Try it on godbolt. From what I understand it would not violate the strict aliasing rule, since a pointer to a struct is always equal to a pointer to its first member, and in this case the first member is a char. Difference between "2-year community or junior colleges" and "4 year undergraduate program" in USA. 사용법 #include void *memcpy(void *dest, const void *src, size_t n); 정의 memcpy() 함수는 src 메모리 영역에서 dest 메모리 영역으로 n byte 만큼 복사합니다.

Has anyone tested the effect of allowing cantrips to be repeatedly cast between battles? What's the verdicts on hub-less circle bicycle wheels? 뭔가 위의 실험들이 강한 삽질의 기운으로 다가오지만 이쯤에서 마무리를 지으면서~! 10 :      memset(src,69,sizeof(int)*SIZE); //src에 내용 채움 What are “size” and the returned value? C 2018 6.3.2.3 7 says any pointer to an object type may be converted to any other pointer to an object type as long as the alignment is correct, but 6.5 7 says an object shall be accessed only by lvalue expressions with certain types.

This is incorrect because the rule in 6.5 7 does not say that, if you can access something using type X, and type Y is compatible with X or is one of the other cases listed with respect to X, then you can access the thing with type Y. Making statements based on opinion; back them up with references or personal experience. Podcast 286: If you could fix any software, what would you change? 기능 src 번지의 데이터를 dest로 size n 만큼 복사 3. Elixir Cross Referencer - Explore source code in your browser - Particularly useful for the Linux kernel and other low-level projects in C/C++ (bootloaders, C libraries...) Live … Feedback for The Loop, August 2020: Community-a … 9 :   Whatever type your DST or SRC is, it is presumably not a struct tmp. The C standard states rules for the effective type of objects. In case you are worried about any possible padding added by a hypothetical alien compiler, adding a _Static_assert() will make the macro very safe: Discussion about the legitimacy of the code. c++ / 6. memcpy에 대해서 실험해봤습니다. Another problem is that a C implementation is allowed to include padding at the end of an array, regardless of whether it is needed for alignment. copy_to_user() 함수는 커널과 어플의 메모리 복사 함수입니다. In a comment moved to chat, the OP argued “If I can access a series of adjacent bytes as if they were of type char, I can access all of them collectively as if they were of type char []. To learn more, see our tips on writing great answers. Is it safe to cast an array to a struct with one array as member? How can I seal a gap between floor joist boxes and foundation? Any way to watch Netflix on an 1stGen iPad Air (MD788LL/A)? Mathematica integrates too well using the "code" I wrote. This is incorrect. Can a druid use Wild Shape in mid-air to survive being dropped? Sum of digits of sum of digits of sum of digits. See 6.7.2.1p15: A pointer to a structure object, suitably converted, points to its initial member (or if that member is a bit-field, then to the unit in which it resides), and vice versa. Georgia doing "hand recount" of 2020 Presidential Election Ballots. Here the new versions of the macros that will work also with optimizations enabled: I'm not enough of a language lawyer to be able to opine on whether the code is correct as far as strict aliasing.

아, 이걸 정리하다가 혹시나 해서 본 memcpy의 설명에 이런 문구가 있네요. (For convenience herein, “may” or “may not” refers to whether the behavior of the action is or is not defined by the C standard. ptr로 시작하는 메모리 주소로부터 num 개의 바이트를 value 값으로 채운다. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy.

It is enough to tell explicitly the compiler about our aliasing by adding a simple *((char *) DST) = 0 before copying the data to DST. Is my Homebrew Born-Lycanthrope Race balanced with other playable races? 7 :        int *src = (int*)malloc(sizeof(int)*SIZE); //메모리 할당

Browse other questions tagged c linux memcpy or ask your own question.

16 :      free(src); “Edit #3” further argues that “it is also legal to map collectively all the bytes above to a single char array, since we are mapping adjacent bytes.” At this point, “Edit #3” presents code defining a pointer to an array of the bytes.

11 :      memcpy(dest,src,sizeof(int)*SIZE); //dest로 복사, 12 :      for(i=0;i
For dynamically allocated objects (which have no declared type; they are created by allocating memory and assigning values to them through pointers), you can change its effective type by assigning a value to it using any lvalue other than a character type (6.5 6). 아쉽게도 s3c2440 에서는 메모리 to 메모리 dma 가 없는 것으로 알고 있습니다.

However, two problems remain: Thanks for contributing an answer to Stack Overflow! Why is "hand recount" better than "computer rescan"?

2 : #include

memcpy나 mem의 함수에 대해서 많이 모르기 때문에 설명서를 읽고 실습을 해보기록 했습니다. 댓글 However, in order to be a compatible type, the two arrays—the one in struct tmp and the one that is SRC or DST—must be the same size. Further, the rules for pointer conversion do not say that converting a pointer to char to a pointer to an array of char results in a pointer value that actually points to the memory where the array of char would be. sizeof(int)만큼 복사를 하게되면 단순히 2만 씌워진 후에 끝났을 것입니다. If that was not the case, we could not do char * newmem = malloc(1000); newmem[whatever] = 'a';, since we would be illegitimately casting a char * to a char [].”. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. 첫번째 실험은 int에 값을 준 후, array에 한번은 부족하게, 다른 한번은 넉넉하게 값을 준 경우입니다. But I was wondering, how legitimate would this code be for implementing memcpy() for a bare-metal environment? For example, if a function is passed a float * and an int *, and it copies elements from one into the other, the aliasing rule allows the compiler to conclude the arrays they point to do not overlap, and therefore it can optimize the loop by copying several elements at a time. 3 : #include , 6 :        int *dest = (int*)malloc(sizeof(int)*SIZE); //메모리 할당 So clearly the fact that a pointer may be converted does not mean the pointed-to object may be accessed with the new type.

17 :      return 0; It says an object shall have its stored value accessed only by certain lvalue expressions. 0, 댓글 Compiling an application for use in highly radioactive environments. 이 때 value는 unsigned char 로 … 결과는 아래와 같이 나왔는데, b의 값을 a에 복사할 때 스택에 담긴 대로 메모리를 침범(? 이번엔 int[2]와 int[3]을 두고, int[3]에 값을 할당한 다음 int[2]에 넣는 실험으로 코드는 아래와 같습니다. What did Pete Stewart think he knew about efficient implementation of floating point denormals?

; 회원 가입은 주민등록번호가 필요 없으며, 메일 주소만 있으면 간단하게 가입하실 수 있습니다. Stack Overflow for Teams is a private, secure spot for you and Job offers - how to negotiate higher salary due to higher costs of living at the new location. Furthermore, 6.5 7 is phrased as an absolute prohibition. Featured on Meta Thank you, Geoff. Of course the destination still needs to have enough memory to hold the new string. This is more of a curiosity than anything else. I have added to the question a discussion about the legitimacy of the code. memcpy는 복사를 하는 함수인데, 단순히 드는 궁금증. memcpy_fromio(&buf[6], &priv->rsp[6], expected - 6); which really should never have worked in the first place, but back before commit 170d13c it *happened* to work, because the memcpy_fromio() would be expanded to a regular memcpy, and (a) gcc would expand the first memcpy in-line, and turn it into a 4-byte and a 2-byte read, and they happened to be in the right order, and the alignment was right. Does a bronze dragon's wing attack work underwater? Why are red and blue light refracted differently if they travel at the same speed in the same medium? 리턴값 dest의 포인터 4.

And in case of very very stupid compilers we would still have the, @madmurphy: For instance, maybe you have a complicated sequence of operations on, It is possible that there are still some risks by using a macro. The Loop, August 2020: Community-a-thon. 따라서 memcpy를 하게 되면 주어진 범위만큼을 메모리 카피로 채워버리게 되는 것입니다. In other words 6.5 7 is not transitive. However, with gcc -O2 on x86-64, it instead outputs 12. So the valued that was copied into a[0] is overwritten by its old value plus 2, instead of being incremented. A purpose of C aliasing rule is to allow a compiler to conclude that accesses to two different types of things access different memory. If that rule is violated, the behavior is undefined, and that is conclusive. 함수원형 void *memcpy(void *dest, const void *src, size_t n); 2. “EDIT #3” argues that any object can be written using a char *. One of those special cases may save you if DST or SRC is an array of char.

rev 2020.11.13.38000, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide, Comments are not for extended discussion; this conversation has been, But why should it? There is no basis for this inference in the C standard; no rule in the C standard says that the bytes of an object in memory, which may be accessed individually via a character lvalue, may be treated as an array of characters. What is this tool called and what is it used for?

두번째 실험은 array에 값을 가득 채운 후, int에 값을 주는 경우입니다. Literally, a program “may” try essentially anything but, for this discussion, we are only interested in what the C standard defines.). Clone existing structs with different alignment in Visual C++.

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