How do omputers negate binary?
145688
Computers represent signed integers using something called twos-complement arithmetic.  For numbers of a certain size, the first bit is reserved for negative numbers.  A negative number is represented as the one that, when added to it's positive, would give zero.\n\nSo in your example (where numbers are 3 bits long):\n\n-1 is represented as 111 because 111 + 001 = 000 (ignoring the carry to the nonexisting 4th bit).\n-2 is represented as 110 because 110 + 010 = 000.\n\nYou can think of it like an odometer.  To see how to represent the positive number n, it's like running the odometer forward n units starting from zero.  To see the negative number, run it backward n units starting from zero.
268949
It's just like the decimal system, only based on the number 2 instead of the number 10..........\n\n1  10  100  1000  10000  100000  1000000  etc... - Decimal\n\n1  2  4  8  16  32  64  128  256  512  1024  etc.. - Binary
307451
Try\n\nhttp://www.purplemath.com/modules/negative.htm
370603
Nor gate\n\nRemember this fact\n\nNOR gate is the exact opposite of OR gate.\n\nSince OR gate has Low (0) output only when both the inputs are low =>NOR gate has High (1) output only when both inputs are low.So with this we can say that for NOR gate the following holds true.\n\n0 nor 0 gives 1\n0 nor 1 gives 0\n1 nor 0 gives 0\n1 nor 1 gives 0\n\nHope this answres ur question.
895848
The opposite\n\ne.g. \nto cancel 5, add -5 to both sides\nto cancel x², add -x² to both sides\nto cancel -12x, add 12x to both sides\netc.
453934
You don't use binary in this way, and compilator embedded into IDE (interface of development environment) does it for you. \n\nAny program is the combination of bytes, and each byte (in majority of systems) consists of 8 bits of information. Each bit is 0 or 1.\n\nActually, you can write in hex code (almost binary format) and it's hard and inconvenient, or in assembler, or more advanced programming languages.\n\nFor example, In assembler, command, say,  "JMP 280000"\nfor example, means go to address 280000, and translated into\ncombination of bytes.\n\nIt's very easy to transform the program written in C or assembler, into binary format, automatically.\nAlso, if you already have somebody's else binary, you could desassemble it into assembler or C code, with some limitations because there is no 1 to 1 mapping all the time.\n\nTherefore, binary is not a programming language.
904666
Here is a super quick guide to learning to write numbers in binary...\n\nbinary uses 1/0 (on and off) to give value to numbers. This is why it is best used in computer language, because the computer can understand on and off. Now look at the following...\n\n1     \n2      \n4       \n8        \n16    \n32    \n64      \n128   \n\nNow,... notice the above... each number is equil to the number above it x2. this is the basis in learning binary...\n\nlets say we want to write the number  37....  we do the following...(read it from the bottom up)\n\n\n1        (there is one 1 in 1, this leaves us with 0 left over...)\n2        0\n4        (there is one 4 in 5, this leaves us with 1 left over....)\n8        0\n16      0\n32      (there is one 32 in 37, this leaves us with 5 left over...)\n64      0\n128    0\n\n\nso,... in binary, 37 would be   101001  (one 1, one 4, one32)\n\n\nusing the above we can learn to count to 10 !!!\n\n1   (1=1)\n01  (2=2)\n11  (2+1=3)\n001 (4=4)\n101 (4+1=5)\n011 (4+2=6)\n111 (4+2+1=7)\n0001 (8=8)\n1001 (8+1=9)\n0101 (8+2=10)\n\n\n\nthis is a very simple view of binary...\n\n\nhope this helps!!\n\n\n=)
751233
I'm not sure why you are avoiding the loop, but you could use recursion instead. If I understand your question correctly this function in c++ should give you your desired results.\n\nint sum(int nbr){\n    if(nbr > 0){\n           return(nbr%10 + sum(nbr/10));       \n    }\n    else{\n         return(0);\n    }\n}\n\nIf you called it with a number such as 256, it would return an answer of 13, or the sum of each digit in the integer.
49715
use bitwise operators. something like:\n\nnum_bits = 0;\nfor(i = 0; i < 32; i++) {\nif number & (1<<i) num_bits++;\n}\n\nnow num_bits holds the number of bits that were set in number. The "<<" operator does bit shifting. So "1<<i" equals a number that's all 0's except for a 1 at position i. If you bitwise and that with something, that tells you if that bit is set in the number. So we do this for each of the bits and add up the results.\n\nThis is assuming 32-bit integers. Adjust accordingly for 64-bit integers.
698756
First conversion the numbers to binary, then make the OR operation.\n\nExp.:\n8 OR 3.\n\nis: 1000 OR 11\n\n= 1011.\n\nWhich mean: 11 (Dec.)\n\nYou can use the Windows Calc.
378548
I_binmid = (I_binmax + I_binmin)/2
204914
I only need four 0.\nbinary:0!+0!+0!+0!=100
912485
there is no "code" for it. The compiler evaluates it at the compile time, and replaces with the actual number.\nThere is no way to write a function that would did that at run time, because at run time, there is not type information available - only addresses of variables not their sizes
248038
I will give your in binary :\n0000\n0001\n0010\n0011\n0100\n0101\n0110\n0111\n1000\n1001\n1010\n1011\n1100\n1101\n1110\n1111
224118
everything that interprets 1s and 0s into something you can use is an operating system, the difference is the architechture, ex:windows is x86
739075
I code almost exclusively in ASM and binary for most CPU architectures out (very old through new), and haven't gone insane because of it. Thats not to say I wasnt insane before I started programming. :)
591222
0.\n...and the counting stops, because we are counting in binary....lol.