I've been thinking about this problem for hours. Here it is:
Write an expression that returns 1 if a given integer "x" has any bits equal to 1. return 0 otherwise.
I understand that I'm essentially just trying to figure out if x == 0 because that is the only int that has no 1 bits, but I can't figure out a solution. You may not use traditional control structures. You may use bitwise operators, addition, subtraction, and bit shifts. Suggestions?
Here's the best I could come up with:
y = (((-x) | x) >> (BITS - 1)) & 1;
where BITS = 32 for 32 bit ints, ie BITS = sizeof(int) * CHAR_BIT;
Here's a test program:
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main(int argc, char *argv[])
{
const int BITS = sizeof(int) * CHAR_BIT;
if (argc == 2)
{
int x = atoi(argv[1]);
int y = (((-x) | x) >> (BITS - 1)) & 1;
printf("%d -> %d\n", x, y);
}
return 0;
}
Using !!x will give you the right answer. Since !0 = 1 and !(any nonzero number) = 0.
分别屏蔽每个位,将它们全部向下移动到lsb位置,或者将它们一起移位。
对于32位值,以下内容适用于所有位模式。
return (a | -a) >> 31;
How about !(x&&~x)&&x ?
#include <stdio.h>
void main(){
int x;
scanf("%d",&x);
printf("%d\n",(!(x&&~x)&&x));
}
It seems work, but I'm not sure when overflow happens.
0 || number - this will return 0 only if the number is 0 and will return 1 if the number is any other number than 0. Since a number without any bit as 1 will be equal to 0, we need to check it with 0.
You could just cast your int
to a bool
. But I doubt that's the purpose of your homework ;-)
int any_bits_to_one(unsigned int n) {
int result = 0, i;
for (i=0; !result && i < sizeof(unsigned int) * 8; i++)
result |= (n & (1<<i)) ? 1 : 0;
return result;
}
For 32 bit integers
int return_not_zero(int v)
{
r=v;
r=(r&0xFFFF) | (r>>16);
r=(r&0xFF) | (r>>8);
r=(r&0x0F) | (r>>4);
r=(r&0x03) | (r>>2);
r=(r&0x01) | (r>>1);
return r;
}
untested, that's the first thing that came to my mind:
while(n & pow(2, e) == 0 && e++ <= 16) ; // 16 or 32
if e == 16
after the loop n is 0.
Bitwise AND with 0 and any number must equal zero, but the only foolproof test would be with 0xFFFF, or every bit being set. To get all bits set, you should have a signed int, and assign it -1. You will then have an int with all bits set to 1, regardless of size.
So my answer would be to bitwise AND it with -1
I believe this is the simplest way.
return !!(0|x);
The only time your x will not have a 1 in it is when all bits are 0, or x == 0. So 0|0 -> 0 else 0|x -> non zero.
In C language, any value other than ZERO (either positive or negative) is treated as TRUE. And there should be a condition to check either your question's solution returns a ZERO or ONE (or other than ZERO). Therefore this answer is perfectly as per your requirement. This uses only bit-wise operators.
return (x & 0xFFFF);
This line returns ZERO when neither of any bit in "x" is 1, and returns Non-Zero (TRUE in a sense) when any of the bit is 1 in "x".
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