Is it an error or am i wrong

This is a discussion on Is it an error or am i wrong within the C++ Programming forums, part of the General Programming Boards category; Hey guys im novice c prog and i am reading the book "jumping into c++". Im in the first few ...

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    Lightbulb Is it an error or am i wrong

    Hey guys im novice c prog and i am reading the book "jumping into c++".
    Im in the first few pages and i came across a problem.
    attached is a screenshot of my problem.
    Is it a printing mistake or have i messed up?
    This is pg 59 of the " jumping into C++"
    Im really enjoying this book.
    Isn't it supposed to be 264 because a double is 8 bytes (8bytes x 8bits=64bits)
    please guide me!
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    But those 64 bits are split up into a significand and exponent.

    Yes it can store numbers up to 10^308, but it's only the first 15 digits which are actually stored within those 8 bytes.
    The rest of the number isn't stored at all (hence significant digits).

    It's a trade-off between a very large dynamic range and a small compact storage format.

    It does however mean you have to be really careful when combining numbers with very different exponents.
    10^100 + 1 will still be 10^100. You see, 1 is just far too small to be of any significance to 10^100.

    If you really wanted to store ALL the digits of 10^308 in complete accuracy, you would need a number containing 128 bytes of memory, not 8 bytes.
    Such things are possible with a multi-precision library, but you have to do some work yourself.
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    oh okok i get it now !!

    thank you.

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    You are right, though, there is a typo in the book.

    There are 2^64 possible bit patterns that can be stored in 8 bytes, so there cannot be more than 2^64 different double precision floating point numbers. In fact there are slightly less since not every bit pattern represents a legal value.

    The number they print out, though, IS 2^64 - they had the expanded form correct but the simpler form mistyped.
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