Doubt in C programming

[navenmou]

#include<stdio.h>
main()
{
printf("%X",-1>>1);
}


i am getting the output is FFFFFFFF in linux..but actual output is 7FFFFFFF...

Please give me explanation why the output is FFFFFFFF

 

[venkateshem]

-1 is internally represented as all 1's. When right shifted the msb bit has been filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value as7FFFFFFF

 

[luben111]

-1 is 0xFFFFFFFF so if you shift it right by one you'll get 0x7FFFFFFF. I guess you're aware that you shift the result right by one

 

[navenmou]

but when iam compiling this using GCC Compiler i am ggetting the output as ffffffff

 

[luben111]

May I suggest you define the constant properly as 0xFFFFFFFF instead of -1 and try again

-1 is not totally defined - it could be 0xFF, or 0xFFFF or 0xFFFFFFFF. If you leave some ambiguity for the compiler it could do crazy things for you

 

[alexan_e]

I agree with all the previous posts,
-1 is internally represented as all 1's, when you shift right one then the MSB will become a 0 so the four MSB bits will be 0111 which is represented by 0x7,
depending on the bit width of the number this will be followed by "F"s 0x7FFF....

Alex

 

[horace1]

#include<stdio.h>
main()
{
printf("%X",-1>>1);
}


i am getting the output is FFFFFFFF in linux..but actual output is 7FFFFFFF...

Please give me explanation why the output is FFFFFFFF

the C and C++ standards state for the >> operator
">> the bit pattern is shifted right, the value shifted in from the left may be 0's (logical shift) or a copy of the sign bit (arithmetic shift); it is implementation dependent."
Hence, if the system uses arithmetic shift you get FFFFFFFF, if logical shift 7FFFFFFF.
These system dependent 'features' of C and C++ can cause lots of problems when porting code and watse days of your time if you are not careful with the orginal implementation.