I have a decimal, say 8, which is 1000 in binary. What I want to do is to NOT the sequence so it becomes 0111 and then get 7. Now the manual way is to just convert 8 to binary and then XOR with 1111 to get 0111. Then finally converting 0111 to 7. I was wondering if there was a way to skip the conversion and directly go from 8 to 7?
Inverting ~8 will be fastest due to underlying hardware support for such operations.
Last edited by itCbitC; 09-13-2011 at 12:40 PM.
Code:$ cat foo.c #include <stdio.h> int main ( ) { int a = 8; printf("%d\n",(~a)&0x0F ); return 0; } $ gcc foo.c $ ./a.out 7
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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I think part of your confusion is this concept of "conversion". The computer only deals in binary. When it shows you a decimal number, it's showing you the decimal representation of the underlying binary data. When you type in a decimal number, the functions like scanf and strtol convert that text to a number. That number is stored in binary. All those decimal numbers you type when programming get converted to binary when the compiler generates code. Also, as Salem pointed out, C has a bitwise NOT operator, the ~. It's equivalent to XORing with all 1's, but it's clearer and as fast or faster.
how would i get it working with 8 bit numbers? the not operator works great, but seems like 0x0f only deals with the 4bit to the right most.
Last edited by tianshiz; 09-13-2011 at 01:17 PM.
thanks
Well, your examples only dealt with the rightmost nibble (4-bit chunk), so naturally Salem's suggestion did likewise. The ~ operator will work with signed and unsigned versions of char as well as short, int, long, long long. Just declare an 8-bit variable (char or unsigned char), and use the ~ without the & 0x0f bit.Originally Posted by tianshiz
You can always do little tests like this... it's a great way to learn what does what.Code:#include <stdio.h> int main (void) { unsigned int x = 0; printf("%d %d", x, ~x); return 0; }
