Code Snippet:
int secret_foo(void)
{
int key = get_secret();
/* use the key to do highly privileged stuff */
....
/* Need to clear the value of key on the stack before exit */
key = 0;
/* Any half decent compiler would probably optimize out the statement above */
/* How can I convince it not to do that? */
return result;
}
I need to clear the value of a variable key from the stack before returning (as shown in the code).
In case you are curious, this was an actual customer requirement (embedded domain).
You can use volatile (emphasis mine):
Every access (both read and write) made through an lvalue expression of volatile-qualified type is considered an observable side effect for the purpose of optimization and is evaluated strictly according to the rules of the abstract machine (that is, all writes are completed at some time before the next sequence point). This means that within a single thread of execution, a volatile access cannot be optimized out or reordered relative to another visible side effect that is separated by a sequence point from the volatile access.
volatile int key = get_secret();
volatile might be overkill sometimes as it would also affect all the other uses of a variable.
Use memset_s (since C11): http://en.cppreference.com/w/c/string/byte/memset
memset may be optimized away (under the as-if rules) if the object modified by this function is not accessed again for the rest of its lifetime. For that reason, this function cannot be used to scrub memory (e.g. to fill an array that stored a password with zeroes). This optimization is prohibited for memset_s: it is guaranteed to perform the memory write.
int secret_foo(void)
{
int key = get_secret();
/* use the key to do highly privileged stuff */
....
memset_s(&key, sizeof(int), 0, sizeof(int));
return result;
}
You can find other solutions for various platforms/C standards here: https://www.securecoding.cert.org/confluence/display/c/MSC06-C.+Beware+of+compiler+optimizations
Addendum: have a look at this article Zeroing buffer is insufficient which points out other problems (besides zeroing the actual buffer):
With a bit of care and a cooperative compiler, we can zero a buffer — but that's not what we need. What we need to do is zero every location where sensitive data might be stored. Remember, the whole reason we had sensitive information in memory in the first place was so that we could use it; and that usage almost certainly resulted in sensitive data being copied onto the stack and into registers.
Your key value might have been copied into another location (like a register or temporary stack/memory location) by the compiler and you don't have any control to clear that location.
If you go with dynamic allocation you can control wiping that memory and not be bound by what the system does with the stack.
int secret_foo(void)
{
int *key = malloc(sizeof(int));
*key = get_secret();
memset(key, 0, sizeof(int));
// other magical things...
return result;
}
One solution is to disable compiler optimizations for the section of the code that you dont want optimizations:
int secret_foo(void) {
int key = get_secret();
#pragma GCC push_options
#pragma GCC optimize ("O0")
key = 0;
#pragma GCC pop_options
return result;
}
Related
I have a few questions regarding memory barriers.
Say I have the following C code (it will be run both from C++ and C code, so atomics are not possible) that writes an array into another one. Multiple threads may call thread_func(), and I want to make sure that my_str is returned only after it was initialized fully. In this case, it is a given that the last byte of the buffer can't be 0. As such, checking for the last byte as not 0, should suffice.
Due to reordering by compiler/CPU, this can be a problem as the last byte might get written before previous bytes, causing my_str to be returned with a partially copied buffer. So to get around this, I want to use a memory barrier. A mutex will work of course, but would be too heavy for my uses.
Keep in mind that all threads will call thread_func() with the same input, so even if multiple threads call init() a couple of times, it's OK as long as in the end, thread_func() returns a valid my_str, and that all subsequent calls after initialization return my_str directly.
Please tell me if all the following different code approaches work, or if there could be issues in some scenarios as aside from getting the solution to the problem, I'd like to get some more information regarding memory barriers.
__sync_bool_compare_and_swap on last byte. If I understand correctly, any memory store/load would not be reordered, not just the one for the particular variable that is sent to the command. Is that correct? if so, I would expect this to work as all previous writes of the previous bytes should be made before the barrier moves on.
#define STR_LEN 100
static uint8_t my_str[STR_LEN] = {0};
static void init(uint8_t input_buf[STR_LEN])
{
for (int i = 0; i < STR_LEN - 1; ++i) {
my_str[i] = input_buf[i];
}
__sync_bool_compare_and_swap(my_str, 0, input_buf[STR_LEN - 1]);
}
const char * thread_func(char input_buf[STR_LEN])
{
if (my_str[STR_LEN - 1] == 0) {
init(input_buf);
}
return my_str;
}
__sync_bool_compare_and_swap on each write. I would expect this to work as well, but to be slower than the first one.
static void init(char input_buf[STR_LEN])
{
for (int i = 0; i < STR_LEN; ++i) {
__sync_bool_compare_and_swap(my_str + i, 0, input_buf[i]);
}
}
__sync_synchronize before each byte copy. I would expect this to work as well, but is this slower or faster than (2)? __sync_bool_compare_and_swap is supposed to be a full barrier as well, so which would be preferable?
static void init(char input_buf[STR_LEN])
{
for (int i = 0; i < STR_LEN; ++i) {
__sync_synchronize();
my_str[i] = input_buf[i];
}
}
__sync_synchronize by condition. As I understand it, __sync_synchronize is both a HW and SW memory barrier. As such, since the compiler can't tell the value of use_sync it shouldn't reorder. And the HW reordering will be done only if use_sync is true. is that correct?
static void init(char input_buf[STR_LEN], bool use_sync)
{
for (int i = 0; i < STR_LEN; ++i) {
if (use_sync) {
__sync_synchronize();
}
my_str[i] = input_buf[i];
}
}
GNU C legacy __sync builtins are not recommended for new code, as the manual says.
Use the __atomic builtins which can take a memory-order parameter like C11 stdatomic. But they're still builtins and still work on plain types not declared _Atomic, so using them is like C++20 std::atomic_ref. In C++20, use std::atomic_ref<unsigned char>(my_str[STR_LEN - 1]), but C doesn't provide an equivalent so you'd have to use compiler builtins to hand-roll it.
Just do the last store separately with a release store in the writer, not an RMW, and definitely not a full memory barrier (__sync_synchronize()) between every byte!!! That's way slower than necessary, and defeats any optimization to use memcpy. Also, you need the store of the final byte to be at least RELEASE, not a plain store, so readers can synchronize with it. See also Who's afraid of a big bad optimizing compiler? re: how exactly compilers can break your code if you try to hand-roll lockless code with just barriers, not atomic loads or stores. (It's written for Linux kernel code, where a macro would use *(volatile char*) to hand-roll something close to __atomic_store_n with __ATOMIC_RELAXED`)
So something like
__atomic_store_n(&my_str[STR_LEN - 1], input_buf[STR_LEN - 1], __ATOMIC_RELEASE);
The if (my_str[STR_LEN - 1] == 0) load in thread_func is of course data-race UB when there are concurrent writers.
For safety it needs to be an acquire load, like __atomic_load_n(&my_str[STR_LEN - 1], __ATOMIC_ACQUIRE) == 0, since you need a thread that loads a non-0 value to also see all other stores by another thread that ran init(). (Which did a release-store to that location, creating acquire/release synchronization and guaranteeing a happens-before relationship between these threads.)
See https://preshing.com/20120913/acquire-and-release-semantics/
Writing the same value non-atomically is also UB in ISO C and ISO C++. See Race Condition with writing same value in C++? and others.
But in practice it should be fine except with clang -fsanitize=thread. In theory a DeathStation9000 could implement non-atomic stores by storing value+1 and then subtracting 1, so temporarily there's be a different value in memory. But AFAIK there aren't real compilers that do that. I'd have a look at the generated asm on any new compiler / ISA combination you're trying, just to make sure.
It would be hard to test; the init stuff can only race once per program invocation. But there's no fully safe way to do it that doesn't totally suck for performance, AFAIK. Perhaps doing the init with a cast to _Atomic unsigned char* or typedef _Atomic unsigned long __attribute__((may_alias)) aliasing_atomic_ulong; as a building block for a manual copy loop?
Bonus question: if(use_sync) __sync_synchronize() inside the loop.
Since the compiler can't tell the value of use_sync it shouldn't reorder.
Optimization is possible to asm that works something like if(use_sync) { slow barrier loop } else { no-barrier loop }. This is called "loop unswitching": making two loops and branching once to decide which to run, instead of every iteration. GCC has been able to do that optimization (in some cases) since 3.4. So that defeats your attempt to take advantage of how the compiler would compile to trick it into doing more ordering than the source actually requires.
And the HW reordering will be done only if use_sync is true.
Yes, that part is correct.
Also, inlining and constant-propagation of use_sync could easily defeat this, unless use_sync was a volatile global or something. At that point you might as well just make a separate _Atomic unsigned char array_init_done flag / guard variable.
And you can use it for mutual exclusion by having threads try to set it to 1 with int old = guard.exchange(1), with the winner of the race being the one to run init while they spin-wait (or C++20 .wait(1)) for the guard variable to become 2 or -1 or something, which the winner of the race will set after finishing init.
Have a look at the asm GCC makes for non-constant-initialized static local vars; they check a guard variable with an acquire load, only doing locking to have one thread do the run_once init stuff and the others wait for that result. IIRC there's a Q&A about doing that yourself with atomics.
I see the list of builtins at https://gcc.gnu.org/onlinedocs/gcc-4.1.0/gcc/Atomic-Builtins.html. But for an atomic set, do you need to use the pair __sync_lock_test_and_set and __sync_lock_release?
I have seen this example of this on https://attractivechaos.wordpress.com/2011/10/06/multi-threaded-programming-efficiency-of-locking/.
volatile int lock = 0;
void *worker(void*)
{
while (__sync_lock_test_and_set(&lock, 1));
// critical section
__sync_lock_release(&lock);
}
But if I use this example, and do my atomic set inside the critical section, then atomic sets to different variables will be unnecessarily serialized.
Appreciate any input on how to do an atomic set where I have multiple atomic variables.
As per definition need to use both
__sync_synchronize (...)
This builtin issues a full memory barrier. type
__sync_lock_test_and_set (type *ptr, type value, ...)
This builtin, as described by Intel, is not a traditional test-and-set
operation, but rather an atomic exchange operation. It writes value
into *ptr, and returns the previous contents of *ptr. Many targets
have only minimal support for such locks, and do not support a full
exchange operation. In this case, a target may support reduced
functionality here by which the only valid value to store is the
immediate constant 1. The exact value actually stored in *ptr is
implementation defined.
This builtin is not a full barrier, but rather an acquire barrier.
This means that references after the builtin cannot move to (or be
speculated to) before the builtin, but previous memory stores may not
be globally visible yet, and previous memory loads may not yet be
satisfied.
void __sync_lock_release (type *ptr, ...)
This builtin releases the
lock acquired by __sync_lock_test_and_set. Normally this means writing
the constant 0 to *ptr. This builtin is not a full barrier, but rather
a release barrier. This means that all previous memory stores are
globally visible, and all previous memory loads have been satisfied,
but following memory reads are not prevented from being speculated to
before the barrier.
I came up with this solution. Please reply if you know a better one:
typedef struct {
volatile int lock; // must be initialized to 0 before 1st call to atomic64_set
volatile long long counter;
} atomic64_t;
static inline void atomic64_set(atomic64_t *v, long long i)
{
// see https://attractivechaos.wordpress.com/2011/10/06/multi-threaded-programming-efficiency-of-locking/
// for an explanation of __sync_lock_test_and_set
while (__sync_lock_test_and_set(&v->lock, 1)) { // we don't have the lock, so busy wait until
while (v->lock); // it is released (i.e. lock is set to 0)
} // by the holder via __sync_lock_release()
// critical section
v->counter = i;
__sync_lock_release(&v->lock);
}
As everybody who has looked into this, I've read the paper http://www.aristeia.com/Papers/DDJ_Jul_Aug_2004_revised.pdf
I have a question about the barriers when DCLP is implemented on a C structure. Here is the code:
typedef struct _singleton_object {
int x;
int y;
} sobject;
static sobject *singleton_object = NULL;
sobject *get_singleton_instance()
{
sobject *tmp = singleton_object;
/* Insert barrier here - compiler or cpu specific or both? */
if (tmp == NULL) {
mutex_lock(&lock); /* assume lock is declared and initialized properly*/
tmp = singleton_object;
if (tmp == NULL) {
tmp = (sobject *)malloc(sizeof(sobject)); /* assume malloc succeeds */
tmp->x = 5;
tmp->y = 7;
/* Insert barrier here - compiler or cpu specific or both ?*/
singleton_object = tmp;
}
mutex_unlock(&lock);
}
return tmp;
}
The first question is as in the comments: When the paper describes insert barriers does it mean just the compiler, CPU or both? I assume both.
My second question is: what prevents the compiler from replacing tmp with singleton_object in the code? What forces the load of singleton_object into tmp, which could be in a register or stack in compiler generated code ? what if the compiler, at every reference to tmp, actually does a load into register from &singleton_object and discard that value?
It seems like the solution in the paper referenced below depends on the fact that we are using the local variable. if the compiler does not load the value in the pointer variable to the local variable tmp, we are back to the original problem described in the paper.
My third question is: Assuming, the compiler does copy the value of singleton_object locally into a register or stack(i.e. variable tmp), Why do we need the first barrier? There should be no reordering of tmp = singleton_object and if (tmp == NULL) in the beginning of the function, since there is an implicit read after write dependency with tmp. Also, even if we read a stale value from the CPU's cache in the first load to tmp, it should be read as NULL. If it is not NULL, then the object construction should be complete, since the thread/CPU that constructs it should execute the barrier, which ensures that the stores to x and y are visible to all CPU's before singleton_object has a non NULL value.
When the paper describes insert barriers does it mean just the compiler, CPU or both?
Both barriers should be CPU-barriers (which implies compiler barriers).
what prevents the compiler from replacing tmp with singleton_object in the code?
The barrier after assignment
sobject *tmp = singleton_object;
among other things means (both for CPU and compiler):
** All read accesses issued before the barrier should be completed before the barrier.
Because of that, compiler is not allowed to read singleton_object variable instead of tmp after the barrier.
If it (singleton_object) is not NULL, then the object construction should be complete, since the thread/CPU that constructs it should execute the barrier, which ensures that the stores to x and y are visible to all CPU's before singleton_object has a non NULL value.
You need to perform barrier for actual use these "visible" variables x and y. Without the barrier read thread may use stale values.
As a rule, every syncrhonization between different threads requires some sort of "barrier" on both sides: read and write.
This question already has answers here:
Why is volatile needed in C?
(18 answers)
Closed 9 years ago.
I am writing program for ARM with Linux environment. its not a low level program, say app level
Can you clarify me what is the difference between,
int iData;
vs
volatile int iData;
Does it have hardware specific impact ?
Basically, volatile tells the compiler "the value here might be changed by something external to this program".
It's useful when you're (for instance) dealing with hardware registers, that often change "on their own", or when passing data to/from interrupts.
The point is that it tells the compiler that each access of the variable in the C code must generate a "real" access to the relevant address, it can't be buffered or held in a register since then you wouldn't "see" changes done by external parties.
For regular application-level code, volatile should never be needed unless (of course) you're interacting with something a lot lower-level.
The volatile keyword specifies that variable can be modified at any moment not by a program.
If we are talking about embedded, then it can be e.g. hardware state register. The value that it contains may be modified by the hardware at any unpredictable moment.
That is why, from the compiler point of view that means that compiler is forbidden to apply optimizations on this variable, as any kind of assumption is wrong and can cause unpredictable result on the program execution.
By making a variable volatile, every time you access the variable, you force the CPU to fetch it from memory rather than from a cache. This is helpful in multithreaded programs where many threads could reuse the value of a variable in a cache. To prevent such reuse ( in multithreaded program) volatile keyword is used. This ensures that any read or write to an volatile variable is stable (not cached)
Generally speaking, the volatile keyword is intended to prevent the compiler from applying any optimizations on the code that assume values of variables cannot change "on their own."
(from Wikipedia)
Now, what does this mean?
If you have a variable that could have its contents changed at any time, usually due to another thread acting on it while you are possibly referencing this variable in a main thread, then you may wish to mark it as volatile. This is because typically a variable's contents can be "depended on" with certainty for the scope and nature of the context in which the variable is used. But if code outside your scope or thread is also affecting the variable, your program needs to know to expect this and query the variable's true contents whenever necessary, more than the normal.
This is a simplification of what is going on, of course, but I doubt you will need to use volatile in most programming tasks.
In the following example, global_data is not explicitly modified. so when the optimization is done, compiler thinks that, it is not going to modified anyway. so it assigns global_data with 0. And uses 0, whereever global_data is used.
But actually global_data updated through some other process/method(say through ptrace ). by using volatile you can force it to read always from memory. so you can get updated result.
#include <stdio.h>
volatile int global_data = 0;
int main()
{
FILE *fp = NULL;
int data = 0;
printf("\n Address of global_data:%x \n", &global_data);
while(1)
{
if(global_data == 0)
{
continue;
}
else if(global_data == 2)
{
;
break;
}
}
return 0;
}
volatile keyword can be used,
when the object is a memory mapped io port.
An 8 bit memory mapped io port at physical address 0x15 can be declared as
char const ptr = (char *) 0x15;
Suppose that we want to change the value at that port at periodic intervals.
*ptr = 0 ;
while(*ptr){
*ptr = 4;//Setting a value
*ptr = 0; // Clearing after setting
}
It may get optimized as
*ptr = 0 ;
while(0){
}
Volatile supress the compiler optimization and compiler assumes that tha value can
be changed at any time even if no explicit code modify it.
Volatile char *const ptr = (volatile char * )0x15;
Used when the object is a modified by ISR.
Sometimes ISR may change tha values used in the mainline codes
static int num;
void interrupt(void){
++num;
}
int main(){
int val;
val = num;
while(val != num)
val = num;
return val;
}
Here the compiler do some optimizations to the while statement.ie the compiler
produce the code it such a way that the value of num will always read form the cpu
registers instead of reading from the memory.The while statement will always be
false.But in actual scenario the valu of num may get changed in the ISR and it will
reflect in the memory.So if the variable is declared as volatile the compiler will know
that the value should always read from the memory
volatile means that variables value could be change any time by any external source. in GCC if we dont use volatile than it optimize the code which is sometimes gives unwanted behavior.
For example if we try to get real time from an external real time clock and if we don't use volatile there then what compiler do is it will always display the value which is stored in cpu register so it will not work the way we want. if we use volatile keyword there then every time it will read from the real time clock so it will serve our purpose....
But as u said you are not dealing with any low level hardware programming then i don't think you need to use volatile anywhere
thanks
volatile char* sevensegment_char_value;
void ss_load_char(volatile char *digits) {
...
int l=strlen(digits);
...
}
ss_load_char(sevensegment_char_value);
In the above example I've got warning from avr-gcc compiler
Warning 6 passing argument 1 of 'strlen' discards 'volatile' qualifier from pointer target type [enabled by default]
So I have to somehow copy the value from volatile to non-volatile var? What is the safe workaround?
There is no such thing like a "built in" Workaround in C. Volatile tells the compiler, that the contents of a variable (or in your case the memory the variable is pointing at) can change without the compiler noticing it and forces the compiler to read the data direct from the data bus rather than using a possibly existing copy in the registers.
Therefore the volatile keyword is used to avoid odd behaviour induced through compiler optimizations. (I can explain this further if you like)
In your case, you have a character buffer declared as volatile. If your program changes the contents of this buffer in a different context like an ISR for example, you have to implement sort of a synchronisation mechanism (like disabling the particular interrupt or so) to avoid inconsistency of data. After aquiring the "lock" (disabling the interrupt) you can copy the data byte by byte to a local (non-volatile) buffer and work on this buffer for the rest of the routine.
If the buffer will not change "outside" of the context of your read accesses I suggest to omit the volatile keyword as there is no use for it.
To judge the correct solution, a little bit more information about your exact use case would be needed.
Standard library routines aren't designed to work on volatile objects. The simplest solution is to read the volatile memory into normal memory before operating on it:
void ss_load_char(volatile char *digits) {
char buf[BUFSIZE];
int i = 0;
for (i = 0; i < BUFSIZE; ++i) {
buf[i] = digits[i];
}
int l=strlen(buf);
...
}
Here BUFSIZE is the size of the area of volatile memory.
Depending on how the volatile memory is configured, there may be routines you are supposed to call to copy out the contents, rather than just using a loop. Note that memcpy won't work as it is not designed to work with volatile memory.
The compiler warning only means that strlen() will not treat your pointer as volatile, i.e. it will maybe cache the pointer in a register when computing the length of your string. I guess, that's ok with you.
In general, volatile means that the compiler will not cache the variable. Look at this example:
extern int flag;
while (flag) { /* loop*/ }
This would loop forever if flag != 0, since the compiler assumes that flag is not changed "from the outside", like a different thread. If you want to wait on the input of some other thread, you must write this:
extern volatile int flag;
while (flag) { /* loop*/ }
Now, the compiler will really look at flag each time the loop loops. This may be must more what we intended in this example.
In answer to your question: if you know what you're doing, just cast the volatile away with int l=strlen((char*)digits).