# How to modify a created byte?

I have created a byte for 16x2 LCD. It is basically a custom character which I will have to change later.

byte char[8]{
B10000,
B01000,
B00100,
B00010,
B00001,
B11000,
B11100
};


I want to edit the third number from fourth row. So, the fourth row will become B00110 from B00010. Is this possible? Are there any alternatives for this?

Well first of all you are trying to name the variable char? which is already a keyword for the variable type of char. But you can access it by the index of the array. Also not sure if you intended to make the array 8 bytes in length, and then only use 7, but thats what you had done in your example

//define it all at once
// byte byteArray[8] = {B10000,B01000,B00100,B00010,B00001,B11000,B11100};
//or by index
byte byteArray[8];
byteArray[0]=B10000;
byteArray[1]=B01000;
byteArray[2]=B00100;
byteArray[3]=B00010;
byteArray[4]=B00001;
byteArray[5]=B11000;
byteArray[6]=B11100;

//if you need to change one in the code elsewhere
byteArray[3]=B00110;
//or
byteArray[3]=6;
}

• In the second last line, why did you set the byteArray[3] equal to 6? Why not 3 because we want to change the third element? – Black Thunder Jun 19 at 11:34
• B00110 = B110 = 1 (first 1) * 2^2 + 1 (second 1) * 2^1 + 0 (last 0) * 2^0 = 1 * 4 + 1 * 2 + 0 * 1 = 6 – Michel Keijzers Jun 19 at 12:05
• @MichelKeijzers Well, i didn't get it. What is the purpose of it? Is there simpler way? I just don't want to involve other elements. – Black Thunder Jun 19 at 12:10
• E.g. in the normal (10 or decimal system) number 76 = 7 * 10^1 + 6 * 10^0 = 7 * 10 + 6 * 1 = 70 + 6 = 76. In the binary (2 system) number 110 = 1 * 2^2 + 1 * 2^1 + 0 * 2^0 = 1 * 4 + 1 * 2 + 1 * 0 = 4 + 2 + 0 = 6. – Michel Keijzers Jun 19 at 12:30
• Was just showing that it could be set using decimal rather then binary x=B00110(Binary), x=6(Dec),x=0x06(Hex) Are all equal. It doesn't really pertain to your question, so I should have just left it out to not be confusing. – Chad G Jun 19 at 16:08

You can use the following array initialization:

In the setup I shows how to set a bit. For this, the bit operator or (|) is used. To reset a bit, you can use &. You can set/reset multiple bits this way.

Also you can use 1 << 2 which means 1 (most right bit) shifted left two places (thus B100).

Btw, it is common practice to initialize all values.

byte lcd[8] =
{
B10000,
B01000,
B00100,
B00010,
B00001,
B11000,
B11100,
B00000  // Also initialize last element
};

void setup()
{
lcd[3] |= B100; // Set 4th row (element 3), 3th bit (from the right)
lcd[3] |= 1 << 2; // Alternative
}

void loop()
{
}


Explanation about the or arithmetics: OR means: if at least one bit is 1, the result is 1, otherwise 0.

Truth table:

A | B | A or B
--+---+-------
0 | 0 |   0
0 | 1 |   1
1 | 0 |   1
1 | 1 |   1


Original value of lcd[3]: B00010

• Why have we done this? lcd[3] |= B100; .Where did the other elements go? – Black Thunder Jun 19 at 9:21
• Can I do directly like lcd[3][3] = B1; where first 3 is the 4th row and second is the 3rd element? – Black Thunder Jun 19 at 11:21