Printable View
I've tried messing with the code below. I'm not sure why it's returning -7. I have a vague idea and that is that there is a overflow since the range of a char is between -127 and 127 but i don't get how c = -7 is calculated.
when i change j=1 i get
c = 35
when c is 4 i get -114 (this is when the minus numbers start to happen). I just haven't made sense of it.
Code:
int main()
{
int i = 10, j = 7;
double q = 3.56;
char c;
c = q * i * j;
printf("c = %d\n", c);
return 0;
}
You are correct in thinking that you have caused an overflow (not a buffer overflow though)
q * i * j = 3.56 * 10 * 7 = 249.2 which will be a double. When you try and store that in a char, the value will get cast to an integer (249) and then the computer will try to store that number in the 8-bit char. 249 in decimal is F9 in hex or 11111001 in binary so will fit ok into the variable c.
The problems start because of the way computers store signed numbers (its called two's complement if you want to do some research). Essentially this is how it works:
So.. in your case 249 stored as 0xF9 looks to the computer like -7; but 35 stored as 0x23 still looks like 35.Code:Number stored in Signed Number
Register (hex)
0x00 0
0x01 1
0x02 2
etc
etc
0x7E 126
0x7F 127
0x80 -128
0x81 -127
etc
etc
0xFE -2
0xFF -1
Be warned though that even if you make c an unsigned char the printf function will cast it to a signed int and you will see exactly the same result
No the result will be 249Quote:
Originally Posted by DavT
thanks davt
just another question why isn't the following casted to a char and then stored as a two's complement?
Code:
printf("%c", q * i * j);
becouse you dont have explicit cast like this
printf("%c\n", (char) (q * i * j));