wried integer addition

[tikviva]

Hi, I have a question regarding to integer addition.

Here is the problem.

I have an integer, x = 0x00000001
and I also have another integer, y = 0xffffff94

Is there any method to concatenate x & y, so that x = 0x00000194 ?
Thanks for the help. =)

God bless,
tikviva

 

[Andrey Tarasevich]

...
I have an integer, x = 0x00000001
and I also have another integer, y = 0xffffff94

Is there any method to concatenate x & y, so that x = 0x00000194 ?
...

I don't see how concatenating 0x00000001 and 0xffffff94 might produce
0x00000194. What kind of "concatenation" are you talking about? Why did
you call it "concatenation" in the first place?

 

[Patrik Stellmann]

Hi, I have a question regarding to integer addition.

Here is the problem.

I have an integer, x = 0x00000001
and I also have another integer, y = 0xffffff94

Is there any method to concatenate x & y, so that x = 0x00000194 ?
Thanks for the help. =)

So you're looking for a function that takes 2 integers and returns
0x00000194 if the input is 0x00000001 and 0xffffff94?

int f(int x, int y)
{
return 0x00000194;
}

Ok, probably not what you were looking for but it's difficult to create
a reasonable function when given only 1 input-output pair...

Maybe you want to take the 1 of x and place at infront of the 94 of y?
Then the function would look like this (i think):
return (x << 8) | (y & 0xFF);

 

[Karl Heinz Buchegger]

So you're looking for a function that takes 2 integers and returns
0x00000194 if the input is 0x00000001 and 0xffffff94?

int f(int x, int y)
{
return 0x00000194;
}

Ok, probably not what you were looking for but it's difficult to create
a reasonable function when given only 1 input-output pair...

Maybe you want to take the 1 of x and place at infront of the 94 of y?
Then the function would look like this (i think):
return (x << 8) | (y & 0xFF);

You forgot about the 0xfffff part, which has to be masked away (if
your theory of what the function should do holds)

return ( (x & ~0xff) << 8 ) | ( y & 0xff );

 

[Karl Heinz Buchegger]

You forgot about the 0xfffff part, which has to be masked away (if
your theory of what the function should do holds)

return ( (x & ~0xff) << 8 ) | ( y & 0xff );

grr.
That's the problem with bit operations: It is easy to read something
into a source code which is not written there.

Your function was fine.

 

[Patrik Stellmann]

You forgot about the 0xfffff part, which has to be masked away (if
your theory of what the function should do holds)

return ( (x & ~0xff) << 8 ) | ( y & 0xff );

Don't think so! The 0xffffff was only in y, not in x. Your function
would return 0x00000094 because (x & ~0xff) equals 0

 

[David Harmon]

grr.
That's the problem with bit operations: It is easy to read something
into a source code which is not written there.

Perhaps you would prefer

const int rad = 256;
return (x * rad) + (y % rad);

 

[Geoffrey Mortimer]

Hi, I have a question regarding to integer addition.

Here is the problem.

I have an integer, x = 0x00000001
and I also have another integer, y = 0xffffff94

Is there any method to concatenate x & y, so that x = 0x00000194 ?
Thanks for the help. =)

God bless,
tikviva

Homework?

(x + 1) << 8 + y