64 bit right? but when I try to assign a value with more than 32 bit,
it fails. To illustrate:

Maybe your 64 bit "int long long" holds positive and negative numbers.
Is there an "unsigned" type you can use?

[Victor Bazarov]

Oliver Graeser wrote:

I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte =
64 bit right? but when I try to assign a value with more than 32 bit,
it fails. To illustrate:

for (i=0; i<64; i++){
long long int k = 1<<i;

Try changing this to

long long int k = 1LL << i; // note the suffix after '1'.

You were shifting a simple int (1 is of type int), and its range is more
limited than that of 'long long'.

>
cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}
results in

1 2 8
2 4 8
3 8 8
4 16 8
5 32 8
6 64 8
7 128 8
8 256 8
9 512 8
10 1024 8
11 2048 8
12 4096 8
13 8192 8
14 16384 8
15 32768 8
16 65536 8
17 131072 8
18 262144 8
19 524288 8
20 1048576 8
21 2097152 8
22 4194304 8
23 8388608 8
24 16777216 8
25 33554432 8
26 67108864 8
27 134217728 8
28 268435456 8
29 536870912 8
30 1073741824 8
31 -2147483648 8
32 1 8
33 2 8
34 4 8
35 8 8
36 16 8
.....

and so on.

I'm beyond confused - can anyone here help me out? Machine is a Core 2
Duo Macbook Pro with OSX10.5, gcc, Xcode.

Thanks

Oliver

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Richard Herring]

In message <vd******************************@speakeasy.net> , A. W.
Dunstan <[e-mail]

>Oliver Graeser wrote:

>I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte =
64 bit right? but when I try to assign a value with more than 32 bit,
it fails. To illustrate:

for (i=0; i<64; i++){
long long int k = 1<<i;

cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}
results in

1 2 8

....

>30 1073741824 8
31 -2147483648 8
32 1 8
33 2 8
34 4 8
35 8 8
36 16 8
.....

and so on.

I'm beyond confused - can anyone here help me out? Machine is a Core 2
Duo Macbook Pro with OSX10.5, gcc, Xcode.

1 is an int, so when you shift it left by more than 32 you wrap around.

Worse. When you shift it left by 32 or more, you get UB.

(assuming 'int' is 32 bits on this machine.)

5.8/1:
"The behavior is undefined if the right operand is negative, or
greater than or equal to the length in bits of the promoted left
operand."

The
(int) result fits nicely into a 'long long', so the compiler has no reason
to complain. Try this:

const long long int ONE=1L;

That would be a 'long' literal, not 'long long'.

>for (i=0; i<64; i++){
long long int k = ONE << i;
cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

--
Richard Herring

[blargg]

In article <gb**********@ijustice.itsc.cuhk.edu.hk>, Oliver Graeser
<gr*****@phy.cuhk.edu.hkwrote:

I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte =
64 bit right?

long long isn't part of C++. Since you're relying on compiler extensions,
you might as well use int64_t from <stdint.hif you want a 64-bit
integer, or int_least64_t if 64 bits or greater is acceptable.

[Victor Bazarov]

Richard Herring wrote:

In message <vd******************************@speakeasy.net> , A. W.
Dunstan <[e-mail]
[..]

> The
(int) result fits nicely into a 'long long', so the compiler has no
reason
to complain. Try this:

const long long int ONE=1L;

That would be a 'long' literal, not 'long long'.

That actually doesn't matter. It could be

const long long int ONE = 1;

or

const long long int ONE = '\1';

the conversion should still yield a suitable value in 'ONE'.

>for (i=0; i<64; i++){
long long int k = ONE << i;
cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Juha Nieminen]

Richard Herring wrote:

>1 is an int, so when you shift it left by more than 32 you wrap around.

Worse. When you shift it left by 32 or more, you get UB.

(assuming 'int' is 32 bits on this machine.)

5.8/1:
"The behavior is undefined if the right operand is negative, or
greater than or equal to the length in bits of the promoted left operand."

That's actually something which should be taken seriously. I remember
back when I was developing on some UltraSparc architecture that if you
shifted an integer by more bits than there was in the integer, the
result would be, IIRC, the original value of that integer (rather than
zero, like AFAIK happens with intel CPUs). In other words, if the amount
of shift was invalid, the UltraSparc would simply not execute it
(leaving the original value of the integer intact). This can come as a
surprise to many.

[James Kanze]

On Sep 25, 6:41 pm, Richard Herring <ju**@[127.0.0.1]wrote:

In message <[e-mail]> , A. W.
Dunstan <[e-mail]

Oliver Graeser wrote:

I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte=
64 bit right? but when I try to assign a value with more than 32 bit,
it fails. To illustrate:

for (i=0; i<64; i++){
long long int k = 1<<i;

cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

results in

1 2 8

...

30 1073741824 8
31 -2147483648 8
32 1 8
33 2 8
34 4 8
35 8 8
36 16 8
.....

and so on.

I'm beyond confused - can anyone here help me out? Machine
is a Core 2 Duo Macbook Pro with OSX10.5, gcc, Xcode.

1 is an int, so when you shift it left by more than 32 you
wrap around.

Worse. When you shift it left by 32 or more, you get UB.

And it certainly doesn't wrap.

(But am I the only one who absolutely refuses to use shift
operators on signed types.)

(assuming 'int' is 32 bits on this machine.)

5.8/1:
"The behavior is undefined if the right operand is negative, or
greater than or equal to the length in bits of the promoted left
operand."

The
(int) result fits nicely into a 'long long', so the compiler has no reason
to complain. Try this:

const long long int ONE=1L;

That would be a 'long' literal, not 'long long'.

But ONE would be an integral constant of type long long.

for (i=0; i<64; i++){
long long int k = ONE << i;
cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

--
James Kanze (GABI Software) email:ja*********@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]

[James Kanze]

On Sep 25, 7:44 pm, Juha Nieminen <[e-mail]:

Richard Herring wrote:

1 is an int, so when you shift it left by more than 32 you wrap around..

Worse. When you shift it left by 32 or more, you get UB.

(assuming 'int' is 32 bits on this machine.)

5.8/1:
"The behavior is undefined if the right operand is negative, or
greater than or equal to the length in bits of the promoted left operand."

That's actually something which should be taken seriously. I
remember back when I was developing on some UltraSparc
architecture that if you shifted an integer by more bits than
there was in the integer, the result would be, IIRC, the
original value of that integer (rather than zero, like AFAIK
happens with intel CPUs). In other words, if the amount of
shift was invalid, the UltraSparc would simply not execute it
(leaving the original value of the integer intact). This can
come as a surprise to many.

I think that on a lot of architectures, i << n will actually
execute as i << (n % B), where B is the number of bits in i. So
that on a 32 bit machine, i << 32 shifts 0, i << 33 shifts 1,
etc.

Of course, if the shift count is a constant, the compiler should
warn.

--
James Kanze (GABI Software) email:ja*********@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]

[farseerfc]

that is because i and 1 is not long long, and therefore the result of 1<<i
is not long long, not enough to hold a long long value.
just try this:

long long int i,one=1;
for (i=0; i<64; i++){
long long int k = one<<i;
cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}
"Oliver Graeser" <gr*****@phy.cuhk.edu.hk写入消�
news:gb**********@ijustice.itsc.cuhk.edu.hk...

I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte =
64 bit right? but when I try to assign a value with more than 32 bit, it
fails. To illustrate:

for (i=0; i<64; i++){
long long int k = 1<<i;

cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

results in

1 2 8
2 4 8
3 8 8
4 16 8
5 32 8
6 64 8
7 128 8
8 256 8
9 512 8
10 1024 8
11 2048 8
12 4096 8
13 8192 8
14 16384 8
15 32768 8
16 65536 8
17 131072 8
18 262144 8
19 524288 8
20 1048576 8
21 2097152 8
22 4194304 8
23 8388608 8
24 16777216 8
25 33554432 8
26 67108864 8
27 134217728 8
28 268435456 8
29 536870912 8
30 1073741824 8
31 -2147483648 8
32 1 8
33 2 8
34 4 8
35 8 8
36 16 8
.....

and so on.

I'm beyond confused - can anyone here help me out? Machine is a Core 2 Duo
Macbook Pro with OSX10.5, gcc, Xcode.

Thanks

Oliver

[Stefan Rondinelli]

There's no need to use a variable for the constant 1 just to make the
type long long. You can also use a suffix (ll for long long):

long long int i;
for (i=0ll; i<64ll; i++){
long long int k = 1ll << i;
cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

----- Original Message -----
From: farseerfc
Date: 26.09.2008 09:26

that is because i and 1 is not long long, and therefore the result of 1<<i
is not long long, not enough to hold a long long value.
just try this:

long long int i,one=1;
for (i=0; i<64; i++){
long long int k = one<<i;
cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}
"Oliver Graeser" <gr*****@phy.cuhk.edu.hk写入消�
news:gb**********@ijustice.itsc.cuhk.edu.hk...

>I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte
= 64 bit right? but when I try to assign a value with more than 32
bit, it fails. To illustrate:

for (i=0; i<64; i++){
long long int k = 1<<i;

cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

results in

1 2 8
2 4 8
3 8 8
4 16 8
5 32 8
6 64 8
7 128 8
8 256 8
9 512 8
10 1024 8
11 2048 8
12 4096 8
13 8192 8
14 16384 8
15 32768 8
16 65536 8
17 131072 8
18 262144 8
19 524288 8
20 1048576 8
21 2097152 8
22 4194304 8
23 8388608 8
24 16777216 8
25 33554432 8
26 67108864 8
27 134217728 8
28 268435456 8
29 536870912 8
30 1073741824 8
31 -2147483648 8
32 1 8
33 2 8
34 4 8
35 8 8
36 16 8
.....

and so on.

I'm beyond confused - can anyone here help me out? Machine is a Core 2
Duo Macbook Pro with OSX10.5, gcc, Xcode.

Thanks

Oliver

[farseerfc]

;-p, thank you telling me the suffix from
but 1ll seems too much like 111, 1LL should be better

and is the suffix of unsigned long long int ULL? maybe we could write 0ULL
to mean NULL, how beauty C++ is!

Just a joke~

"Stefan Rondinelli" <ro***@polizisten-duzer.dewrote
news:9e**************************@news.cyberlink.c h...

There's no need to use a variable for the constant 1 just to make the type
long long. You can also use a suffix (ll for long long):

long long int i;
for (i=0ll; i<64ll; i++){
long long int k = 1ll << i;
cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

----- Original Message -----
From: farseerfc
Date: 26.09.2008 09:26

>that is because i and 1 is not long long, and therefore the result of
1<<i
is not long long, not enough to hold a long long value.
just try this:

long long int i,one=1;
for (i=0; i<64; i++){
long long int k = one<<i;
cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}
"Oliver Graeser" <gr*****@phy.cuhk.edu.hk写入消�
news:gb**********@ijustice.itsc.cuhk.edu.hk...

>>I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte =
64 bit right? but when I try to assign a value with more than 32 bit, it
fails. To illustrate:

for (i=0; i<64; i++){
long long int k = 1<<i;

cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

results in

1 2 8
2 4 8
3 8 8
4 16 8
5 32 8
6 64 8
7 128 8
8 256 8
9 512 8
10 1024 8
11 2048 8
12 4096 8
13 8192 8
14 16384 8
15 32768 8
16 65536 8
17 131072 8
18 262144 8
19 524288 8
20 1048576 8
21 2097152 8
22 4194304 8
23 8388608 8
24 16777216 8
25 33554432 8
26 67108864 8
27 134217728 8
28 268435456 8
29 536870912 8
30 1073741824 8
31 -2147483648 8
32 1 8
33 2 8
34 4 8
35 8 8
36 16 8
.....

and so on.

I'm beyond confused - can anyone here help me out? Machine is a Core 2
Duo Macbook Pro with OSX10.5, gcc, Xcode.

Thanks

Oliver

[Oliver Graeser]

Victor Bazarov wrote:

Oliver Graeser wrote:

>I need a >49 bit integer type. tried sizeof(long long), says 8. 8 byte
= 64 bit right? but when I try to assign a value with more than 32
bit, it fails. To illustrate:

for (i=0; i<64; i++){
long long int k = 1<<i;

Try changing this to

long long int k = 1LL << i; // note the suffix after '1'.

You were shifting a simple int (1 is of type int), and its range is more
limited than that of 'long long'.

> cout<<i<<"\t"<<k<<"\t"<<sizeof(k)<<"\n";
}

Indeed this works. Thanks a lot, wasn't aware that directly typed
numbers are always int. Also tried k=100000000000000 but it wouldn't
even compile.

[Pete Becker]

On 2008-09-26 07:41:11 -0400, Oliver Graeser <gr*****@phy.cuhk.edu.hksaid:

>
Indeed this works. Thanks a lot, wasn't aware that directly typed
numbers are always int.

Just to complicate matters, that's true if the value will fit in an
int. If it's too large, it has a type that's large enough to hold it.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)