warning: left shift count >= width of type

问题:

I'm very new to dealing with bits and have got stuck on the following warning when compiling:

 7: warning: left shift count >= width of type

My line 7 looks like this

unsigned long int x = 1 << 32;

This would make sense if the size of long on my system was 32 bits. However, sizeof(long) returns 8 and CHAR_BIT is defined as 8 suggesting that long should be 8x8 = 64 bits long.

What am I missing here? Are sizeof and CHAR_BIT inaccurate or have I misunderstood something fundamental?

回答1:

long may be a 64-bit type, but 1 is still an int. You need to make 1 a long int using the L suffix:

unsigned long x = 1UL << 32;

(You should also make it unsigned using the U suffix as I've shown, to avoid the issues of left shifting a signed integer. There's no problem when a long is 64 bits wide and you shift by 32 bits, but it would be a problem if you shifted 63 bits)

回答2:

unsigned long is 32 bit or 64 bit which depends on your system. unsigned long long is always 64 bit. You should do it as follows:

unsigned long long x = 1ULL << 32

回答3:

unsigned long x = 1UL << 31;

Not show the error message. Because before you specify the 32, is not true because only limited to 0-31.

回答4:

You can't shift a value to its max bit

int x;         // let int be 4 bytes so max bits : 32 
x <<= 32; 

So, this generates the warning

left shift count >= width of type (i.e type = int = 32 )

回答5:

You can use something like that:

unsigned long x = 1;
x = x << 32;