问题
Working on Mac OS X 10.6.2, Intel, with i686-apple-darwin10-g++-4.2.1, and compiling with the -arch x86_64 flag, I just noticed that while...
std::numeric_limits<long double>::max_exponent10 = 4932
...as is expected, when a long double is actually set to a value with exponent greater than 308, it becomes inf--ie in reality it only has 64bit precision instead of 80bit.
Also, sizeof() is showing long doubles to be 16 bytes, which they should be.
Finally, using <limits.h> gives the same results as <limits>.
Does anyone know where the discrepancy might be?
long double x = 1e308, y = 1e309;
cout << std::numeric_limits<long double>::max_exponent10 << endl;
cout << x << '\t' << y << endl;
cout << sizeof(x) << endl;
gives
4932
1e+308 inf
16
回答1:
It's because 1e309 is a literal that gives a double. You need to use a long-double literal 1e309L.
来源：https://stackoverflow.com/questions/2565512/actual-long-double-precision-does-not-agree-with-stdnumeric-limits