|之间的区别 当a或b为零时b和a + b

Question

I am trying to implement an if statement in C using bit manipulation. the following code seems to work:

// if (x) then y else z
int conditional(int x, int y, int z) {
    int v = (!!x); //v is 1 or 0
    return ((v<<31)>>31)&y | ((~v<<31)>>31)&z;
}

however, when I replace "|" with "+" in the last line, the code fails when x = 0x80000000,y = 0x80000000,z = 0x7fffffff. It returns 0 instead of 0x80000000.

so my question is: why does the following fail? thanks!

((v<<31)>>31)&y + ((~v<<31)>>31)&z

edit: I understand that | is bitwise OR and + is addition. but in this case, v is either 1 or zero, so (v<<31)>>31) is either all ones or all zeros. y | 0x00000000 should be the same as y + 0x00000000, right? specifically, v is 1. so the result with the above input should be y, but returns 0 when "+" is used.

Answer 1

I am afraid we have to use math.

The highest bits of your variables are set, thus doubling equals a bitshift of 1 left, which equals an overflow. All remaining bits are 0. Thus you get 0 as result.

Keep in mind that | is a bitwise || and does not carry over values to other binary digits. + does.

So speaking, if a and b are orthogonal base functions/vectors of your bitvectorspace (which in your case is not true) a|b and a+b are equal – but only then.

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Update:

The root cause here is operator precedence: (decreasing order) + & | .

So

((v<<31)>>31)&y + ((~v<<31)>>31)&z

becomes

((v<<31)>>31) & (y + ((~v<<31)>>31)) & z

Answer 2

In binary, 10+01 = 11 and 10 | 01 = 11. But 10 + 10 = 100, and 10 | 10 = 10.