i want to change the a to the mask. How can I do that? The code is wrong in the bolded line.Code:#include<stdio.h> #include<limits.h> int main() { unsigned char a=0; int i=0; int masca[]={0,0,1,1,1,0,0,1}; for(i=CHAR_BIT-1;i>=0;i--) { (a>>i)=(a>>i)|(masca[8-i]); } for(i=CHAR_BIT-1;i>=0;i--) printf("%d",(a>>i)&1); printf("\n"); return 0; }
Thank you!
Maybe
Code:a = a | ( masca[8-i] << i );
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Also, don't forget you can use the shortcut operators with bitwise operations (e.g. a = a | b; is equivalent to a |= b;)
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may be this?
Code:a = (a>>i)|(masca[8-i]);
This?
Code:#include <stdio.h> #include <limits.h> int main() { unsigned char a = 0; int i, masca[] = {0,0,1,1,1,0,0,1}; for ( i = 0; i < CHAR_BIT; ++i ) { a |= masca[i] << ((CHAR_BIT - 1) - i); } printf("a = 0x%X\n", (unsigned)a); return 0; } /* my output a = 0x39 */
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
