Here's the problem: your program temporarily uses some sensitive data and wants to erase it when it's no longer needed. Using std::fill()
on itself won't always help - the compiler might decide that the memory block is not accessed later, so erasing it is a waste of time and eliminate erasing code.
User ybungalobill suggests using volatile
keyword:
{
char buffer[size];
//obtain and use password
std::fill_n( (volatile char*)buffer, size, 0);
}
The intent is that upon seeing the volat开发者_如何学Goile
keyword the compiler will not try to eliminate the call to std::fill_n()
.
Will volatile
keyword always prevent the compiler from such memory modifying code elimination?
The compiler is free to optimize your code out because buffer
is not a volatile object.
The Standard only requires a compiler to strictly adhere to semantics for volatile objects. Here is what C++03 says
The least requirements on a conforming implementation are:
- At sequence points, volatile objects are stable in the sense that previous evaluations are complete and subsequent evaluations have not yet occurred. [...]
and
The observable behavior of the abstract machine is its sequence of reads and writes to volatile data and calls to library I/O functions
In your example, what you have are reads and writes using volatile lvalues to non-volatile objects. C++0x removed the second text I quoted above, because it's redundant. C++0x just says
The least requirements on a conforming implementation are:
- Access to volatile objects are evaluated strictly according to the rules of the abstract machine.[...]
These collectively are referred to as the observable behavior of the program.
While one may argue that "volatile data" could maybe mean "data accessed by volatile lvalues", which would still be quite a stretch, the C++0x wording removed all doubts about your code and clearly allows implementations to optimize it away.
But as people pointed out to me, It probably does not matter in practice. A compiler that optimizes such a thing will most probably go against the programmers intention (why would someone have a pointer to volatile otherwise) and so would probably contain a bug. Still, I have experienced compiler vendors that cited these paragraphs when they were faced with bugreports about their over-aggressive optimizations. In the end, volatile
is inherent platform specific and you are supposed to double check the result anyway.
From the last C++0x draft [intro.execution]:
8 The least requirements on a conforming implementation are:
— Access to volatile objects are evaluated strictly according to the rules of the abstract machine.
[...]
12 Accessing an object designated by a volatile glvalue (3.10), modifying an object, calling a library I/O function, or calling a function that does any of those operations are all side effects, [...]
So even the code you provided must not be optimized.
The memory content you wish to remove may have already been flushed out from your CPU/core's inner cache to RAM, where other CPUs can continue to see it. After overwriting it, you need to use a mutex / memory barrier instruction / atomic operation or something to trigger a sync with other cores. In practice, your compiler will probably do this before calling any external functions (google Dave Butenhof's post on volatile's dubious utility in multi-threading), so if you thread does that soon afterwards anyway then it's not a major issue. Summarily: volatile isn't needed.
A conforming implementation may, at its leisure, defer the actual performance of any volatile reads and writes until the result of a volatile read would affect the execution of a volatile write or I/O operation.
For example, given something like:
volatile unsigned char vol1,vol2;
extern unsigned char res[1000];
void test(int scale)
{
unsigned char ch;
for (int 0=0; i<10000; i++)
{
res[i] = i*vol1*scale;
vol2 = res[i];
}
}
a conforming compiler could, at its option, check whether scale
is a multiple of 128 and--if so--clear out all even-indexed values of res
before doing any reads from vol1
or writes to vol2
. Even though the compiler would need to do each reads from vol1
before it could do the following write to vol2
, a compiler may be able to defer both operations until after it has run an essentially unlimited amount of code.
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