=BITXOR() ☆

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◧ Arguments

▣ Examples

§ Remarks

• BITXOR returns a decimal number that is the result of the sum of a bitwise 'XOR' (exclusive XOR) of its parameters.
• If either argument is outside its constraint, BITXOR returns the #NUM! error value.
• If either argument is greater than (2^48)-1, BITXOR returns the #NUM! error value.
• If either argument is a non-numeric value, BITXOR returns the #VALUE! error value.
• In the result, each bit position is 1 if the values of the parameters at that bit position are not equal; in other words, one value is 0 and the other is 1. For example, using BITXOR(5,3), 5 is expressed as 101 in binary and 3 as 11 in binary. To help with comparison, you can consider 3 as 011. From right to left, the bit values at the three positions in this example are the same (1) only at the rightmost position. A 'not equal' result returns a 1 for the second and third positions from the right, and an ‘equal’ result returns 0 for the rightmost position.
• Values of 1 returned from the bit positions progress from right to left as powers of 2. The rightmost bit returns 1 (2^0), the bit to the left returns 2 (2^1), and so on.
• Using the same example, 0 is returned for the rightmost bit position because it is a 0, 2 (2^1) is returned for the second bit position from the right (a 1 value), and 4 (2^2) is returned for the leftmost bit (also a 1 value). The total is 6, in decimal representation.

✦ Tips

• BITXOR returns a decimal number that is the result of the sum of a bitwise 'XOR' (exclusive XOR) of its parameters.
• In the result, each bit position is 1 if the values of the parameters at that bit position are not equal; in other words, one value is 0 and the other is 1. For example, using BITXOR(5,3), 5 is expressed as 101 in binary and 3 as 11 in binary. To help with comparison, you can consider 3 as 011. From right to left, the bit values at the three positions in this example are the same (1) only at the rightmost position. A 'not equal' result returns a 1 for the second and third positions from the right, and an ‘equal’ result returns 0 for the rightmost position.
• Values of 1 returned from the bit positions progress from right to left as powers of 2. The rightmost bit returns 1 (2^0), the bit to the left returns 2 (2^1), and so on.
• Using the same example, 0 is returned for the rightmost bit position because it is a 0, 2 (2^1) is returned for the second bit position from the right (a 1 value), and 4 (2^2) is returned for the leftmost bit (also a 1 value). The total is 6, in decimal representation.

⚠ Common Errors

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