Chapter 3. Using Mathematical Operators

3.0. Introduction

Almost every sketch uses mathematical operators to manipulate the value of variables. This chapter provides a brief overview of the most common mathematical operators. As the preceding chapter is, this summary is primarily for nonprogrammers or programmers who are not familiar with C or C++. For more details, see one of the C reference books mentioned in the Preface.

3.1. Adding, Subtracting, Multiplying, and Dividing

Problem

You want to perform simple math on values in your sketch. You want to control the order in which the operations are performed and you may need to handle different variable types.

Solution

Use the following code:

int myValue;
myValue = 1 + 2;  // addition
myValue = 3 - 2;  // subtraction
myValue = 3 * 2;  // multiplication
myValue = 3 / 2;  // division (the result is 1)

Discussion

Addition, subtraction, and multiplication for integers work much as you expect.

Note

Make sure your result will not exceed the maximum size of the destination variable. See Recipe 2.2.

Integer division truncates the fractional remainder in the division example shown in this recipe’s Solution; myValue will equal 1 after the division (see Recipe 2.3 if your application requires fractional results):

int value =   1 + 2 * 3 + 4;

Compound statements, such as the preceding statement, may appear ambiguous, but the precedence (order) of every operator is well defined. Multiplication and division have a higher precedence than addition and subtraction, so the result will be 11. It’s advisable to use brackets in your code to make the desired calculation precedence clear. int value = 1 + (2 * 3) + 4; produces the same result but is easier to read.

Use parentheses if you need to alter the precedence, as in this example:

  int value =   ((1 + 2) * 3) + 4;

The result will be 13. The expression in the inner parentheses is calculated first, so 1 gets added to 2, this then gets multiplied by 3, and finally is added to 4, yielding 13.

3.2. Incrementing and Decrementing Values

Problem

You want to increase or decrease the value of a variable.

Solution

Use the following code:

int myValue = 0;

myValue = myvalue + 1;  // this adds one to the variable myValue
myValue += 1;           // this does the same as the above

myValue = myvalue - 1;  // this subtracts one from the variable myValue
myValue -= 1;           // this does the same as the above

myValue = myvalue + 5;  // this adds five to the variable myValue
myValue += 5;           // this does the same as the above

Discussion

Increasing and decreasing the values of variables is one of the most common programming tasks, and the Arduino board has operators to make this easy. Increasing a value by one is called incrementing, and decreasing it by one is called decrementing. The longhand way to do this is as follows:

myValue = myvalue + 1;  // this adds one to the variable myValue

But you can also combine the increment and decrement operators with the assign operator, like this:

myValue += 1;           // this does the same as the above

See Also

Recipe 3.1

3.3. Finding the Remainder After Dividing Two Values

Problem

You want to find the remainder after you divide two values.

Solution

Use the % symbol (the modulus operator) to get the remainder:

int myValue0 =  20 % 10;  // get the modulus(remainder) of 20 divided by 10
int myValue1 =  21 % 10;  // get the modulus(remainder) of 21 divided by 10

myValue0 equals 0 (20 divided by 10 has a remainder of 0). myValue1 equals 1 (21 divided by 10 has a remainder of 1).

Discussion

The modulus operator is surprisingly useful, particularly when you want to see if a value is a multiple of a number. For example, the code in this recipe’s Solution can be enhanced to detect when a value is a multiple of 10:

int myValue;
//... code here to set the value of myValue
if (myValue % 10 == 0)
{
  Serial.println("The value is a multiple of 10");
}

The preceding code takes the modulus of the myValue variable and compares the result to zero (see Recipe 2.17). If the result is zero, a message is printed saying the value is a multiple of 10.

Here is a similar example, but by using 2 with the modulus operator, the result can be used to check if a value is odd or even:

int myValue;
//... code here to set the value of myValue
if (myValue % 2 == 0)
{
  Serial.println("The value is even");
}
else
{
  Serial.println("The value is odd");
}

This example calculates the hour on a 24-hour clock for any given number of hours offset:

void printOffsetHour( int hourNow, int offsetHours)
{
   Serial.println((hourNow + offsetHours) % 24);
}

void printOffsetHour(int hourNow, int offsetHours)
{
    Serial.println((hourNow + offsetHours) % 24);
}

See Also

Arduino reference for % (the modulus operator): http://www.arduino.cc/en/Reference/Modulo

3.4. Determining the Absolute Value

Problem

You want to get the absolute value of a number.

Solution

abs(x) computes the absolute value of x. The following example takes the absolute value of the difference between readings on two analog input ports (see Chapter 5 for more on analogRead()):

int x = analogRead(0);
int y = analogRead(1);

if (abs(x-y) > 10)
{
  Serial.println("The analog values differ by more than 10");
}

Discussion

abs(x-y); returns the absolute value of the difference between x and y. It is used for integer (and long integer) values. To return the absolute value of floating-point values, see Recipe 2.3.

See Also

Arduino reference for abs: http://www.arduino.cc/en/Reference/Abs

3.5. Constraining a Number to a Range of Values

Problem

You want to ensure that a value is always within some lower and upper limit.

Solution

constrain(x, min, max) returns a value that is within the bounds of min and max:

myConstrainedValue  = constrain(myValue, 100, 200);

Discussion

myConstrainedValue is set to a value that will always be greater than or equal to 100 and less than or equal to 200. If myValue is less than 100, the result will be 100; if it is more than 200, it will be set to 200.

Table 3-1 shows some example output values using a min of 100 and a max of 200.

Table 3-1. Output from constrain with min = 100 and max = 200

myValue (the input value)

constrain(myValue, 100, 200)

99

100

100

100

150

150

200

200

201

200

See Also

Recipe 3.6

3.6. Finding the Minimum or Maximum of Some Values

Problem

You want to find the minimum or maximum of two or more values.

Solution

min(x,y) returns the smaller of two numbers. max(x,y) returns the larger of two numbers:

myValue = analogRead(0);
myMinValue = min(myValue, 200);  // myMinValue will be the smaller of
                                 // myVal or 200

myMaxValue = max(myValue, 100);  // myMaxValue will be the larger of
                                 // myVal or 100

Discussion

Table 3-2 shows some example output values using a min of 200. The table shows that the output is the same as the input (myValue) until the value becomes greater than 200.

Table 3-2. Output from min(myValue, 200)

myValue (the input value)

min(myValue, 200)

99

99

100

100

150

150

200

200

201

200

Table 3-3 shows the output using a max of 100. The table shows that the output is the same as the input (myValue) when the value is greater than or equal to 100.

Table 3-3. Output from max(myValue, 100)

myValue (the input value)

max(myValue, 100)

99

100

100

100

150

150

200

200

201

201

Use min when you want to limit the upper bound. That may be counterintuitive, but by returning the smaller of the input value and the minimum value, the output from min will never be higher than the minimum value (200 in the example).

Similarly, use max to limit the lower bound. The output from max will never be lower than the maximum value (100 in the example).

If you want to find the min or max value from more than two values, you can cascade the values as follows:

// myMinValue will be the smaller of the three analog readings:
int myMinValue = min(analogRead(0), min(analogRead(1), analogRead(2)) );

In this example, the minimum value is found for analog ports 1 and 2, and then the minimum of that and port 0. This can be extended for as many items as you need, but take care to position the parentheses correctly. The following example gets the maximum of four values:

int myMaxValue = max(analogRead(0), max(analogRead(1), max(analogRead(2),
                                                             analogRead(3))));

See Also

Recipe 3.5

3.7. Raising a Number to a Power

Problem

You want to raise a number to a power.

Solution

pow(x, y) returns the value of x raised to the power of y:

myValue =  pow(3,2);

This calculates 32, so myValue will equal 9.

Discussion

The pow function can operate on integer or floating-point values and it returns the result as a floating-point value:

Serial.print(pow(3,2)); // this prints 9.00
int z = pow(3,2);
Serial.println(z);      // this prints 9

The first output is 9.00 and the second is 9; they are not exactly the same because the first print displays the output as a floating-point number and the second treats the value as an integer before printing, and therefore displays without the decimal point. If you use the pow function, you may want to read Recipe 2.3 to understand the difference between these and integer values.

Here is an example of raising a number to a fractional power:

  float s = pow(2, 1.0 / 12); // the twelfth root of two

The twelfth root of two is the same as 2 to the power of 0.083333. The resultant value, s, is 1.05946 (this is the ratio of the frequency of two adjacent notes on a piano).

3.8. Taking the Square Root

Problem

You want to calculate the square root of a number.

Solution

The sqrt(x) function returns the square root of x:

Serial.print( sqrt(9) ); // this prints 3.00

Discussion

The sqrt function returns a floating-point number (see the pow function discussed in Recipe 3.7).

3.9. Rounding Floating-Point Numbers Up and Down

Problem

You want the next smallest or largest integer value of a floating-point number (floor or ceil).

Solution

floor(x) returns the largest integral value that is not greater than x. ceil(x) returns the smallest integral value that is not less than x.

Discussion

These functions are used for rounding floating-point numbers; use floor(x) to get the largest integer that is not greater than x. Use ceil to get the smallest integer that is greater than x.

Here is some example output using floor:

  Serial.println( floor(1) );    // this prints  1.00
  Serial.println( floor(1.1) );  // this prints  1.00
  Serial.println( floor(0) );    // this prints  0.00
  Serial.println( floor(.1) );   // this prints  0.00
  Serial.println( floor(-1) );   // this prints -1.00
  Serial.println( floor(-1.1) ); // this prints -2.00

Here is some example output using ceil:

  Serial.println( ceil(1) );     // this prints  1.00
  Serial.println( ceil(1.1) );   // this prints  2.00
  Serial.println( ceil(0) );     // this prints  0.00
  Serial.println( ceil(.1) );    // this prints  1.00
  Serial.println( ceil(-1) );    // this prints -1.00
  Serial.println( ceil(-1.1) );  // this prints -1.00

You can round to the nearest integer as follows:

result = round(floatValue);

Note

You can truncate a floating-point number by casting (converting) to an int, but this does not round correctly. Negative numbers such as –1.9 should round down to –2, but when cast to an int they are rounded up to –1. The same problem exists with positive numbers: 1.9 should round up to 2 but will round down to 1. Use floor and ceil to get the correct results.

3.10. Using Trigonometric Functions

Problem

You want to get the sine, cosine, or tangent of an angle given in radians or degrees.

Solution

sin(x) returns the sine of angle x. cos(x) returns the cosine of angle x. tan(x) returns the tangent of angle x.

Discussion

Angles are specified in radians and the result is a floating-point number (see Recipe 2.3). The following example illustrates the trig functions:

  float deg = 30;                 // angle in degrees
  float rad  = deg * PI / 180;    // convert to radians
  Serial.println(rad);            // print the radians
  Serial.println (sin(rad));      // print the sine
  Serial.println (cos(rad));      // print the cosine

This converts the angle into radians and prints the sine and cosine. Here is the output with annotation added:

0.52   30 degrees is 0.5235988 radians, print only shows two decimal places
0.50   sine of 30 degrees is .5000000, displayed here to two decimal places
0.87   cosine is .8660254, which rounds up to 0.87

Although the sketch calculates these values using the full precision of floating-point numbers, the Serial.print routine shows the values of floating-point numbers to two decimal places.

The conversion from radians to degrees and back again is textbook trigonometry. PI is the familiar constant for π (3.14159265...). PI and 180 are both constants, and Arduino provides some precalculated constants you can use to perform degree/radian conversions:

rad = deg * DEG_TO_RAD;  // a way to convert degrees to radians
deg = rad * RAD_TO_DEG;  // a way to convert radians to degrees

Using deg * DEG_TO_RAD looks more efficient than deg * PI / 180, but it’s not, since the Arduino compiler is smart enough to recognize that PI / 180 is a constant (the value will never change), so it substitutes the result of dividing PI by 180, which happens to be the same value as the constant DEG_TO_RAD (0.017453292519...). Use whichever approach you prefer.

3.11. Generating Random Numbers

Problem

You want to get a random number, either ranging from zero up to a specified maximum or constrained between a minimum and maximum value you provide.

Solution

Use the random function to return a random number. Calling random with a single parameter sets the upper bound; the values returned will range from zero to one less than the upper bound:

random(max);      // returns a random number between 0 and max -1

Calling random with two parameters sets the lower and upper bounds; the values returned will range from the lower bound (inclusive) to one less than the upper bound:

random(min, max); // returns a random number between min and max -1

Discussion

Although there appears to be no obvious pattern to the numbers returned, the values are not truly random. Exactly the same sequence will repeat each time the sketch starts. In many applications, this does not matter. But if you need a different sequence each time your sketch starts, use the function randomSeed(seed) with a different seed value each time (if you use the same seed value, you’ll get the same sequence). This function starts the random number generator at some arbitrary place based on the seed parameter you pass:

randomSeed(1234);  // change the starting sequence of random numbers.

Here is an example that uses the different forms of random number generation available on Arduino:

// Random
// demonstrates generating random numbers

int randNumber;

void setup()
{
  Serial.begin(9600);

  // Print random numbers with no seed value
  Serial.println("Print 20 random numbers between 0 and 9");
  for(int i=0; i < 20; i++)
  {
    randNumber = random(10);
    Serial.print(randNumber);
    Serial.print(" ");
  }
  Serial.println();
  Serial.println("Print 20 random numbers between 2 and 9");
  for(int i=0; i < 20; i++)
  {
    randNumber = random(2,10);
    Serial.print(randNumber);
    Serial.print(" ");
  }

  // Print random numbers with the same seed value each time
  randomSeed(1234);
  Serial.println();
  Serial.println("Print 20 random numbers between 0 and 9 after constant seed ");
  for(int i=0; i < 20; i++)
  {
    randNumber = random(10);
    Serial.print(randNumber);
    Serial.print(" ");
  }

  // Print random numbers with a different seed value each time
  randomSeed(analogRead(0));  // read from an analog port with nothing connected
  Serial.println();
  Serial.println("Print 20 random numbers between 0 and 9 after floating seed ");
  for(int i=0; i < 20; i++)
  {
    randNumber = random(10);
    Serial.print(randNumber);
    Serial.print(" ");
  }
  Serial.println();
  Serial.println();
}

void loop()
{
}

Here is the output from this code:

Print 20 random numbers between 0 and 9
7 9 3 8 0 2 4 8 3 9 0 5 2 2 7 3 7 9 0 2
Print 20 random numbers between 2 and 9
9 3 7 7 2 7 5 8 2 9 3 4 2 5 4 3 5 7 5 7
Print 20 random numbers between 0 and 9 after constant seed
8 2 8 7 1 8 0 3 6 5 9 0 3 4 3 1 2 3 9 4
Print 20 random numbers between 0 and 9 after floating seed
0 9 7 4 4 7 7 4 4 9 1 6 0 2 3 1 5 9 1 1

If you press the reset button on your Arduino to restart the sketch, the first three lines of random numbers will be unchanged. Only the last line changes each time the sketch starts, because it sets the seed to a different value by reading it from an unconnected analog input port as a seed to the randomSeed function. If you are using analog port 0 for something else, change the argument to analogRead to an unused analog port.

See Also

Arduino references for random (http://www.arduino.cc/en/Reference/Random) and randomSeed (http://arduino.cc/en/Reference/RandomSeed)

3.12. Setting and Reading Bits

Problem

You want to read or set a particular bit in a numeric variable.

Solution

Use the following functions:

bitSet(x, bitPosition)

Sets (writes a 1 to) the given bitPosition of variable x

bitClear(x, bitPosition)

Clears (writes a 0 to) the given bitPosition of variable x

bitRead(x, bitPosition)

Returns the value (as 0 or 1) of the bit at the given bitPosition of variable x

bitWrite(x, bitPosition, value)

Sets the given value (as 0 or 1) of the bit at the given bitPosition of variable x

bit(bitPosition)

Returns the value of the given bit position: bit(0) is 1, bit(1) is 2, bit(2) is 4, and so on

In all these functions, bitPosition 0 is the least significant (rightmost) bit.

Here is a sketch that uses these functions to manipulate the bits of an 8-bit variable called flags:

// bitFunctions
// demonstrates using the bit functions


byte flags = 0; // these examples set, clear or read bits in a variable called flags.

// bitSet example
void setFlag( int flagNumber)
{
   bitSet(flags, flagNumber);
}

// bitClear example
void  clearFlag( int flagNumber)
{
   bitClear(flags, flagNumber);
}

// bitPosition example

int  getFlag( int flagNumber)
{
   return  bitRead(flags, flagNumber);
}

void setup()
{
  Serial.begin(9600);
}

void loop()
{
    showFlags();
    setFlag(2);  // set some flags;
    setFlag(5);
    showFlags();
    clearFlag(2);
    showFlags();

    delay(10000); // wait a very long time
}

// reports flags that are set
void showFlags()
{
    for(int flag=0; flag < 8; flag++)
    {
      if (getFlag(flag) == true)
         Serial.print("* bit set for flag ");      else
         Serial.print("bit clear for flag ");

      Serial.println(flag);
    }
    Serial.println();
}

This code will print the following:

bit clear for flag 0
bit clear for flag 1
bit clear for flag 2
bit clear for flag 3
bit clear for flag 4
bit clear for flag 5
bit clear for flag 6
bit clear for flag 7

bit clear for flag 0
bit clear for flag 1
* bit set for flag 2
bit clear for flag 3
bit clear for flag 4
* bit set for flag 5
bit clear for flag 6
bit clear for flag 7

bit clear for flag 0
bit clear for flag 1
bit clear for flag 2
bit clear for flag 3
bit clear for flag 4
* bit set for flag 5
bit clear for flag 6
bit clear for flag 7

Discussion

Reading and setting bits is a common task, and many of the Arduino libraries use this functionality. One of the more common uses of bit operations is to efficiently store and retrieve binary values (on/off, true/false, 1/0, high/low, etc.).

Note

Arduino defines the constants true and HIGH as 1 and false and LOW as 0.

The state of eight switches can be packed into a single 8-bit value instead of requiring eight bytes or integers. The example in this recipe’s Solution shows how eight values can be individually set or cleared in a single byte.

The term flag is a programming term for values that store the state of some aspect of a program. In this sketch, the flag bits are read using bitRead, and they are set or cleared using bitSet or bitClear. These functions take two parameters: the first is the value to read or write (flags in this example), and the second is the bit position indicating where the read or write should take place. Bit position 0 is the least significant (rightmost) bit; position 1 is the second position from the right, and so on. So:

bitRead(2, 1); // returns 1 : 2 is binary 10 and bit in position 1 is 1
bitRead(4, 1); // returns 0 : 4 is binary 100 and bit in position 1 is 0

There is also a function called bit that returns the value of each bit position:

bit(0)  is equal to 1;
bit(1)  is equal to 2;
bit(2)  is equal to 4;
...
bit(7)  is equal to 128

3.13. Shifting Bits

Problem

You need to perform bit operations that shift bits left or right in a byte, int, or long.

Solution

Use the << (bit-shift left) and >> (bit-shift right) operators to shift the bits of a value.

Discussion

This fragment sets variable x equal to 6. It shifts the bits left by one and prints the new value (12). Then that value is shifted right two places (and in this example becomes equal to 3):

int x = 6;
int result = x << 1;  // 6  shifted left 1 is 12
Serial.println(result);
int result = x >> 2;   // 12 shifted right 2 is 3;
Serial.println(result);

Here is how this works: 6 shifted left one place equals 12, because the decimal number 6 is 0110 in binary. When the digits are shifted left, the value becomes 1100 (decimal 12). Shifting 1100 right two places becomes 0011 (decimal 3). You may notice that shifting a number left by n places is the same as multiplying the value by 2 raised to the power of n. Shifting a number right by n places is the same as dividing the value by 2 raised to the power of n. In other words, the following pairs of expressions are the same:

x << 1 is the same as x * 2.
x << 2 is the same as x * 4.
x << 3 is the same as x * 8.
x >> 1 is the same as x / 2.
x >> 2 is the same as x / 4.
x >> 3 is the same as x / 8.

The Arduino controller chip can shift bits more efficiently than it can multiply and divide, and you may come across code that uses the bit shift to multiply and divide:

int c = (a << 1) + (b >> 2); //add (a times 2) plus ( b divided by 4)

The expression (a << 1) + (b >> 2); does not look much like (a * 2) + (b / 4);, but both expressions do the same thing. Indeed, the Arduino compiler is smart enough to recognize that multiplying an integer by a constant that is a power of two is identical to a shift and will produce the same machine code as the version using shift. The source code using arithmetic operators is easier for humans to read, so it is preferred when the intent is to multiply and divide.

See Also

Arduino references for bit and byte functions: lowByte, highByte, bitRead, bitWrite, bitSet, bitClear, and bit (see Recipe 3.12)

3.14. Extracting High and Low Bytes in an int or long

Problem

You want to extract the high byte or low byte of an integer; for example, when sending integer values as bytes on a serial or other communication line.

Solution

Use lowByte(i) to get the least significant byte from an integer. Use highByte(i) to get the most significant byte from an integer.

The following sketch converts an integer value into low and high bytes:

//ByteOperators

int intValue = 258; // 258 in hexadecimal notation is 0x102


void setup()
{
  Serial.begin(9600);
}

void loop()
{
  int loWord,hiWord;
  byte loByte, hiByte;


  hiByte = highByte(intValue);
  loByte = lowByte(intValue);

  Serial.println(intValue,DEC);
  Serial.println(intValue,HEX);
  Serial.println(loByte,DEC);
  Serial.println(hiByte,DEC);

  delay(10000); // wait a very long time
}

Discussion

The example sketch prints intValue followed by the low byte and high byte:

258     // the integer value to be converted
102     // the value in hexadecimal notation
2       // the low byte
1       // the high byte

To extract the byte values from a long, the 32-bit long value first gets broken into two 16-bit words that can then be converted into bytes as shown in the earlier code. At the time of this writing, the standard Arduino library did not have a function to perform this operation on a long, but you can add the following lines to your sketch to provide this:

#define highWord(w) ((w) >> 16)
#define lowWord(w) ((w) & 0xffff)

These are macro expressions: highWord performs a 16-bit shift operation to produce a 16-bit value, and lowWord masks the lower 16 bits using the bitwise And operator (see Recipe 2.20).

Note

The number of bits in an int varies on different platforms. On Arduino it is 16 bits, but in other environments it is 32 bits. The term word as used here refers to a 16-bit value.

This code converts the 32-bit hex value 0x1020304 to its 16-bit constituent high and low values:

  loWord = lowWord(longValue);
  hiWord = highWord(longValue);
  Serial.println(loword,DEC);
  Serial.println(hiword,DEC);

This prints the following values:

772   // 772 is 0x0304 in hexadecimal
258   // 258 is 0x0102 in hexadecimal

Note that 772 in decimal is 0x0304 in hexadecimal, which is the low-order word (16 bits) of the longValue 0x1020304. You may recognize 258 from the first part of this recipe as the value produced by combining a high byte of 1 and a low byte of 2 (0x0102 in hexadecimal).

See Also

Arduino references for bit and byte functions: lowByte, highByte, bitRead, bitWrite, bitSet, bitClear, and bit (see Recipe 3.12)

3.15. Forming an int or long from High and Low Bytes

Problem

You want to create a 16-bit (int) or 32-bit (long) integer value from individual bytes; for example, when receiving integers as individual bytes over a serial communication link. This is the inverse operation of Recipe 3.14.

Solution

Use the word(h,l) function to convert two bytes into a single Arduino integer. Here is the code from Recipe 3.14 expanded to convert the individual high and low bytes back into an integer:

//ByteOperators

int intValue = 0x102;  // 258


void setup()
{
  Serial.begin(9600);
}

void loop()
{
  int loWord,hiWord;
  byte loByte, hiByte;


  hiByte = highByte(intValue);
  loByte = lowByte(intValue);

  Serial.println(intValue,DEC);
  Serial.println(loByte,DEC);
  Serial.println(hiByte,DEC);

  loWord = word(hiByte, loByte);   // convert the bytes back into a word
  Serial.println(loWord,DEC);
  delay(10000); // wait a very long time
}

Discussion

The word(high,low) expression assembles a high and low byte into a 16-bit value. The code in this recipe’s Solution takes the low and high bytes formed as shown in Recipe 3.14, and assembles them back into a word. The output is the integer value, the low byte, the high byte, and the bytes converted back to an integer value:

258
2
1
258

Arduino does not have a function to convert a 32-bit long value into two 16-bit words (at the time of this writing), but you can add your own makeLong() capability by adding the following line to the top of your sketch:

#define makeLong(hi, low)  ((hi) << 16 & (low))

This defines a command that will shift the high value 16 bits to the left and add it to the low value:

#define makeLong(hi, low)  (((long) hi) << 16 | (low))
#define highWord(w) ((w) >> 16)
#define lowWord(w) ((w) & 0xffff)

// declare a value to test
long longValue = 0x1020304;  // in decimal: 16909060
                             // in binary : 00000001 00000010 00000011 00000100

void setup()
{
  Serial.begin(9600);
}

void loop()
{
  int loWord,hiWord;

  Serial.println(longValue,DEC);  // this prints  16909060
  loWord = lowWord(longValue);    // convert long to two words
  hiWord = highWord(longValue);
  Serial.println(loWord,DEC);     // print the value 772
  Serial.println(hiWord,DEC);     // print the value 258
  longValue = makeLong( hiWord, loWord);  // convert the words back to a long
  Serial.println(longValue,DEC);  // this again prints  16909060

  delay(10000); // wait a very long time
}

The output is:

16909060
772
258
16909060

See Also

Arduino references for bit and byte functions: lowByte, highByte, bitRead, bitWrite, bitSet, bitClear, and bit (see Recipe 3.12)

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