code cat

데비안 계열에서만 동작할려는지 모르겠는데...
어쨌든 다음과 같이 해서 현재 사용중인 java에 대한 설정을 확인 할 수 있다. (옵션은 소문자 'L' 이다 )

그러면 위와 같이 jdk 버전과 패스가 나온다.  이를 보고 필요에 따라   JAVA_HOME, JRE_HOME 을 설정하면 되겠다.

출처:https://developer.gnome.org/glib/unstable/glib-The-Main-Event-Loop.html

The GMainLoop data type represents a main event loop. 

A GMainLoop is created with g_main_loop_new(). After adding the initial event sources, g_main_loop_run() is called. This continuously checks for new events from each of the event sources and dispatches them. Finally, the processing of an event from one of the sources leads to a call to g_main_loop_quit() to exit the main loop, andg_main_loop_run() returns.

하면 간단하게 다음과 같이 나온다.

text/x-c; charset=utf-8

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