问题
What is the maximum value for a UInt32?
Is there a way I can use the sizeof operator to get the maximum value (as it is unsigned)? So I don't end up with #defines or magic numbers in my code.
回答1:
There's a macro UINT32_MAX
defined in stdint.h
which you can use
#include <stdint.h>
uint32_t max = UINT32_MAX;
More about the relevant header <stdint.h>
:
http://pubs.opengroup.org/onlinepubs/009695299/basedefs/stdint.h.html
回答2:
The maximum value for UInt32 is 0xFFFFFFFF
(or 4294967295 in decimal).
sizeof(UInt32)
would not return the maximum value; it would return 4, the size in bytes of a 32 bit unsigned integer.
回答3:
Just set the max using standard hexadecimal notation and then check it against whatever you need. 32-bits is 8 hexadecimals bytes, so it'd be like this:
let myMax: UInt32 = 0xFFFFFFFF
if myOtherNumber > myMax {
// resolve problem
}
回答4:
4.294.967.295 is the maximal value or in hexadecimal 0xFFFFFFFF
回答5:
The portable way:
std::numeric_limits<uint32_t>::max()
回答6:
An alternative for any unsigned in C or C++ is:
anUnsigned = -1;
This is useful since it works for them all, so if you change from unsigned int
to unsigned long
you don't need to go through your code. You will also see this used in a lot of bit fiddling code:
anUnsigned |= -(aBoolOrConditionThatWhenTrueCausesAnUnsignedToBeSetToAll1s)
anUnsigned |= -(!aValueThatWhenZeroCausesAnUnsignedToBeSetToAll1s)
anUnsigned |= -(!!aValueThatWhenNonZeroCausesAnUnsignedToBeSetToAll1s)
The downside is that it looks odd, assigning a negative number to an unsigned!
来源:https://stackoverflow.com/questions/20070374/how-do-i-set-a-uint32-to-its-maximum-value