In embedded systems where interrupts are handled from user space (or the code you show is in the kernel), then it is guaranteed that your stack will be obliterated if an interrupt occurs at the exact right time, so your value will not be there.
This is not academic: I have debugged code that relied on that behavior, and would fail once in a million runs. Very frustrating to figure out.
Of course, the exact details depend on the architecture. For example, I've been doing a bunch of PIC24 microcontroller programming recently, in both C and assembly. On these, an interrupt will store the return address on the stack, obliterating the variable. However, if more dead variables are on the stack, they may be left intact, since PIC24 uses shadow-registers to stash away the users register values temporarily so that the interrupt may reuse them. That is, they are not temporarily pushed onto the stack (obliterating more dead variables), but rather stashed in shadow registers.
Actually, interrupts store two 16bit values from the stack - a 24bit instruction pointer and the lower 8bit of the STATUS register, so would destroy 32 bits worth of dead variables' values.
This is not academic: I have debugged code that relied on that behavior, and would fail once in a million runs. Very frustrating to figure out.