Why are structures copied via memcpy in embedded system code? - c

Question

In embedded software domain for copying structure of same type people don't use direct assignment and do that by memcpy() function or each element copying.
lets have for example
struct tag
{
int a;
int b;
};
struct tag exmple1 = {10,20};
struct tag exmple2;
for copying exmple1 into exmple2..
instead of writing direct
exmple2=exmple1;
people use
memcpy(exmple2,exmple1,sizeof(struct tag));
or
exmple2.a=exmple1.a;
exmple2.b=exmple1.b;
why ????

Answer 1

One way or the other there is nothing specific about embedded systems that makes this dangerous, the language semantics are identical for all platforms.
C has been used in embedded systems for many years, and early C compilers, before ANSI/ISO standardisation did not support direct structure assignment. Many practitioners are either from that era, or have been taught by those that were, or are using legacy code written by such practitioners. This is probably the root of the doubt, but it is not a problem on an ISO compliant implementation. On some very resource constrained targets, the available compiler may not be fully ISO compliant for a number of reasons, but I doubt that this feature would be affected.
One issue (that applies to embedded and non-embedded alike), is that when assigning a structure, an implementation need not duplicate the value of any undefined padding bits, therefore if you performed a structure assignment, and then performed a memcmp() rather than member-by-member comparison to test for equality, there is no guarantee that they will be equal. However if you perform a memcpy(), any padding bits will be copied so that memcmp() and member-by-member comparison will yield equality.
So it is arguably safer to use memcpy() in all cases (not just embedded), but the improvement is marginal, and not conducive to readability. It would be a strange implementation that did not use the simplest method of structure assignment, and that is a simple memcpy(), so it is unlikely that the theoretical mismatch would occur.

Answer 2

In your given code there is no problem even if you write:
example2 = example1;
But just assume if in future, the struct definition changes to:
struct tag
{
int a[1000];
int b;
};
Now if you execute the assignment operator as above then (some of the) compiler might inline the code for byte by byte (or int by int) copying. i.e.
example1.a[0] = example.a[0];
example1.a[1] = example.a[1];
example1.a[2] = example.a[2];
...
which will result in code bloat in your code segment. Such kind of memory errors are not trivial to find. That's why people use memcpy.
[However, I have heard that modern compilers are capable enough to use memcpy internally when such instruction is encountered especially for PODs.]

Answer 3

Copying C-structures via memcpy() is often used by programmers who learned C decades ago and did not follow the standardization process since. They simple don't know that C supports assignment of structures (direct structure assignment was not available in all pre-ANSI-C89 compilers).
When they learn about this feature some still stick to the memcpy() way because it is their custom. There are also motivations that originate in cargo cult programming, e.g. it is claimed that memcpy is just faster - of course - without being able to back this up with a benchmark test case.
Structures are also memcpy()ied by some newbie programmers because they either confuse structure assignment with the assignment of a pointer of a structure - or they simply overuse memcpy() (they often also use memcpy() where strcpy() would be more appropriate).
There is also the memcmp() structure comparison anti-pattern that is sometimes cited by some programmers for using memcpy() instead of structure assignment. The reasoning behind this is the following: since C does not automatically generate a == operator for structures and writing a custom structure comparison function is tedious, memcmp() is used to compare structures. In the next step - to avoid differences in the padding bits of compared structures - memset(...,0,...) is used to initialize all structures (instead of using the C99 initializer syntax or initializing all fields separately) and memcpy() is used to copy the structures! Because memcpy() also copies the content of the padding bits ...
But note that this reasoning is flawed for several reasons:
the use of memcpy()/memcmp()/memset() introduce new error possibilities - e.g. supplying a wrong size
when the structure contains integer fields the ordering under memcmp() changes between big- and little-endian architectures
a char array field of size n that is 0-terminated at position x must also have all elements after position x zeroed out at any time - else 2 otherwise equal structs compare unequal
assignment from a register to a field may also set the neighbouring padding bits to values unequal 0, thus, following comparisons with otherwise equal structures yield an unequal result
The last point is best illustrated with a small example (assuming architecture X):
struct S {
int a; // on X: sizeof(int) == 4
char b; // on X: 24 padding bits are inserted after b
int c;
};
typedef struct S S;
S s1;
memset(&s1, 0, sizeof(S));
s1.a = 0;
s1.b = 'a';
s1.c = 0;
S s2;
memcpy(&s2, &s1, sizeof(S));
assert(memcmp(&s1, &s2, sizeof(S)==0); // assertion is always true
s2.b = 'x';
assert(memcmp(&s1, &s2, sizeof(S)!=0); // assertion is always true
// some computation
char x = 'x'; // on X: 'x' is stored in a 32 bit register
// as least significant byte
// the other bytes contain previous data
s1.b = x; // the complete register is copied
// i.e. the higher 3 register bytes are the new
// padding bits in s1
assert(memcmp(&s1, &s2, sizeof(S)==0); // assertion is not always true
The failure of the last assertion may depend on code reordering, change of the compiler, change of compiler options and stuff like that.
Conclusion
As a general rule: to increase code correctness and portability use direct struct assignment (instead of memcpy()), C99 struct initialization syntax (instead of memset) and a custom comparison function (instead of memcmp()).

Answer 4

In C people probably do that, because they think that memcpy would be faster. But I don't think that is true. Compiler optimizations would take care of that.
In C++ it may also have different semantics because of user defined assignment operator and copy constructors.

Answer 5

On top of what the others wrote some additional points:
Using memcpy instead of a simple assignment gives a hint to someone who maintains the code that the operation might be expensive. Using memcpy in these cases will improves the understanding of the code.
Embedded systems are often written with portability and performance in mind. Portability is important because you may want to re-use your code even if the CPU in the original design is not available or if a cheaper micro-controller can do the same job.
These days low-end micro-controllers come and go faster than the compiler developers can catch up, so it is not uncommon to work with compilers that use a simple byte-copy loop instead of something optimized for structure assignments. With the move to 32 bit ARM cores this is not true for a large part of embedded developers. There are however a lot of people out there who build products that target obscure 8 and 16 bit micro-controllers.
A memcpy tuned for a specific platform may be more optimal than what a compiler can generate. For example on embedded platforms having structures in flash memory is common. Reading from flash is not as slow as writing to it, but it is still a lot slower than a ordinary copy from RAM to RAM. A optimized memcpy function may use DMA or special features from the flash controller to speed up the copy process.

Answer 6

That is a complete nonsense. Use whichever way you prefer. The simplest is :
exmple2=exmple1;

Answer 7

Whatever you do, don't do this:
exmple2.a=exmple1.a;
exmple2.b=exmple1.b;
It poses a maintainability problem because any time that anyone adds a member to the structure, they have to add a line of code to do the copy of that member. Someone is going to forget to do that and it will cause a hard to find bug.

Answer 8

On some implementations, the way in which memcpy() is performed may differ from the way in which "normal" structure assignment would be performed, in a manner that may be important in some narrow contexts. For example, one or the other structure operand may be unaligned and the compiler might not know about it (e.g. one memory region might have external linkage and be defined in a module written in a different language that has no means of enforcing alignment). Use of a __packed declaration would be better if a compiler supported such, but not all compilers do.
Another reason for using something other than structure assignment could be that a particular implementation's memcpy might access its operands in a sequence that would work correctly with certain kinds of volatile source or destination, while that implementation's struct assignment might use a different sequence that wouldn't work. This is generally not a good reason to use memcpy, however, since aside from the alignment issue (which memcpy is required to handle correctly in any case) the specifications for memcpy don't promise much about how the operation will be performed. It would be better to use a specially-written routine which performed the operations exactly as required (for example, if the target is a piece of hardware which needs to have 4 bytes of structure data written using four 8-bit writes rather than one 32-bit writes, one should write a routine which does that, rather than hoping that no future version of memcpy decides to "optimize" the operation).
A third reason for using memcpy in some cases would be the fact that compilers will often perform small structure assignments using a direct sequence of loads and stores, rather than using a library routine. On some controllers, the amount of code this requires may vary depending upon where the structures are located in memory, to the point that the load/store sequence may end up being bigger than a memcpy call. For example, on a PICmicro controller with 1Kwords of code space and 192 bytes of RAM, coping a 4-byte structure from bank 1 to bank 0 would take 16 instructions. A memcpy call would take eight or nine (depending upon whether count is an unsigned char or int [with only 192 bytes of RAM total, unsigned char should be more than sufficient!] Note, however, that calling a memcpy-ish routine which assumed a hard-coded size and required both operands be in RAM rather than code space would only require five instructions to call, and that could be reduced to four with the use of a global variable.

Answer 9

first version is perfect.
second one may be used for speed (there is no reason for your size).
3rd one is used only if padding is different for target and source.