Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
1.1k views
in Technique[技术] by (71.8m points)

c - Why are the contents pointed to by a pointer not changed when memory is deallocated using free()?

I am a newbie when it comes to dynamic memory allocation. When we free the memory using void free(void *ptr) the memory is deallocated but the contents of the pointer are not deleted. Why is that? Is there any difference in more recent C compilers?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Computers don't "delete" memory as such, they just stop using all references to that memory cell and forget that anything of value is stored there. For example:

int* func (void)
{
  int x = 5;
  return &x;
}

printf("%d", *func()); // undefined behavior

Once the function has finished, the program stops reserving the memory location where x is stored, any other part of the program (or perhaps another program) is free to use it. So the above code could print 5, or it could print garbage, or it could even crash the program: referencing the contents of a memory cell that has ceased to be valid is undefined behavior.

Dynamic memory is no exception to this and works in the same manner. Once you have called free(), the contents of that part of the memory can be used by anyone.

Also, see this question.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...