实际的长双精度与std :: numeric

实际的长双精度与std :: numeric_limits不一致(Actual long double precision does not agree with std::numeric_limits)

在Mac OS X 10.6.2，Intel，使用i686-apple-darwin10-g ++ - 4.2.1，并使用-arch x86_64标志进行编译时，我注意到了......

std::numeric_limits<long double>::max_exponent10 = 4932

...正如预期的那样，当long double实际设置为指数大于308的值时，它变为inf - 即实际上它只有64位精度而不是80位。

此外， sizeof()显示长双精度为16字节，它们应该是。

最后，使用<limits.h>给出与<limits>相同的结果。

有谁知道差异可能在哪里？

long double x = 1e308, y = 1e309; cout << std::numeric_limits<long double>::max_exponent10 << endl; cout << x << '\t' << y << endl; cout << sizeof(x) << endl;

给

4932 1e + 308 inf 16

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

最满意答案

这是因为1e309是一个双倍的文字。 你需要使用long-double literal 1e309L 。

It's because 1e309 is a literal that gives a double. You need to use a long-double literal 1e309L.