Dynamic memory allocation on stack

Question

I recently tried this experiment in which instead of going for dynamic memory allocation for memory requirements of unknown size, I did a static allocation. When an array a[i] was declared by me, I kept i (size of the array) variable and dependent on the input that the user gives.

 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
 #include <conio.h>
 void function(int );
 int main(void)
 {
     int i;
     printf("Enter:");
     scanf("%d",&i);
     function(i);
     printf("i = %d\n",i);
     getch();
     return 0;
 }
 void function(int i)
 {
      char a[i];
      char b[4];
      strncpy(a,"hello",i);
      strcpy(b,"world");
      int j = 0;
      char *c = a;
      for( j = 0; j< 20; j++ )
           printf("%c",*c++);
 }

My questions are:

• Is such an operation legal?
• If no, why does the compiler not issue any warning or error?
• Where will this memory be allocated: Stack or heap?
• Why does ANSI C/GCC allow this?

Answer 1

Is such an operation legal?

It's called a variable length array.

VLAs are legal in ANSI C99 and as an extension to some pre-C99 compilers. GCC supports it both as strict C99 and as an extension to non-C99 code. It's also legal in C++0x.

If no, why does the compiler not issue any warning or error?

With gcc:

$ gcc -std=c89 src/vla.c  -Wall -ansi -pedantic
src/vla.c: In function ‘function’:, not dynamic array.
src/vla.c:17: warning: ISO C90 forbids variable length array ‘a’
src/vla.c:21: warning: ISO C90 forbids mixed declarations and code

The presence of 'conio.h' from MSDOS indicates you're probably using a Microsoft Visual C++ compiler, so don't worry about it. MS has worked to make their compiler more conformant to the C++0x standard, but makes no claims about how standard its C compiler mode is. You're asking why Spanish dialect words aren't in the French dictionary.

Where will this memory be allocated: Stack or heap?

It is an automatic object, so most C implementations will put in on the stack for efficiency reasons.

Why does ANSI C/GCC allow this

It is useful for creating temporary arrays of variable size at runtime whose lifetime doesn't extend beyond the function call.

Answer 2

This is valid C99.

Look here for a more detailed explanation in another StackOverflow question.

Answer 3

This is legal, but not all compilers support it. At least Visual Studio <= 2003 afaik do not support it.

I would assume it is not Ansi C++, try gcc -ansi -pedantic.