numeric limits - How double can store 1024 bits in just 8 bytes in C++? -



numeric limits - How double can store 1024 bits in just 8 bytes in C++? -

i can't figure what's going on here.

i find size of next info types in bytes:

char:1 int:4 float:4 double:8 long long int:8 long long int max size 9223372036854775807 whereas double max size 179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368.000000

what!!!

how mr. double storing such big value in 8 bytes ???

how many decimal digits need represent every integer between 0 , 10100 - 1?

the reply should obvious: 100 digits.

how many decimal digits need represent integral values between 0 , 10100 - 1, if need 7 important digits?

you need 9. every number can represented abcdefg * 10hi.

doubles work in same way.

c++ numeric-limits

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