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Is `memcpy((void *)dest, src, n)` with a `volatile` array safe?

I have a buffer that I use for UART, which is declared this way:

union   Eusart_Buff {
    uint8_t     b8[16];
    uint16_t    b9[16];
};

struct  Eusart_Msg {
    uint8_t             msg_posn;
    uint8_t             msg_len;
    union Eusart_Buff   buff;
};

struct  Eusart {
    struct Eusart_Msg   tx;
    struct Eusart_Msg   rx;
};

extern  volatile    struct Eusart   eusart;

And here is the function that fills the buffer (which will be sent using interrupts):

void    eusart_msg_transmit (uint8_t n, void *msg)
{

    if (!n)
        return;

    /*
     * The end of the previous transmission will reset
     * eusart.tx.msg_len (i.e. ISR is off)
     */
    while (eusart.tx.msg_len)
        ;

    if (data_9b) {
        memcpy((void *)eusart.tx.buff.b9, msg,
                sizeof(eusart.tx.buff.b9[0]) * n);
    } else {
        memcpy((void *)eusart.tx.buff.b8, msg,
                sizeof(eusart.tx.buff.b8[0]) * n);
    }
    eusart.tx.msg_len   = n;
    eusart.tx.msg_posn  = 0;

    reg_PIE1_TXIE_write(true);
}

At the moment of using memcpy(), I know no one else is going to use the buffer (atomic), because the while loop ensures that the last message has been sent, and therefore the interrupt is disabled.

Is it safe to cast away volatile this way so that I am able to use memcpy() or should I make a function maybe called memcpy_v() like these to be safe?:

void *memcpy_vin(void *dest, const volatile void *src, size_t n)
{
    const volatile char *src_c  = (const volatile char *)src;
    char *dest_c                = (char *)dest;

    for (size_t i = 0; i < n; i++)
        dest_c[i]   = src_c[i];

    return  dest;
}

volatile void *memcpy_vout(volatile void *dest, const void *src, size_t n)
{
    const char *src_c       = (const char *)src;
    volatile char *dest_c   = (volatile char *)dest;

    for (size_t i = 0; i < n; i++)
        dest_c[i]   = src_c[i];

    return  dest;
}

volatile void *memcpy_v(volatile void *dest, const volatile void *src, size_t n)
{
    const volatile char *src_c  = (const volatile char *)src;
    volatile char *dest_c       = (volatile char *)dest;

    for (size_t i = 0; i < n; i++)
        dest_c[i]   = src_c[i];

    return  dest;
}

Edit:

If I need those new functions, given that I know no one is going to modify the array at the same time, would it make sense to use restrict to (maybe) help the compiler optimize (if it can)? Possibly this way (correct me if I'm wrong):

volatile void *memcpy_v(restrict volatile void *dest,
                        const restrict volatile void *src,
                        size_t n)
{
    const restrict volatile char *src_c = src;
    restrict volatile char *dest_c      = dest;

    for (size_t i = 0; i < n; i++)
        dest_c[i]   = src_c[i];

    return  dest;
}

Edit 2 (add context):

void    eusart_end_transmission (void)
{

    reg_PIE1_TXIE_write(false); /* TXIE is TX interrupt enable */
    eusart.tx.msg_len   = 0;
    eusart.tx.msg_posn  = 0;
}

void    eusart_tx_send_next_c   (void)
{
    uint16_t    tmp;

    if (data_9b) {
        tmp     = eusart.tx.buff.b9[eusart.tx.msg_posn++];
        reg_TXSTA_TX9D_write(tmp >> 8);
        TXREG   = tmp;
    } else {
        TXREG   = eusart.tx.buff.b8[eusart.tx.msg_posn++];
    }
}

void __interrupt()  isr(void)
{

    if (reg_PIR1_TXIF_read()) {
        if (eusart.tx.msg_posn >= eusart.tx.msg_len)
            eusart_end_transmission();
        else
            eusart_tx_send_next_c();
    }
}

Although volatile may not be is needed (I asked it in another question: volatile for variable that is only read in ISR?), this question still should be answered in the assumption that volatile is needed so that future users that really need volatile (for example me when I implement the RX buffer), can know what to do.


EDIT (Related) (Jul/19):

volatile vs memory barrier for interrupts

Basically says that volatile is not needed, and therefore this issue disappears.

like image 303
alx Avatar asked Mar 02 '19 23:03

alx


3 Answers

Is memcpy((void *)dest, src, n) with a volatile array safe?

No. In the general case, memcpy() is not specified to work correctly with volatile memory.
OP's case looks OK to cast away volatile, yet posted code is insufficient to be certain.

If code wants to memcpy() volatile memory, write the helper function.

OP's code has restrict in the wrong place. Suggest

volatile void *memcpy_v(volatile void *restrict dest,
            const volatile void *restrict src, size_t n) {
    const volatile unsigned char *src_c = src;
    volatile unsigned char *dest_c      = dest;

    while (n > 0) {
        n--;
        dest_c[n] = src_c[n];
    }
    return  dest;
}

A singular reason for writing your own memcpy_v() is the a compiler can "understand"/analyze memcpy() and emit code that is very different than expected - even optimize it out, if the compiler thinks the copy is not needed. Remind oneself that the compiler thinks memcpy() manipulated memory is non-volatile.


Yet OP is using volatile struct Eusart eusart; incorrectly. Access to eusart needs protection that volatile does not provide.

In OP's case, code can drop volatile on the buffers and then use memcpy() just fine.

A remaining issue is in the scant code of how OP is using eusart. Using volatile does not solve OP's problem there. OP's does assert "I write to it atomically,", yet without posted atomic code, that is not certain.


Code like the below benefits with eusart.tx.msg_len being volatile, yet that is not sufficient. volatile insures .tx.msg_len is not cached and instead re-reads each time.

while (eusart.tx.msg_len)
    ;

Yet the read of .tx.msg_len is not specified as atomic. When .tx.msg_len == 256 and a ISR fires, decrementing .tx.msg_len, the read of the the LSbyte (0 from 256) and MSbyte (0 from 255), the non-ISR code may see .tx.msg_len as 0, not 255 nor 256, thus ending the loop at the wrong time. The access of .tx.msg_len needs to be specified as indivisible (atomic), else, once in a while code fails mysteriously.

while (eusart.tx.msg_len); also suffers from being an end-less loop. Should the transmission stop for some reason other than empty, the while loop never exits.

Recommend instead to block interrupts while inspecting or changing eusart.tx.msg_len, eusart.tx.msg_posn. Review your compilers support of atomic or

size_t tx_msg_len(void) {
  // pseudo-code
  interrupt_enable_state = get_state();
  disable_interrupts();
  size_t len = eusart.tx.msg_len;
  restore_state(interrupt_enable_state);
  return len;
}

General communication code ideas:

  1. While non-ISR code reads or writes eusart, make sure the ISR cannot ever change eusart.

  2. Do not block ISR for long in step #1.

  3. Do not assume underlying ISR() will chain input/output successfully without a hiccup. Top level code should be prepared to re-start output if it gets stalled.

like image 59
chux - Reinstate Monica Avatar answered Nov 10 '22 18:11

chux - Reinstate Monica


The Standard lacks any means by which programmers can demand that operations that access a region of storage by means of an ordinary pointer are completed before a particular volatile pointer access is performed, and also lacks any means of ensuring that operations which access a region of storage by means of an ordinary pointer are not carried out until after some particular volatile pointer access is performed. Since the semantics of volatile operations are Implementation-Defined, the authors of the Standard may have expected that compiler writers would recognize when their customers might need such semantics, and specify their behavior in a fashion consistent with those needs. Unfortunately, that hasn't happened.

Achieving the semantics you require will either making use of a "popular extension", such as the -fms-volatile mode of clang, a compiler-specific intrinsic, or else replacing memcpy with something that's so horribly inefficient as to swamp any supposed advantage compilers could gain by not supporting such semantics.

like image 40
supercat Avatar answered Nov 10 '22 19:11

supercat


The C17 standard says this in section 6.7.3 "Type qualifiers":

If an attempt is made to refer to an object defined with a volatile-qualified type through use of an lvalue with non-volatile-qualified type, the behavior is undefined.135)

135)This applies to those objects that behave as if they were defined with qualified types, even if they are never actually defined as objects in the program (such as an object at a memory-mapped input/output address).

So no, that's not safe.

like image 41
Joseph Sible-Reinstate Monica Avatar answered Nov 10 '22 17:11

Joseph Sible-Reinstate Monica