how to perform bitwise operation on floating point nubers?
919429
http://computer.howstuffworks.com/boolean.htm check this site out.
6928
When both operands are numeric, ^ in Java is a bitwise exclusive OR (otherwise known as XOR).  In binary terms, 2^4 is the same as 00000010 XOR 00000100 which results in 00000110 (6).  This is unless the ^ operator is overloaded in which case you'd have to refer to the overloaded method for XOR.  If you are not familiar with XOR, XOR returns TRUE when both operands are not equal, and FALSE when both are equal.
779807
Yiu would first decide how many decimal places you are going to consider.  The usual is 3.  Then work it from there...;
808564
printf("$%i.%i",dollars,cents);\n\nor if you have a floating point value you can do\n\nprintf("%5.2f",money);
139131
Something like this would work (syntax may not be exactly correct here...):\n\npublic class FibonacciNumbers {\n    public int getNthNumber(int n)  {\n        if (n <= 1) {\n            return 1;\n        }\n        int curFib = 1;\n        int lastFib = 0;\n        int temp;\n        for (int i = 2; i < n; i++) {\n            temp = curFib;\n            curFib = curFib + lastFib;\n            lastFib = temp;\n        }\n        return curFib;\n    }\n};\n\n\nYou can also write it recursively (i.e. call getNthNumber() inside the function), but this way is a bit faster due to overhead from function calls.
369365
move the decimal point to the left 9 spaces.\n.000000004.
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
790619
you can use the normal calculator of windows for these conversions.\n\nOptions are only available in a scientific view of the calculator.
782532
there is no built-in at-the-nth-power operator in c++,\n\nsimply use \n#include <math.h>\ndouble z = pow(3,2);
163804
Continuously divide the decimal number by 16 and stack up the remainders at each stage, until you finally reach a divisor that is less than 16. Of course, you have to write the remainders above 9 in the alphabetical form - 10=A, 11=B and so on till 15=F. Finally, string the final divisor followed by the stack of remainders read backwards to obtain your hex number.\n\nAlternately, if you are good at converting between decimal and binary, do that and then read off the binary bits from the end in groups of four in the equivalent hex. For example, if after decimal-to-binary conversion, you were to get\n11100011010101, then group them as\n(0011)(1000)(1101)(0101). Now, this is equivalently 38D5 in hex.
743387
Working above is good until this step:\n\n2 + 0.8x >0.9(x + 6)\n2 + 0.8x >0.9x + 5.4)\n0.1x > -3.4 ← Symbol error in this line\n\nShould read 0.1x < -3.4\nie x < -34
415435
nope... 01 (in binary) = 1 (in decimal)\n\nand \n\n10 (on binary) = 2 (in decimal)
1016721
actually, I use BCD for decimals\n\n0.264\n0010 0110 0100 (BCD)
767431
you need to cast the value. try this\n\nDouble.parseDouble(someStingValue);\nyou can use the variable as a double then.
731189
Yes. Shifting the binary point one position right doubles both the top and bottom.
85305
This is 8 bit binary system.\nand u know, there are only two possibilities for a switch (on/off or 0/1).\nin binary syatem, as i have to take 10101000, to convert into decimal.\nfrom right to left:\n0*2^0 + 0*2^1 + 0*2^2 + 1*2^3 + 0*2^4 + 1*2^5 + 0*2^6 + 1*2^7 = 168 (it is the decimal form).\n\nsimilarly as we take 10110001\n\nthen its decimal is :\nfrom right to left : and increase pawer of 2 from 0 step 1.\n1*2^0 + 0*2^1 + 0*2^2 + 0*2^3 + 1*2^4 + 1*2^5 + 0*2^6 + 1*2^7 =  177
6190
alternative 1)\nUse int/short instead. Make sure you mask potential values with "& 0xff" so they don't overflow.\n\nalternative 2)\nUse (signed) byte for storage, but when doing comparisons, assignment to other integer types, cast to int/short and mask with 0xff