Why use a function parameter 'foo' in this way: *(&foo)?

Question

A code snippet in Linux kernel 0.12 use a function parameter like this:

int do_signal(int signr, int eax /* other parameters... */) {     /* ... */      *(&eax) = -EINTR;      /* ... */ } 

The purpose of the code is to put -EINTR to the memory where eax lives, but I can not tell why it won't work if just assigning to eax:

eax = -EINTR 

How would the compiler make a difference between eax and *(&eax)?

Answer 1

The old Linux you posted was trying to perform an very fragile hack. The function was defined like this:

int do_signal(long signr,long eax,long ebx, long ecx, long edx, long orig_eax,     long fs, long es, long ds,     long eip, long cs, long eflags,     unsigned long * esp, long ss) 

The function arguments don't actually represent arguments to the function (except signr), but the values that calling function (a kernel interrupt/exception handler written in assembly) preserved on the stack before calling do_signal. The *(&eax) = -EINTR statement is meant to modify the preserved value of EAX on the stack. Similarly the statement *(&eip) = old_eip -= 2 is meant to modify the return address of the calling handler. After do_signal returns the handler pops the first 9 "arguments" off the stack restoring them to the named registers. It then executes an IRETD instruction that pops the remaining arguments off the stack and return to user mode.

Needless to say this hack is incredibly unreliable. It's dependent on the compiler generating code exactly the way they expected it to. I'm surprised it even worked the GCC compiler of the era, I doubt it was long before GCC introduced an optimization that broke it.