C++ 使用 printf 打印 size_t 的正确方法是什么？
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What's the correct way to use printf to print a size_t?
提问by bradtgmurray
Size_tis defined as an unsignedinteger, but the size of it depends on whether you're on a 32 or 64-bit machine. What's the correct and portable way to print out a size_t?
Size_t定义为unsigned整数，但它的大小取决于您使用的是 32 位还是 64 位机器。打印输出的正确且便携的方法是size_t什么？
回答by JaredPar
Try using the %zuformat string
尝试使用%zu格式字符串
size_t val = get_the_value();
printf("%zu",val);
The z portion is a length specifier which says the argument will be size_t in length.
z 部分是一个长度说明符，表示参数的长度为 size_t。
Source - http://en.wikipedia.org/wiki/Printf#printf_format_placeholders
来源 - http://en.wikipedia.org/wiki/Printf#printf_format_placeholders
回答by David Thornley
There's a C++ tag on this, so cout <<is another possible answer.
这里有一个 C++ 标签，所以cout <<是另一个可能的答案。
This is surprisingly hard to get right in all versions of C. In C90, casting to unsigned longshould work, but that may well not work in C99, and the C99 solutions won't necessarily work in C90. The ability to reliably distinguish between C90 and C99 was introduced in the 1995 changes (specifying the allowable values for __STDC__). I don't think there is a completely portable way that works for C90, C99, and C++, although there are solutions for any individual one of those.
在所有版本的 C 中都很难做到这一点。在 C90 中，强制转换unsigned long应该可以工作，但这在 C99 中很可能不起作用，而且 C99 解决方案不一定在 C90 中工作。1995 年的更改中引入了可靠区分 C90 和 C99 的能力（指定 的允许值__STDC__）。我不认为有一种完全可移植的方式适用于 C90、C99 和 C++，尽管有针对其中任何一个的解决方案。
回答by D.Shawley
I think that the C++ answer is:
我认为 C++ 的答案是：
std::size_t n = 1;
std::cout << n;
For C-style IO it's a little more complicated. In C99 they added the zlength modifier for size_tvalues. However, previous to TR1 this is not supported so you are left with casting to a specific size like:
对于 C 风格的 IO，它有点复杂。在 C99 中，他们z为size_t值添加了长度修饰符。但是，在 TR1 之前，不支持此功能，因此您只能将其转换为特定大小，例如：
std::size_t n = 1;
std::printf("%lu\n", static_cast<unsigned long>(n));
Then again, unsigned long longisn't really supported by C++ anyway so the above will work fine since unsigned longis the largest legal integral type. After TR1 you can use %zusafely for size_tvalues.
再说一次，unsigned long long无论如何 C++ 并没有真正支持，因此上述内容可以正常工作，因为unsigned long它是最大的合法整数类型。在 TR1 之后，您可以%zu安全地使用size_t值。
