How can i put together two int value in a new int?
int one = 1;
int two = 2;
int twelve;
void setup() {
Serial.begin (9600);
twelve = twelve + one;
twelve = twelve & two; //here i want the int "twelve" get the value 12 but i can't find the way
Serial.println (twelve);
}
void loop() {
}
Did you sleep through math class all the way through school?
twelve = 10 * one + two;
twelve = 10 * one + two;
I can't understand the logic behind but thank u very much.
(deleted)
Did you sleep through math class all the way through school?
Huh? What is a "math class"?
Does anyone know a good const variable name for value 10?
GrOnThOs:
Does anyone know a good const variable name for value 10?
const int ten = 10;
GrOnThOs:
Does anyone know a good const variable name for value 10?
Slept through English class also 
...R
Robin2:
Slept through English class also...R
Bet he woke up for Sex Ed.
I was never present on an English classroom.
But this seems to me more like as personal attack rather than answer to a programming question.
(deleted)
But this seems to me more like as personal attack rather than answer to a programming question.
We are trying to make light of the fact that someone who doesn't even understand the decimal number system is attempting to program a computer.
Be warned: you have an extremely long, dusty road to travel, on foot.
GrOnThOs:
Does anyone know a good const variable name for value 10?
How about "tenToTheFirstPower" ?
We are talking about computers aren't we? So I'm going for Two.
Steve
Maybe we could get GolamMustafa to drop by and give a real complicated explanation of how this works involving 8-bit adders and registers and primitive assembly commands from long extinct processors
To get 12 (a decimal value) from the discrete positional factors 1 (Ten Positional Factor) and 2 (Unit Positional factor), the following approach is being presented which is still practiced in hard computer science classes!
12 = Value
Value = 1x101 + 2x100
===> Value = Ten Positional Factor (TPF) x Ten Positional Weight (TPW) + Unit Positional Factor (UPF) x Unit Positional Weight (UPW)
===> Value = TPF x TPW + UPF x UPW
while (UPF != 0)
{
IPR1 = IPR1 + UPW //IPR1 = 0x02
}
while (TPF !=0)
{
IPR2 = IPR2 + TPW //IPR2 = 0x0A
}
Value = IPR1 + IPR2 //Value = 0x0C
Serial.print(Value, DEC); //12
8085 Assembly Codes:
MVI A, 00H ; IPR1 = A = 00H
MVI C, 02H ; UPF = 02H
L1: ADD A, 01H ; UPW = 01H
DEC C
JNZ L1 ; IPR1 = A
MOV B, A ; IPR1 = B = 02H
MOV A, 00H ; IPR2 = A = 00H
MOV C, 01H ; TPF = 01H
L2: ADD A, 0AH ; TPW = 10 = 0AH
DEC C
JNZ L2
L3: ADD A, B ; A = 0CH = 00001100
L4: ;converting Binary to Decimal (BCD) using Horner Rule
for (n = 7; n >=0; x--)
{
IPBCD x 2 + bn ------> IPBCD //IPBCD = 12
DAA
}
Delta_G:
It may just be that the OP is expecting it to be harder than that. You know, overthinking it.Maybe we could get GolamMustafa to drop by and give a real complicated explanation of how this works involving 8-bit adders and registers and primitive assembly commands from long extinct processors
With flowcharts...
GrOnThOs:
Does anyone know a good const variable name for value 10?
[C++ Programming Style Guidelines](C++ Programming Style Guidelines Conventions)
+1 Karma for GolamMustafa for the good sense of humor and hard work
GolamMostafa:
8085 Assembly Codes:
MVI A, 00H ; IPR1 = A = 00H
MVI C, 02H ; UPF = 02H
L1: ADD A, 01H ; UPW = 01H
DEC C
JNZ L1 ; IPR1 = AMOV B, A ; IPR1 = B = 02H
MOV A, 00H ; IPR2 = A = 00H
MOV C, 01H ; TPF = 01H
L2: ADD A, 0AH ; TPW = 10 = 0AH
DEC C
JNZ L2L3: ADD A, B ; A = 0CH = 00001100
8085 assembly!!
brain memories of youth.....back in1981
Thank you so much to you all!