Im trying to figure our how to perform a logical Not operation on a bit string.
ex. from 1101001 to the negation 00101100.
I was thinking of making an if statment for each value of the array, but this would be a lot of code. I assume there is a more efficient way to do this.
How about this?
Code:#include <stdio.h> int main(void) { char str[] = "1101001"; int i; for(i = 0; str[i]; ++i) putchar('0' + !(str[i] - '0')); putchar('\n'); return 0; }
If you understand what you're doing, you're not learning anything.
Use the bitwise logical not operator (~)
e.g.
Couldn't be simpler.Code:int notx = ~x;
Although you talk of a "bit string" and an array - if it is encoded in a string then you would need to extract the bits first.
I'm assuming it is an ASCII string that only contains '0's and/or '1's.
Code:char * bitstrnot(char *const string) { register char *pointer; for (pointer = string; *pointer != '\0'; pointer++) *pointer ^= 1; return string; }
