in below code when i run it i get y=-124 and z=4294967172
can you explain me?? (tested that if x<128 there isnt any problem)
char x=132;
signed y=x;
unsigned z=x;
cout<<y<<endl;
cout<<z<<endl;
3 Answers
char is 8-bits, so when you write char x = 132, you actually do this:
x = 1000 0100
signed int and unsigned int both are 32-bits, and whenever you assign the value of a 'smaller' variable into a 'larger' one, the system uses sign extension, i.e. copies the sign bit to every bit to the left. So the value becomes:
1111 1111 1111 1111 1111 1111 1000 0100
If you interpret it as a signed value, it's -124, and as an unsigned value it's 4294967172.
Moreover, if you define the char as unsigned in the first line, you would always get 132, as sign extension is not done for unsigned values.
1 Comment
Arash
wow thanks,very interesting thing i got here.also thanks to James McNellis to complete the answer and showed me the way to solve this
chars are only 8-bit, which means they can only hold 256 values including 0. You're using a signed char, which means half the values are reserved for negative numbers, making the maximum positive value 127.
After first line x is 10000100 (132 is 10000100b) and it's the -124 value in machine (see here).
So after second line y is -124.
And after 3rd line z is 111...1110000100b. There are 1s on the higher bits because you're setting a negative number but when you print compiler prints it's binary value because z is unsigned.
x, they cannot in any way affect the semantics of the program.