从 C 中的二进制“组件”构造一个 int

Question

I have several equations that will return the binary value for each of the three bits in my number.

I am programing this in C which is a new language for me.

So let's say my equations return Y0 = 1 , Y1 = 0 and Y2 = 1 ( 101 ); I want to store that value as 5 .

Is this possible in C if these values are returned by different equations?

I know how to do it by multiplying, but I am looking for a function or something which I imagine is already built into C.

Answer 1

No such luck. You have to multiply (or shift, which is the same)

unsigned Y0 = 1, Y1 = 0, Y2 = 1, val;

val = (Y0 * 4)  + (Y1 * 2)  + (Y2 * 1);  /* parens */
val = (Y0 << 2) + (Y1 << 1) + (Y2 << 0); /* redundant */

Answer 2

Just bitshift:

Y0 | (Y1 << 1) | (Y2 << 2)

Answer 3

There is no such function in the C standard library.

You will need to do:

int x = (Y2 << 2) | (Y1 << 1) | (Y0 << 0);

Answer 4

Although not very popular, you do have an alternative, in that you can use bit fields to perform the type of operation you're talking about. So, on ideone , the following code would perform the action you've suggested.

union combine {
    struct bits{
        int x: 1;
        int y: 1;
        int z: 1;
        int pad : 29;
    } bitvalues;
    int value;
};

int main() {
    union combine test;
    test.bitvalues.x = 1;
    test.bitvalues.y = 0;
    test.bitvalues.z = 1;
    test.bitvalues.pad = 0;
    printf("result: %d\n", test.value);
    return 0;
}

It's important to note however, that because you're using a combination of a bitfields and a union, this is not a portable solution . It is dependent on the order that you define your bits in, matching the bit/byte order for integer on that machine. So, there is a way to do it without bit shifting, but you're almost certainly better off going with the bit shifting solution.