Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

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

How C strings are allocated in memory?

Say I have a simple function that returns a C string this way:

const char * getString()
{
  const char * ptr = "blah blah";
  return ptr; 
}

and I call getString() from main() this way:

  const char * s = getString();

1) According to gdb, the variable ptr is stored on the stack, but the string pointed by ptr is not:

(gdb) p &ptr
$1 = (const char **) 0x7fffffffe688

(gdb) p ptr
$2 = 0x4009fc "blah blah"

Does this mean that "blah blah" is not a local variable inside getString()?

I guess that if it were a local variable, I would not be able to pass it to my main() function... But if it's not, where is it stored? On the heap? Is that a "kind of" dynamically memory allocation implemented by the OS every time it hits on a string, or what?

2) If I use an array instead of a pointer, this way:

const char *getString2()
{
  const char a[] = "blah blah blah";
  return a;
}

the compiler warns me that:

warning: address of local variable ‘a’ returned

(and of course the program compiles, but it doesn't work).

Actually, if I ask gdb, I get

(gdb) p &a
$2 = (const char (*)[15]) 0x7fffffffe690

But I thought that const char * ptr and const char a[] were basically the same thing. Looks like they're not.

Am I wrong? What is exactely the difference between the two versions?

Thank you!

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 Answer

0 votes
by (71.8m points)

When you write

const char *ptr = "blah blah";

then the following happens: the compiler generates a constant string (of type char []) with the contents "blah blah" and stores it somewhere in the data segment of the executable (it basically has a similar storage duration to that of variables declared using the static keyword).

Then, the address of this string, which is valid throughout the lifetime of the program, is stored in the ptr pointer, which is then returned. All is fine.

Does this mean that "blah blah" is not a local variable inside getString()?

Let me respond with a broken English sentence: yes, it isn't.

However, when you declare an array, as in

const char a[] = "blah blah";

then the compiler doesn't generate a static string. (Indeed, this is a somewhat special case when initializing strings.) It then generates code that will allocate a big enough piece of stack memory for the a array (it's not a pointer!) and will fill it with the bytes of the string. Here a is actually a local variable and returning its address results in undefined behavior.

So...

But I thought that const char *ptr and const char a[] were basically the same thing.

No, not at all, because arrays are not pointers.


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 OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...