noo

https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html

any of the data type has a specific range. take byte! its range is from -128 to 127; if you raise - 2 to the power of 7 i.e. -128. (this number is in the required range!). but if you raise 2 to the power of 7 i.e. 128. (this number is not in the required range(127)). So, all you need is subtract 1 from 128 and hence the code :

// sample code.
if(x> -Math.pow(2,7) && x < Math.pow(2,7) - 1){
    System.out.println("* byte");
    System.out.println("* short");
    System.out.println("* int");
    System.out.println("* long");
}

bacause, like byte range is 8 so last in 0 to 7 meanse 8-1=7 hope you got it

It's to do with the range that data type can have.

For example short can have minimum -32,768 to maximum of 32,767.

Same case for int & long, which is ONE less number for it's maximum.

It's easier to write (2^31 - 1) than it is to write 2147483648 - 1 = 2147483647

long 64 –9,223,372,036,854,775,808 to 9 ,223,372,036,854,775,807

int 32 –2,147,483,648 to 2,147,483,647

short 16 –32,768 to 32,767

byte 8 –128 to 127

                            if you recognize the positive number of the primitive variable is alway less with 1, check the Byte the minimum  is -128 and the positive is not 128, its 128 -1 = 127, so we add -1 so we can have the right positive range and have the right value.

                            wish i was helpfull

Because in range of every databytes from negative to positive(postive is (-1) less than to negative ) as in byte -128 to +127.