Bit-Shift array of Bits into Int Variable

Question

So I'm not sure if this is the easiest way to accomplish what my end goal is, so if anyone has a better suggestion, feel free to throw it at me. I've tried searching both on here, and with Google, but I can't seem to figure out how to do this.

I have an array of int values: int bitVal[8]

In my program, each of these values is 1 bit, within a byte. I need to create a loop which goes through each value, and bit shift the value into the int variable.

I tried this as a simple test:

int t = 0;
int e = 1;
for(int i = 0; i < 3; i++) {
    t <<= e;
    printf("%d\n", t);
}

Now I know that if I had a variable that had a byte value of 0000 0010 and I do var <<= 1; the result will be 0000 0100. Is there a way to set it so the shifted bit is a 1 instead of a 0?

Since t is 0 shifting it remains 0. The question is not very clear. int bitVal[8] is not in the code snippet, and the loop constraint is the curious i < 3;. — Weather Vane
– Weather Vane, Commented Feb 6, 2018 at 21:34

Jongware · Accepted Answer · 2018-02-06 21:34:59Z

2

You're probably overthinking it. You can shift in one bit at a time, regardless whether it is a 0 or 1. All it needs is a loop such as this:

result = 0;
for (i=0; i<8; i++)
{
    result = (result<<1) | bitVal[i];
}

answered Feb 6, 2018 at 21:34

Jongware

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1 Comment

Skitzafreak Over a year ago

Thank you. You're right, I was completely overthinking it T_T

John Bollinger · Accepted Answer · 2018-02-06 21:35:37Z

1

Is there a way to set it so the shifted bit is a 1 instead of a 0?

No, not as such. Instead, what you do is shift the target value to make room, and then combine with the value you want to add. For example, if x is your accumulation variable and y contains a value that you are certain is either 0 or 1, then you might do this:

x = (x << 1) | y;

answered Feb 6, 2018 at 21:35

John Bollinger

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