Check the following table:
a b c a+b+c a^b^c
0 0 0 0 0 =
0 0 1 1 1 =
0 1 0 1 1 =
0 1 1 1 0
1 0 0 1 1 =
1 0 1 1 0
1 1 0 1 0
1 1 1 1 1 =
So they are equal if the number of ones between a, b, and c is zero or an odd number.
The or operator will evaluate to true if either side of the operation is true.The xor operator will evaluate to true only if exactly one side of the operation is true.This means that these two operators will evaluate equally for all cases except when both sides of the operations are true.true or true -> truetrue xor true -> false
The XOR operator is the carat character: ^Example:// Create characterchar ch = 'a';System.out.println(ch);// Perform XORch = (char)(ch ^ 86);System.out.println(ch);// Perform XOR...againch = (char)(ch ^ 86);System.out.println(ch);
The bitwise XOR operator is ^, or shift 6. The bitwise XOR assignment operator is ^=.
a XOR b is a^b in C language
If you're allowed to prove this the easy way (by showing you can use XOR and AND to create the set of AND, OR, and NOT), this is pretty straightforward. x AND y = x AND y (of course) x OR y = (x XOR y) XOR (x AND y) NOT x = x XOR 1 Also, (x AND y) XOR 1 is equivalent to x NAND y, which is a universal gate.
The or operator will evaluate to true if either side of the operation is true.The xor operator will evaluate to true only if exactly one side of the operation is true.This means that these two operators will evaluate equally for all cases except when both sides of the operations are true.true or true -> truetrue xor true -> false
The XOR operator is the carat character: ^Example:// Create characterchar ch = 'a';System.out.println(ch);// Perform XORch = (char)(ch ^ 86);System.out.println(ch);// Perform XOR...againch = (char)(ch ^ 86);System.out.println(ch);
XOR 0000111111110000
An XOR (exclusive OR gate) has two inputs and one output. If only one of the inputs is at level 1, then the output is 1 otherwise the output is 0. The truth table looks like this: A B Out0 0 00 1 11 0 11 1 0 Exclusive OR represents in logic what "or" means in English; for example, if asked if you want tea or coffee it's usually meant that you can have one or the other - not both.
xor
The bitwise XOR operator is ^, or shift 6. The bitwise XOR assignment operator is ^=.
a XOR b is a^b in C language
Truth table of 'NOR' is 0 0 - 1 0 1 - 0 1 0 - 0 1 1 - 0 NOR is just opposite of OR as the name itself suggest NOR is the not of OR. Whole XOR is 1 for different outputs and 0 for same outputs.
yes... xor is derived gate from primary gates
Three 2-input XOR gates and one 3-input NOR gate will do the work. Connect each output of each XOR gate to one input of the 3-input NOR gate and apply the two 3-bit words to the inputs of the XOR gates. If X (X2X1X0) and Y(Y2Y1Y0) are two 3-bit words, X2 and Y2 will connect to one XOR gate, X1 and Y1 to the next XOR gate and X0 and Y0 to the last XOR gate. You could see the result of the operation on a LED connected to the output of the NOR gate. Other implementations are also possible of course. The solution above is absolutely correct, but includes a 3 input gate. If the task is to use only two input gates, then a small change will be needed. Take the outputs from any two XOR gates into a 2 input OR gate. Then take the output of the OR gate and the output of the third XOR gate into a 2 input NOR gate. The operation remains identical to the first solution but adheres to the brief of using gates with 2 inputs. In the real world, there is probably no reason to impose such a limitation on a design so the first solution would normally be the preferred route to take.
If you're allowed to prove this the easy way (by showing you can use XOR and AND to create the set of AND, OR, and NOT), this is pretty straightforward. x AND y = x AND y (of course) x OR y = (x XOR y) XOR (x AND y) NOT x = x XOR 1 Also, (x AND y) XOR 1 is equivalent to x NAND y, which is a universal gate.
To convert Gray code to binary code you must be familiar with the logical XOR operator. XOR outputs a 1 bit if either of two input bits is 1, but not both. The truth table for XOR, for all possible inputs p and q, is as follows:p q output0 0 00 1 11 0 11 1 0The algorithm to convert from Gray code to binary code is as follows:Step 1: Fix the most-significant bit, the MSB, which is always the same for both codes. If there are no more bits, we're done, otherwise proceed to step 2.Step 2: XOR the most recently fixed binary bit with the next available Gray bit. Fix the result as the next binary bit.Step 3: If there is at least one more Gray bit available, go to step 2. Otherwise we're done.Therefore, to convert 10101111 from Gray to binary, we proceed as follows:Gray = 10101111Fix MSB = 11 XOR 0 = 11 XOR 1 = 00 XOR 0 = 00 XOR 1 = 11 XOR 1 = 00 XOR 1 = 11 XOR 1 = 1Thus: Binary = 11010101Note that we carry the fixed bit (the bold bit) onto the next line as the l-value (left operand) of XOR. The r-value (right operand) of XOR is always the next available Gray bit after the MSB. Reading the fixed bits from top to bottom reveals the binary code.We can also write this as follows:Gray = 10101111Binary = 1 XOR 0 = 1 XOR 1 = 0 XOR 0 = 0 XOR 1 = 1 XOR 1 = 0 XOR 1 = 1 XOR 1 = 1Reading the fixed (bold) bits left to right reveals the binary code.