问题
My understanding is that numeric_limits::max_digits10 gives the max number of digits after the decimal point that are available. But if I setprecision() to a value that is greater than this, it still gives me nonzero digits beyond this max value :
assert(std::numeric_limits<double>::max_digits10 == 17);
std::cout << std::setprecision(30) << double(.1) << '\n';
This prints out:
0.100000000000000005551115123126
Are the digits beyond 17 not to be trusted to be accurate?
回答1:
Converting the 53 bit (leading 1 implied) mantissa of the double to binary fixed point:
0.00011001100110011001100110011001100110011001100110011010
This equals the decimal value
0.1000000000000000055511151231257827021181583404541015625
which matches the question's result to 30 digits
0.100000000000000005551115123126
However what could be a 54th bit is unknown, and if you consider this as a range of possible values, the binary fixed point numbers just below and above would be
0.000110011001100110011001100110011001100110011001100110001
0.000110011001100110011001100110011001100110011001100110101
representing decimal values:
0.099999999999999984734433411404097569175064563751220703125
0.100000000000000012490009027033011079765856266021728515625
which implies 16 or 17 digits of precision. So the 30 digits of precision are only accurate if you consider the double precision number to be an exact representation rather than the closest representation.
来源：https://stackoverflow.com/questions/44739874/print-30-digits-of-a-double-value-in-c