Convert Long Type Variables into Int in Haskell

In Haskell, we can use fromIntegral function along with toInteger and truncate function, div and mod functions to convert long type variable into int. In the first example, we are going to use ( let intVar = fromIntegral longVar :: Int) and in the second example, we are going to use (let highPart = fromIntegral (longVar `div` 2^32) :: Int and let lowPart = fromIntegral (longVar `mod` 2^32) :: Int).

Algorithm

• Step 1 ? The program execution will be started from main function. The main() function has whole control of the program. It is written as main = do.

• Step 2 ? The variable named, "longVar" is being initialized. It will hold the long type variable that is to be converted to respective integer value.

• Step 3 ? The function, fromIntegral is being called and longVar is passed to it.

• Step 4 ? The resultant integer value is printed to the console using ?putStrLn' statement after the function is called.

Example 1

In this example, the function is defined with the help of fromIntegral function to convert the long type variables to Int.

main :: IO ()
main = do
   let longVar = 123456789123456789
   let intVar = fromIntegral longVar :: Int
   putStrLn $ "The long value is " ++ show longVar ++ "."
   putStrLn $ "The corresponding integer value is " ++ show intVar ++ "."

Output

The long value is 123456789123456789.
The corresponding integer value is 123456789123456789.

Example 2

In this example, the function is defined with the help of fromIntegral and truncate function to convert the long type variables to Int.

main :: IO ()
main = do
   let longVar = 123456789123456789
   let intVar = truncate (fromIntegral longVar :: Double) :: Int
   putStrLn $ "The long value is " ++ show longVar ++ "."
   putStrLn $ "The corresponding integer value is " ++ show intVar ++ "."

Output

The long value is 123456789123456789.
The corresponding integer value is 123456789123456789.

Example 3

In this example, the function is defined with the help of fromIntegral and toInteger function to convert the long type variables to Int.

main :: IO ()
main = do
   let longVar = 123456789123456789
   let intVar = fromIntegral (toInteger longVar) :: Int
   putStrLn $ "The long value is " ++ show longVar ++ "."
   putStrLn $ "The corresponding integer value is " ++ show intVar ++ "."

Output

The long value is 123456789123456789.
The corresponding integer value is 123456789123456789.

Example 4

In this example, the function is defined with the help of fromIntegral function to convert the long type variables to Int.

main :: IO ()
main = do
   let longVar = 123456789123456789
   let highPart = fromIntegral (longVar `div` 2^32) :: Int
   let lowPart = fromIntegral (longVar `mod` 2^32) :: Int
   let intVar = highPart * 2^32 + lowPart
   putStrLn $ "The long value is " ++ show longVar ++ "."
   putStrLn $ "The corresponding integer value is " ++ show intVar ++ "."

Output

The long value is 123456789123456789.
The corresponding integer value is 123456789123456789.

Conclusion

In Haskell, a Long type variable represents a 64-bit signed integer, while an Int type variable represents a machine-dependent signed integer, which is typically either 32 or 64 bits depending on the platform. Therefore, converting a Long type variable to an Int type variable may result in a loss of precision if the Long value is too large to fit in an Int. In Haskell, a long type variables is converted to int using fromIntegral function along with truncate and toInteger function and also by using div & mod function.