code cat

새로운 노드의 주소는 항상 같은 offset을 두고 생성될까? 적어도 주소의 절대값이 줄지는 않은 거 처럼 보이는데....

출처: C++ 에스프레소

생성자 호출 방법 중 실수할 수 있는 경우가 있다.

2번째의 경우 아무생각없이 하다가 쓸 수 있는 방법인데, 잘못된 방법이다.

error: aggregate value used where an integer was expected

위와 같은 에러를 본다면, 아마도 구조조체의 멤버 대신 구조체를 직접 integer로 캐스팅하려는 실수를 했을 때 나오는 것이다.

엑셀은 포스팅 안하는데 아래팁은 좀 많이 편해서...

우선 원하는 셀에 가서 초기값을 x 만큼 입력한다.

그리고 마지막 값의 위치(A3)에 가서

=IF(A3=A1, A3+1, A3)

이라고 입력 뒤, 쭈욱 드래그 하면 된다.

시간날때 마져 다 읽어볼려고 남기는데, 재밌다.  물론 나도 C에는 call-by-reference는 없다고 생각하는 사람이지만, 읽어보면 재밌다.

https://kldp.org/node/119844

sys-filesystem을 설치하는데 에러가 났다.

이건 다음과 같이 해결하면 된다.

이후 다시 gem install sys-filesystem을 하면 잘 된다.

출처: http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc

In this question, someone suggested in a comment that I should not cast the results of malloc, i.e:

int *sieve = malloc(sizeof(int)*length);

rather than:

int *sieve = (int *)malloc(sizeof(int)*length);

Why would this be the case?

No; you don't cast the result, since:

• It is unnecessary, as void * is automatically and safely promoted to any other pointer type in this case.
• It can hide an error, if you forgot to include <stdlib.h>. This can cause crashes, in the worst case.
• It adds clutter to the code, casts are not very easy to read (especially if the pointer type is long).
• It makes you repeat yourself, which is generally bad.

As a clarification, note that I said "you don't cast", not "you don't need to cast". In my opinion, it's a failure to include the cast, even if you got it right. There are simply no benefits to doing it, but a bunch of potential risks, and including the cast indicates that you don't know about the risks.

Also note, as commentators point out, that the above changes for straight C, not C++. I very firmly believe in C and C++ as separate languages.

To add further, your code needlessly repeats the type information (int) which can cause errors. It's better to dereference the pointer being used to store the return value, to "lock" the two together:

int *sieve = malloc(length * sizeof *sieve);

This also moves the length to the front for increased visibility, and drops the redundant parentheses with sizeof; they are only needed when the argument is a type name. Many people seem to not know (or ignore) this, which makes their code more verbose. Remember: sizeof is not a function! :)

@sirgeorge - without the prototype/decl (in stdlib.h), a C compiler might/will assume that malloc returns an int. Then the compiler generates code for the call in question as if it is getting an int. Then, say, you are compiling for a platform where pointers and ints do not have the same size, and you cast the returning int into a pointer. Bad juju. And the compiler, upon seeing the casting, won't give you a warning in this case. But if you don't cast it, then the compiler will complain abou the int-to-ptr assignment (and you will be able to detect the missing include for stdlib.h) –  luis.espinal Mar 14 '12 at 19:28