On Tue, Feb 9, 2010 at 1:09 PM, Hitoshi Mitake
<mitake@xxxxxxxxxxxxxxxxxxxxx> wrote:
> +void base64_encode(char *out, const char *in, int inlen)
> +{
> + const char *inp = in;
> + char *outp = out;
...Why? It's copying the pointers to pointers of identical type with
different names, and never using the originals again... Looks like a
sloppy extraction from another code-base to me.
> +
> + while (inlen >= 3) {
> + *outp++ = base64char[(inp[0] >> 2) & 0x3f];
> + *outp++ = base64char[((inp[0] & 0x03) << 4) |
> + ((inp[1] >> 4) & 0x0f)];
> + *outp++ = base64char[((inp[1] & 0x0f) << 2) |
> + ((inp[2] >> 6) & 0x03)];
> + *outp++ = base64char[inp[2] & 0x3f];
> +
> + inp += 3;
> + inlen -= 3;
> + }
> +
> + if (inlen > 0) {
> + *outp++ = base64char[(inp[0] >> 2) & 0x3f];
> + if (inlen == 1) {
> + *outp++ = base64char[(inp[0] & 0x03) << 4];
> + *outp++ = '=';
> + } else {
> + *outp++ = base64char[((inp[0] & 0x03) << 4) |
> + ((inp[1] >> 4) & 0x0f)];
> + *outp++ = base64char[((inp[1] & 0x0f) << 2)];
> + }
> + *outp++ = '=';
> + }
> +
> + *outp = '\0';
> +}
If inlen is 0, a single '=' should be emitted (plus the obvious zero
termination). It could be that the code deals with that by making sure
that inlen never is zero, though.
--
Erik "kusma" Faye-Lund
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