How can I exchange two bits?

I can't help but think you are assuming that you can utilize the ACCumulator from C, which of course you can, but it is extremely ill-advised to do so. "ACC" is an SFR defined in regxx.h for the microcontroller's accumulator register -- the single most frequently used register! To assume that ACC contains anything predictable after a call to send_string() is sheer lunacy. In this case, it may happen to always be zero if only because (and this is a pure guess on my part) send_string() may return after encountering the NUL-terminator of the string you are passing it and having checked for zero using the accumulator before returning.

Well, for starters, do you really want to do this...

ACC0 = out[0] & 1 == 1 ? 1 : 0;

send_byte(ACC);

PS Is something up with the source code highlighting stuff?

It was highlighted in the preview, but not in the final post!

Well, there's the straightforward way:

bit temp;

// save bit 0
temp = out[0] & 1;

// bit 0 = bit 1
out[0] &= ~1;
out[0] |= (out[0] & 2) ? 1 : 0;

// bit 1 = temp
out[0] &= ~2;
out[0] |= temp ? 2 : 0;

bit tempBit;

// temporary swap area
bdata temp;
sbit temp0 = temp^0;
sbit temp1 = temp^1;

temp = out[0];

tempBit = temp0;
temp0 = temp1;
temp1 = tempBit;

out[0] = temp;

temp0 = temp0 ^ temp1;
temp1 = temp0 ^ temp1;
temp0 = temp0 ^ temp1;

Thank you, I've solved it this way:

temp1 = out[0] & 0xFC;
switch (out[0] & 3) {
case 1: out[0] = temp1 | 2; break;
case 2: out[0] = temp1 | 1;
};