Is' long unsigned 'as valid as' unsigned long' in C?
是无符号的长一样有效，在C无符号长吗？
This is true for unsigned short, unsigned long, and unsigned long long.
对于unsignedshort , unsigned long ,以及 unsigned long long 来说也是这样。
What happens when unsigned long has more bits is that not all of those are 1.
会发生什么事无符号长具有更多的位是，不是所有的都是1。
Note that the pointer you used is declared an unsigned long long rather than just a pointer.
注意所使用的指针被声明为unsigned long long 而非指针。
If you in your expression used UL (said unsigned long integer), then you have a good starting point.
如果你在你的表达式中用到ul(表示无符号长整型)，那么你有了一个好的起点。
Offset Unsigned long integer giving the IL offset from the input string, or zero if there was no IL offset or the string was not well formed.
无符号长整型数，它给出输入字符串的IL偏移量，或在没有IL偏移量或字符串格式不良时给出零。
These arguments are canonically of type unsigned long long, although some programs engage in mild type punning, treating them as 64-bit Pointers or other types.
这些参数按照规定都是unsigned long long类型的，不过有些程序加入了适度的类型双关，将它们当作64位的指针或其他类型。
There are lots of flavors of this, but the basic idea is that you put a tag at the beginning of a variable name saying what type it is. (So all unsigned long variables begin with ul, etc.)
其种类很多，但基本的理念是在变量名的开始添加一个标记以表示其类型（例如，所有无符号长型变量都以ul 开头）。
In cases where unsigned and signed 32-bit integers are mixed in an expression and assigned to a signed long, cast one of the operands to its 64-bit type.
如果在表达式中混合使用无符号和有符号的32位整数，并将其赋值给一个有符号的long类型，那么将其中一个操作数转换成64位的类型。
The problem arises when passing the sum of signed and unsigned ints as long.
在将有符号整型和无符号整型的和作为long 类型传递时就会出现问题。
C-89 data types (include signed and unsigned) short, int, long, long long, float, and double.
C- 89数据类型(包括有符号和无符号)short、int、long、long long、float和double。
C-89 data types (include signed and unsigned) short, int, long, long long, float, and double.
C- 89数据类型(包括有符号和无符号)short、int、long、long long、float和double。
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