I want to get the low 32 bit of a int64 as int32

问题

I have an Int64 value, but I only need the lower 32 bits. Thus I want a quick way to get the Int32 value from the lower 32 bits of the Int64 value.

Thanks

回答1:

Do something like this:

long tempLong = ((yourLong >> 32) << 32); //shift it right then left 32 bits, which zeroes the lower half of the long
int yourInt = (int)(yourLong - tempLong);

This may not be the most compact way to do it, but it seems to be the most readable to me. The following code will extract the high half of the long:

long tempLong = (int)(yourLong >> 32);

回答2:

If you assign a int64 value into a int32 value, the compiler will automatically do that for you
(as Steven Sudit mentioned):

int64 val64 = ...;
int32 val32 = ...;
...

val32 = val64; // get the low 32 bits
// or
val32 = (val64 >> 32); // get the high 32 bits

and because the compiler may display warnings you can specify the cast

val32 = (int32)val64;

回答3:

You didn't specify the language, but in many, all you need to do is cast it to an Int32. The top bits will be discarded.

回答4:

You could let the compiler handle the endian-ness, and hide all the bit shifting, pointer manipulation etc.

DWORD CUtility::ConvertUint64toUint32(unsigned __int64 in64){
    ULARGE_INTEGER uli;
    uli.QuadPart = in64;
    return uli.LowPart;
}

回答5:

In C/C++ the best way (in my opinion) is to use a union. For example the following function uses an anonymous union to extract the lower-order 32 bits from a 64 bit integer.

uint32_t lower_32_bits(uint64_t value) {
    union { uint64_t value; struct { uint32_t high, low; }; } converter;
    converter.value = value;
    return converter.low;
}

This union trick can be used for all sorts of things, like getting the bits of a floating point value into an integer of the same bitlength, e.g. for doing bitwise operations and other hacks.

来源：https://stackoverflow.com/questions/1447798/i-want-to-get-the-low-32-bit-of-a-int64-as-int32

标签

int64