Bitwise >> formula concept...
Ok, so I've been studying bitwise operators tonight, yay!
I came up with the formula (I'm sure I'm not the first) x << y x=x*2^y for a left shift and x >> y x=x/2^y for a right shift, the only problem is that if x = 1 then the 2nd formula doesn't ring true, so I'm just going to live with that even though my OCD doesn't wish it to be so. Am I correct with these formulas or am I in err? As a side not, is their a mathematical notation for an exception. I would just like to know for my own piece of mind so that I can notate it. 
You are spot on if U are considering operation on integers and result to be an integer. Also I assume that x<<y=x*(2^y) and x>>y=x/(2^y).
I think that even if x=1 the formulae still holds.. like 1>>1=1/2^1=1/2=0(considering integer values) 
Doh!
Yes, you are correct in that assumption. That is funny, because I just wrote down the Operator Precedence chart while waiting for a response. You're absolutely right about the formula, I was not thinking about integer division properly. Thank you, 
Also tying this into my learning on pointers, would
unsigned int *c; c++; change the pointer c to the next memory address, and if so, would: unsigned int *c; c<<16; (assuming that int is 16 bit) do the same thing? 
I'm afraid of being wrong .. but if the int is 16 bits .. and you left shift it 16 other bits .. wouldn't that turn the pointer to null?? I mean wouldn't the pointer adress 0x0000?
also c++ changes the pointer to the next memory address but in increments depending on the pointer type .. vor example for chars the increment will be 1 byte .. where as for int it will be 4 bytes (assuming 32bit int) I'm still not sure if I'm correct and am waiting for someone to set me straight :) Cheers Edit And even though the int is a 16 bit value .. most of the time ( I mean % of the values it can take) will fit in less than 16 bit 
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T* c; c += sizeof(T); Quote:
Thanks this has just fixed something I needed. Compile time unsigned range based on type.(for uint8 and uint16) Code:
template<typename T> or simpler still Code:
template<typename T> 
Would you explain that code for me a little bit please.
if sizeof(T) were 4 first one, would it be: 1 * (2^4) =16 * (2^3)= 128  1 = 127 (?) second one, I have no clue what it is saying... I dont' know what I was thinking about the bitwise operator, I was thinking of moving a 1 over 16 bits but yeah, I see now. 
the first one and the second one do the same thing
first one Code:
value = ( ( 1<< (sizeof(T)<<3) ) 1)}; If T is a unsigned char then sizeof is 1, this is then multiplied by 8 (bit shift left 3, 2^3). this then is the total number of bits that this type occupies (ie 8). one is pushed to the high bit and 1 deducted. giving 255.(well that nearly what happens, infact we push it a bit further, that why the deduction of one is there) then second one is better and less work(although its compile time so it doesn't matter). Code:
value = T(~0); 
I'm still not getting you on the first part of code because if you do
1 << 1 that equals two? then 2 << 3 would equal 16? 16 bits 1 = Since bitwise operators are evaluated left to right.' EDIT: IGNORE MISSED PAREN. so 1 << 3 = 8; 1 << 8 = 256 1 = 255 gotcha, shwew! in binary (for my understanding) 1 << 11 = 1000; 1 << 1000 = 100000000 1 = 11111111 
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Thanks for helping me understand that, I realized I had missed the paren shortly after posting. That's usually the way it works. My brain doesn't kick in until after I've spoken. 
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