The code is as follow :
#include <stdlib.h>
int num = 3; // Static external variable
int *ptr = #
int main(void)
{
int num2 = 4; // Automatic variable
int *ptr2 = &num2;
free(ptr); //Free static variable
free(ptr2); //Free automatic variable
return 0;
}
I try to compile the above code and it works, I'm curious does the free() function able to free both the static variable and also automatic variable? Or basically it does nothing?
Calling free() on a pointer not returned by memory allocating functions(malloc,calloc etc) causes Undefined Behavior.
Your code has an Undefined Behavior, So the compiler does not need to give you any diagnostic of it and it can show any behavior, it might work, or crash, or literally do anything.
Just avoid writing code which causes an Undefined Behavior is the only solution.
You shouldn't do that. free is only used for memory dynamically allocated from heap by malloc family of functions.
Memory for num is statically allocated in data segment and can't be released. Memory for num2 is allocated in the main's call stack frame and will be released as soon as main returns.
What actually happens depend on implementation of free. There are usually specific data structures maintained in heap to help malloc/free track the allocated and free memory areas. free expects these data structures to be somewhere around the place its argument points to. An when you pass it a pointer which doesn't point to a malloc-allocated heap area, it'll consider garbage data as some useful information and do some strange things. And you're lucky if the result is just an immediate program crash.
Related
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
char* test() {
char* s = "Hello World";
size_t len = strlen(s);
char* t = malloc(sizeof(char)*(len+1));
strcpy(t, s);
free(t);
return t;
};
int main(void) {
printf("%s\n", test());
return 0;
};
I would like to allocate and de-allocate memory inside the function. I tested this code and works, but I am wondering:
Why does this work?
Is it good practice to use the value of a freed pointer in main ?
Once you call free on a pointer, the memory it pointed to is no longer valid. Attempting to use a pointer to freed memory triggers undefined behavior. In this particular case it happened to work, but there's no guarantee of that.
If the function returns allocated memory, it is the responsibility of the caller to free it:
char* test() {
char* s = "Hello World";
size_t len = strlen(s);
char* t = malloc(sizeof(char)*(len+1));
strcpy(t, s);
return t;
};
int main(void) {
char *t = test();
printf("%s\n", t);
free(t);
return 0;
};
malloc reserves memory for use.
free releases that reservation. In general, it does not make the memory go away, it does not change the contents of that memory, and it does not alter the value of the pointer that held the address.
After free(t), the bytes of t still contain the same bit settings they did before the free. Then return t; returns those bits to the caller.
When main passes those bits to printf, printf uses them as the address to get the characters for %s. Since nothing has changed them, they are printed.
That is why you got the behavior you did with this program. However, none of it is guaranteed. Once free was called with t, the memory reservation was gone. Something else in your program could have used that memory. For example, printf might have allocated a buffer for its own internal use, and that could have used the same memory.
For the most part, malloc and free are just methods of coordinating use of memory, so that different parts of your program do not try to use the same memory at the same time for different purposes. When you only have one part of your program using allocated memory, there are no other parts of your program to interfere with that. So the lack of coordination did not cause your program to fail. If you had multiple routines in your program using allocated memory, then attempting to use memory after it has been released is more likely to encounter problems.
Additionally, once the memory has been freed, the compiler may treat a pointer to it as if it has no fixed value. The return t; statement is not required to return any particular value.
It doesn't matter where do you free() a pointer. Once it is free()d, the pointer is not deferrenciable anymore (neither inside nor ouside the function where it was free()d)
The purpose of free() is to return the memory allocated with malloc() so the semantics are that, once you have freed a chunk of memory, it is not anymore usable.
In C, all parameters are passed by value, so free() cannot change the value expression you passed to it, and this is the reason the pointer is not changed into an invalid pointer value (like NULL) but you are advised that no more uses of the pointer can be done without incurring in Undefined Behaviour.
There could be a solution in the design of free() and it is to pass the pointer variable that holds the pointer by address, and so free() would be able to turn the pointer into a NULL. But this not only takes more work to do, but free() doesn't know how many copies you have made of the value malloc() gave to you... so it is impossible to know how many references you have over there to be nullified. That approach makes it impossible to give free() the responsibility of nullifying the reference to the returned memory.
So, if you think that free doesn't turn the pointer into NULL and for some strange reason you can still use the memory returned, don't do it anymore, because you'll be making mistakes.
You are adviced! :)
The code is as follow :
#include <stdlib.h>
int num = 3; // Static external variable
int *ptr = #
int main(void)
{
int num2 = 4; // Automatic variable
int *ptr2 = &num2;
free(ptr); //Free static variable
free(ptr2); //Free automatic variable
return 0;
}
I try to compile the above code and it works, I'm curious does the free() function able to free both the static variable and also automatic variable? Or basically it does nothing?
Calling free() on a pointer not returned by memory allocating functions(malloc,calloc etc) causes Undefined Behavior.
Your code has an Undefined Behavior, So the compiler does not need to give you any diagnostic of it and it can show any behavior, it might work, or crash, or literally do anything.
Just avoid writing code which causes an Undefined Behavior is the only solution.
You shouldn't do that. free is only used for memory dynamically allocated from heap by malloc family of functions.
Memory for num is statically allocated in data segment and can't be released. Memory for num2 is allocated in the main's call stack frame and will be released as soon as main returns.
What actually happens depend on implementation of free. There are usually specific data structures maintained in heap to help malloc/free track the allocated and free memory areas. free expects these data structures to be somewhere around the place its argument points to. An when you pass it a pointer which doesn't point to a malloc-allocated heap area, it'll consider garbage data as some useful information and do some strange things. And you're lucky if the result is just an immediate program crash.
This is the erroneous code.
#include <stdio.h>
#include<stdlib.h>
int main() {
int* t = (int*) malloc(sizeof(int));
int a = 4;
t = &a;
printf("%d\n",*t);
free(t);
return 0;
}
How can I fix this?
malloc and calloc are used for dynamic memory allocation for arrays. As Blaze said in comments, just removing malloc will solve your problem:
#include <stdio.h>
#include<stdlib.h>
int main() {
int* t;
int a = 4;
t = &a;
printf("%d\n",*t);
return 0;
}
You won't need free as well, because, you did not allocate dynamic memory. Moreover, if you do create dynamic memory for array, I would recommend you to use calloc over malloc, as unlike malloc, calloc initializes memory with zero.
This is what happens in your code:
A pointer t points to an int which is created dynamically
int* t = (int*) malloc(sizeof(int));
Data:
t -> INT
A new variable is created
int a = 4;
Data:
t -> unnamed: integer
a: 4
T is assigned to a
t = &a;
Data:
unnamed: integer
t-> a: 4
This causes a problem, because the dynamically created integer is not referred anymore to. There is also no way to free it now there is no variable pointer to it.
(Trying to) free memory
free(t);
Data:
unnamed: integer
t-> a: 4
Memory cannot be freed because it points to stack space (for in this case local variable), not heap space (for dynamic variables). See comment of Jeremy below.
Solution:
Do not use the malloc and free, and assign t (as pointer) to variable a, after you create a.
Broadly speaking, there are three types of memory available to a C program.
memory for global and static variables
"automatic" memory for function local variables and parameters
dynamic memory which is allocated by the program code itself using functions like malloc().
Automatic memory (I call it that because there used to be a keyword in C automatic you could use to specify it) is allocated automatically from the program stack or the CPU's register file when a function is entered and automatically deallocated when the function is left.
Dynamic memory is allocated from the "heap" which is basically just a heap of memory not in any particular order.
In your code you have automatic and dynamic allocations:
int* t = (int*) malloc(sizeof(int));
This code allocates memory for a pointer on the stack (t) and enough memory for an int from the heap. A pointer to that memory is saved in t
int a = 4;
This code allocates memory for an int on the stack (a) and saves the int 4 in it.
t = &a;
This code gets a pointer to a (which is allocated from the stack) and stores it in t.
Note that the previous content of t the pointer to the dynamically allocated block of memory from the heap has now been lost. There is now no way for your program to deallocate the heap memory. This is called a memory leak and, in larger programs, bugs of this nature cause the memory usage to keep going up until the system grinds to a halt or the program fails because it can't be allocated any more heap memory.
printf("%d\n",*t);
This just prints out what is pointed to by t i.e a which is 4.
free(t);
free() is used by the C library to return memory that was allocated from the heap back to the heap. Unfortunately, t no longer points to memory from the heap, it points to memory on the stack. It's likely that this line corrupts the heap. Were you not about to exit the program, it would be a disaster.
I've read conflicting information on the internet about this. To the best of my knowledge all variables in a function only exist for the life time of the function and so this shouldn't be necessary. But with dynamic pointer arrays I'm not sure.
Thanks
You don't free pointers - they are cleaned up automatically. What you need to free is the memory that was acquired by malloc. You use pointers to access that memory.
Put laconically: Every pointer you get from malloc has to go to free exactly once. Nothing more, nothing less.
It depends on where the memory referenced by the pointer has been allocated.
There are fundamentally 2 ways of allocating space in C.
Using the stack :
void foo(){
int stack_variable = 10;
int *stack_pointer = &stack_variable; //bar shouldn't be freed.
}
Using the heap:
void foo(){
int * heap_pointer = (int *) malloc(sizeof(int)); //heap_pointer need to be freed somewhere
}
In the first case there is no problem: memory allocated in the stack will be released when the function returns. Even if you are using a pointer, and it points to some data in the stack, free isn't necessary. Be careful to not use a pointer to data allocated in the stack when the data itself goes out of scope:
int * foo(){
int stack_variable = 10;
int *ptr = &stack_variable;
return ptr;
}
int * ptr = foo(); // you have a problem here: ptr points to a memory region that no longer exist!
In the second case you are using the heap so you need to use free somewhere, to explicitly release it.
If you have allocated data using malloc/calloc, you need to free the data.
Addition due to curious comment by #Julia Childe:
Well, the point of allocating dynamic memory is that it will remain there till you explicitly free it. This enables you to pass pointers both from and to functions and you are not limited by the scope of a specific function, i.e. main.
Thereby, you can allocate memory for data when you need to and not in advance, thus dynamic memory.
If we did not have this ability, we would have to know how much memory space we would use at compile time.
Hope this clears out some question marks.
1)
For which datatypes must I allocate memory with malloc?
For types like structs, pointers, except basic datatypes, like int
For all types?
2)
Why can I run this code? Why does it not crash? I assumed that I need to allocate memory for the struct first.
#include <stdio.h>
#include <stdlib.h>
typedef unsigned int uint32;
typedef struct
{
int a;
uint32* b;
}
foo;
int main(int argc, char* argv[])
{
foo foo2;
foo2.a = 3;
foo2.b = (uint32*)malloc(sizeof(uint32));
*foo2.b = 123;
}
Wouldn't it be better to use
foo* foo2 = malloc(sizeof(foo));
3)
How is foo.b set? Does is reference random memory or NULL?
#include <stdio.h>
#include <stdlib.h>
typedef unsigned int uint32;
typedef struct
{
int a;
uint32* b;
}
foo;
int main(int argc, char* argv[])
{
foo foo2;
foo2.a = 3;
}
All types in C can be allocated either dynamically, automatically (on the stack) or statically. The issue is not the type, but the lifetime you want - you use malloc when you want an object to exist outside of the scope of the function that created it, or when you don't know in advance how big a thing you need.
Edit to address your numbered questions.
There are no data types you must allocate with malloc. Only if you want a pointer type to point to valid memory must you use the unary & (address-of) operator or malloc() or some related function.
There is nothing wrong with your code - the line:
foo foo2;
Allocates a structure on the stack - then everything works as normal. Structures are no different in this sense than any other variable. It's not better or worse to use automatic variables (stack allocation) or globals or malloc(), they're all different, with different semantics and different reasons to choose them.
In your example in #3, foo2.b's value is undefined. Any automatic variable has an undefined and indeterminate value until you explicitly initialize it.
You must allocate with malloc any memory that you wish to be managed manually, as opposed to automatically. It doesn't matter if what's stored there is an int or a double or a struct or anything; malloc is all about manual memory management.
When you create a variable without malloc, it is stored on the stack and when it falls out of scope, its memory is automatically reclaimed. A variable falls out of scope when the variable is no longer accessible; e.g. when the block or function that the variable was declared in ends.
When you allocate memory with malloc, it is stored on the heap, and malloc returns a pointer to this memory. This memory will not be reclaimed until you call free on it, regardless of whether or not a pointer to it remains accessible (when no pointers remain to heap-allocated memory, this is a memory leak). This means that you gain the ability to continue to use the memory you allocated after the block or function that it was allocated in ends, but on the other hand you now have the responsibility to manually deallocate it when you are finished with it.
In your example, foo2 is on the stack, and will be automatically deallocated when main ends. However, the memory pointed to by foo2.b will not be automatically deallocated, since it is on the heap. This isn't a problem in your example because all memory is returned to the OS when the program ends, but if it were in a function other than main it would have been a memory leak.
foo foo2; automatically allocates the structure on the stack, and it is automatically deallocated when the enclosing function (main in this case) ends.
You only need to allocate memory on the heap, using malloc, if you need the structure to persist after the enclosing scope ends. You may also need to do this when the object is too large to fit on the stack.
2) Why can I run this code? Why does it not crash?
The code never refers to undefined memory or NULL. Why would it crash? (You have a memory leak as written, but it's probably because you're only showing part of the code and in the given program it's not a problem anyway.)
The alternative code you suggest will also work, though memory returned from malloc is also uninitialized by default. (I once worked with a custom memory allocator that filled returned memory blocks with ? characters by default. Still perfectly legal by the rules. Note that calloc returns a pointer to zero-initialized memory; use it if that's what you want.)
3) How is foo.b set? Does is reference random memory or NULL?
Random memory. Stack allocated structures are not initialized for you.
You can do it, but this is not enough.
Because, the second field is a pointer, which has to be set to a valid address. One of the ways to do it is by allocating memory (with malloc).
To your first question -
use malloc when you MUST manage object's lifetime manually.
For instance, the second code could be rewritten as:
int main(int argc, char* argv[])
{
foo foo2; uint32 b;
foo2.a = 3;
foo2.b = &b;
*foo2.b = 123;
}
This is better, because lifetime is the same, and the memory is now on stack - and doesn't need to be freed.
The memory for the struct instance ("foo2") will be allocated on the stack - there's no need to allocate memory for this yourself - if you do allocate using malloc be sure to free off the memory at a later date.