问题
I have a binary file that will be read in as characters. Each character was bit shifted to the left unknown number of times (assuming with wrap) by someone else. I want to be able to read in each character and then wrap shift to the right (the number of times to shift I guess will have to be figured out manually, because I haven't figured out another way).
So, my current idea is that I read in a character, create a copy with temp and then use XOR:
char letter; //will hold the read in letter
char temp; //will hold a copy of the letter
while(file.read(&letter, sizeof(letter)) //letter now holds 00001101
{
temp = letter; //temp now holds 00001101
letter >>= 1; //shift 1 position to the right, letter now holds 00000110
temp <<= 7; //shift to the left by (8-1), which is 7, temp now holds 10000000
letter ^= temp; //use XOR to get the wrap, letter now holds 10000110
cout << letter;
}
That makes sense in my exhausted head, but it doesn't work... and I can't figure out why. Size of char is 1 byte, so I figured I only have to mess around with 8 bits.
Any help would be appreciated.
EDIT: Solved. Thanks so much to everyone. Love this community to death, you guys are awesome!
回答1:
Pay attention to the signess of a char. On many systems it is signed. So your letter >>= 1
is sign filling the shift.
Rotating integers is usually done as follows
letter = ((unsigned char)letter >> 1) | (letter << 7);
As Mark points out in the comments, you can use either OR |
or XOR ^
.
回答2:
The statement temp <<= 7
is losing the bits that you want to wrap.
You will need to loop shifting left one bit at a time. First checking the most significant char bit and if set moving it to the right most bit before doing the shift.
回答3:
I'd be inclined to use a larger integral type:
unsigned val = (unsigned)letter & 0xFF;
val |= val << 8;
Now you just have to shift values in val
, without any additional code to wrap the high bits back in.
来源:https://stackoverflow.com/questions/15648116/bit-shifting-a-character-with-wrap-c