Based on Data types in c++

#1

1.why the max value form from integer 32 bit is (2^32-1)i think it should be 2^32
2.why the range of signed short int is -32768to32767
why 32767 is one lesser side than negative value

#2

Hi @bansalpriyal45, the answer to both of your questions is that:

  1. the counting starts from 0 AND NOT 1. Thus if counting of numbers were to start from 1 then the largest number that will be stored in in an unsigned int will be 2^32 i.e. 4294967296 but since we have to use 0 also thus max value stored is 2^32 - 1 .

  2. Now when we deal with negative numbers, then the first bit of reserved memory is used to determine wether no. is +ve or -ve. 1 is used for -ve and 0 for +ve. Size of short signed int is 2 bytes i.e. 16 bits.

 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ = 16 bits = 2 bytes
:point_up_2: this is either 1 or 0

so then we are left with 15 bits. max value that will be stored is 2^15 = 32768
so we can store 32768 negative numbers and 32768 +ve numbers.
When we store -ve numbers they start from -1 and go on to -32768 hence total 32768 numbers stored.
When we store +ve numbers we start count from 0 instead of 1 hence the max number that will be stores will be 2^15 - 1 i.e. 32767.
Hence the numbers stored are:

-32768, -32767 ,… -2, -1,   0, 1, 2, … 32766, 32767
[ ___ 32768 -ve nos.___ ] , [ __ 32768 +ve nos.__ ]

Hope this helps :slight_smile:

#3

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