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copied!<p>The assumption that the <code>&</code> operator is always irreversible is incorrect.</p> <p>Yes, in general if you have</p> <pre><code>c = a & b </code></pre> <p>and all you know is the value of <code>c</code>, then you cannot know what values <code>a</code> or <code>b</code> had before hand.</p> <p>However it's very common for <code>&</code> to be used to <em>extract</em> certain bits from a longer value, where those bits were previously combined together with the <code>|</code> operator and where each 'bit field' is independent of every other. The fundamental difference with the generic <code>&</code> or <code>|</code> operators that makes this reversible is that the original bits were <em>all zero</em> beforehand, and the other bits in the word are left unchanged. i.e:</p> <pre><code>0xc0 | 0x03 = 0xc3 // combine two nybbles 0xc3 & 0xf0 = 0xc0 // extract the top nybble 0xc3 & 0x0f = 0x03 // extract the bottom nybble </code></pre> <p>In this case your current function appears to be extracting a 1 bit-per-pixel (monochrome image) and converting it to 32-bit RGBA.</p> <p>You'll need something like:</p> <pre><code>int source_image[]; byte dest_image[]; for (int h = 0; h < height; ++h) { for (int w = 0; w < width; ++w) { int offset = (h * width) + w; if (source_image[offset] == 0xffffffff) { int mask = w % 8; // these two lines convert from one int-per-pixel offset /= 8; // offset to one-bit-per-pixel dest_image[offset] |= (1 << mask); // only changes _one_ bit } } } </code></pre> <p>NB: assumes the image is a multiple of 8 pixels wide, that the <code>dest_image</code> array was previously all zeroes. I've used <code>%</code> and <code>/</code> in that inner test because it's easier to understand and the compiler <em>should</em> convert to mask / shift itself. Normally I'd do the masking and shifting myself.</p>

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