Question

Working with a Microchip 18f4620 PIC. This should be a standard ANSI C question, though.

Say I have

unsigned int16 badFlow=65535 //max unsigned int16 value

This has a binary value of 1111 1111 1111 1111.

if I then

badFlow++;

the bit pattern becomes 1 0000 0000 0000 0000
17 bits. Obviously badFlow == 0, but the additional flipped bit either

1. gets discarded
2. or resides at wherever byte* flowPtr = &badFlow+2;
   points.

I’m assuming the latter, but hoping for the former.

My problem: a coworker has written some bad code with a counter that has been overflowing on all produced products for ~2 years. Considering what our customers charge for use of these tools, that’s a few million dollars in peril due to potentially bad data.

Answer 1

Arithmetic in C takes place with values, not bytes in memory. Your expression badFlow++ is equivalent to badFlow = badFlow + 1. The right-hand side is evaluated as type int (due to default promotions, assuming int is larger than 16 bits; if int is only 16 bits then it would be evaluated as unsigned int) resulting in 65536, then when 65536 is assigned into an unsigned 16-bit variable, it is reduced modulo 65536, resulting in 0.

The important thing to get out of this answer is that badFlow++ is not a direct operation on the memory at &badFlow (although it could possibly be implemented as such on some implementations). It’s simply shorthand for an addition and assignment.