Precision of C++ left/right shift operator..

[G Iveco]

I am using this type of code to do right-shifting,

B = 3;
data1 = (data + (1 << (B-1))) >B;

data1 seems incorrect when data = -4-8*i.. which means it
rounds -1.5 to -1 instead of -2.

On the positive side, 1.5 is rounded to 2, which ic correct.
For left-shift, it's simply as follows, no pitfalls, am I right?
data1 = data << B;

Thanks.

[Victor Bazarov]

G Iveco wrote:

I am using this type of code to do right-shifting,

B = 3;
data1 = (data + (1 << (B-1))) >B;

data1 seems incorrect when data = -4-8*i.. which means it
rounds -1.5 to -1 instead of -2.

Do you mean 'data' is declared as 'std::complex<i nt>' ? If not,
how is it declared? What's the type of 'data1'?

On the positive side, 1.5 is rounded to 2, which ic correct.

Are you sure it's -1.5? What if it's -1.49999999999?

For left-shift, it's simply as follows, no pitfalls, am I right?
data1 = data << B;

Please post the complete program (not all of your code, but
something we can just compile and run to see the behaviour).

V
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[G Iveco]

"Victor Bazarov" <v.********@com Acast.netwrote in message
news:f8******** **@news.datemas .de...

>G Iveco wrote:

>I am using this type of code to do right-shifting,

B = 3;
data1 = (data + (1 << (B-1))) >B;

data1 seems incorrect when data = -4-8*i.. which means it
rounds -1.5 to -1 instead of -2.

Do you mean 'data' is declared as 'std::complex<i nt>' ? If not,
how is it declared? What's the type of 'data1'?

>On the positive side, 1.5 is rounded to 2, which ic correct.

Are you sure it's -1.5? What if it's -1.49999999999?

Thank you for reply.

data and data1 are integer. Herer I use -1.5 to mean
-12.0/8 = -1.5, when right-shift, i was expecting -2.0 instead of -1.0..

same goes with 1.5, which means 12.0/8.0 and yield 2, as expected.

>For left-shift, it's simply as follows, no pitfalls, am I right?
data1 = data << B;

Please post the complete program (not all of your code, but
something we can just compile and run to see the behaviour).

V
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I do not respond to top-posted replies, please don't ask

My complete source is simply a test of ideas.. as follows,

#include <cstdlib>
#include <iostream>

using namespace std;

int main(int argc, char *argv[])
{
int i, j;
int data;
double tmp;
int data0;
int data1;
int data2;
FILE *fp;

fp = fopen("./data.txt", "w");
int KK = -153;
int B = 5;
fprintf(fp, "\n\n\nRigh t shifting, division by 2^B");
for(i=0; i<300; i++) {
data = KK + i;
tmp = data * 1.0 / 32.0;
data0 = (int)tmp;
data1 = (data + (1 << (B-1))) >B;
data2 = (data >B) + ((data >B-1) & 0x1);
fprintf(fp, "\ndata = %2d, tmp = %3.2f, data1 = %3d, data2 = %3d", data,
tmp, data1, data2);
}

fprintf(fp, "\n\n\nLeft shifting, multiply by 2^B");
for(i=0; i<300; i++) {
data = KK + i;
tmp = data * 8.0;
data0 = (int)tmp;
data1 = data << 3;
fprintf(fp, "\ndata = %2d, tmp = %3.2f, data1 = %3d, data0 = %3d", data,
tmp, data1, data0);
}
fclose(fp);
system("PAUSE") ;
return EXIT_SUCCESS;
}

[Victor Bazarov]

G Iveco wrote:

"Victor Bazarov" <v.********@com Acast.netwrote in message
news:f8******** **@news.datemas .de...

>G Iveco wrote:

>>I am using this type of code to do right-shifting,

B = 3;
data1 = (data + (1 << (B-1))) >B;

data1 seems incorrect when data = -4-8*i.. which means it

When you wrote 'data = -4-8*i' I took it that it's "complex" and
its real part is -4 and imaginary part is -8 (*i is the notation
to indicate the formula of a compex number 'a+b*i'). Since your
code didn't have any 'i', there was no other way for me to
interpret your "-4-8*i" expression.

>>rounds -1.5 to -1 instead of -2.

Do you mean 'data' is declared as 'std::complex<i nt>' ? If not,
how is it declared? What's the type of 'data1'?

>>On the positive side, 1.5 is rounded to 2, which ic correct.

Are you sure it's -1.5? What if it's -1.49999999999?

Thank you for reply.

data and data1 are integer. Herer I use -1.5 to mean
-12.0/8 = -1.5, when right-shift, i was expecting -2.0 instead of
-1.0..

WHY??? Right shift does not round. It discards the bits that are
shifted "out of existence".

I am guessing you're confused about shifting, division (those are
two separate operations), and how it is all done for integers versus
doubles. You need to read up on integral operations.

V
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[James Kanze]

On Jul 31, 5:36 pm, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:

G Iveco wrote:

"Victor Bazarov" <v.Abaza...@com Acast.netwrote in message
data and data1 are integer. Herer I use -1.5 to mean
-12.0/8 = -1.5, when right-shift, i was expecting -2.0 instead of
-1.0..

WHY??? Right shift does not round. It discards the bits that are
shifted "out of existence".

Actually, right shift isn't well defined for negative numbers.
It's implementation defined whether it shifts in the sign or a
0. And of course, the results (either way) depend on the
representation of negative numbers.

But whatever. As you say, right shift deals with bits (or the
bit representation) and not the numeric values. If you want to
deal with the numeric values, use the numeric operators.
(Except that in the current version of the C++ standard, which
way division rounds when the results are negative is
implementation defined as well.)

I am guessing you're confused about shifting, division (those are
two separate operations), and how it is all done for integers versus
doubles. You need to read up on integral operations.

Exactly.

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