Memory Management : scope and local pointer variable

Question

Q. In terms of Memory Management, What is the error in the following code?

 char* secret_message()
 {
   char message_buffer[100];
   char* text = "Hey man!";
   int n = 0;
   while (text[n] != '\0')
     n++;
   for (int i = 0; i <= n ; i++)
     message_buffer[i] = text[i];
   return message_buffer;
 }

Answer. I think message_buffer is local variable that is automatically reclaimed after function ends. This function returns a reference to an invalid memory location , since message_buffer disappears right after return statement.

Is it correct?

Please let me know. Thanks,

Yes, it is just as you assume. – Nbr44 Apr 23 '13 at 02:36 — Nbr44, Apr 23 '13 at 02:36

score 2 · Answer 1 · answered Apr 23 '13 at 02:36

Answer. I think message_buffer is local variable that is automatically reclaimed after function ends. This function returns a reference to an invalid memory location , since message_buffer disappears right after return statement.

Yes, it is correct. message_buffer is allocated on stack, its memory will be deallocated when the function exits. the pointer will points to release memory.

BTW:

char* text = "Hey man!";

should be

const char* text = "Hey man!";

in modern C++.

score 2 · Answer 2 · answered May 02 '13 at 09:17

2

message_buffer is automatic variable whose scope is within the function only. this variable either should be declared in main function or declared as static variable within the function.

answered May 02 '13 at 09:17

Narendra Bista

279
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6

score 0 · Answer 3 · answered Apr 23 '13 at 02:38

0

You are correct. This produces undefined behaviour.

answered Apr 23 '13 at 02:38

Muscles

471
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Memory Management : scope and local pointer variable

3 Answers3