#include<stdio.h>
#include<stdlib.h>
int main()
{
char *temp;
while(1)
{
temp= malloc(sizeof(char)*10);
/*some code*/
}
}
My question is: I was allocating memory in every loop with same name. What will happen to previous memory allocated; and as we are creating variables with same identifier why we are not getting any errors?
PS: Don't ask what I'm doing in this code. I was doing some other code and by mistake did it. Just re modified it and asking my problem .
It's called a memory leak. You have no way of ever freeing the memory, so it's lost to you.
Your process will keep growing in size until you run out of available memory (or, if it's 64 bit, swap the box into oblivion at which point your local sysadmin will come over and trounce you. If it's a production machine, you're likely to find out what a performance improvement plan is from your boss and HR.)
what will happen to prev memory allocated.
Will be lost forever. That is what is called "memory leak".
and as we are creating variables with same identifier why we are not getting any errors
That's not a compilation error. The compiler doesn't follow your logic, only your syntax. Overwriting values is perfectly fine with the compiler.
The previos memory allocated will be leaked, since its not freed and its address gets lost to the program. You don't get any errors because you are not creating variables with the same identifier, you are allocating different chunks of memory and storing the pointer to them into the same temp variable.
Consider reading some basic C programming material as you seem to be confusing variables with memory.
The compiler will in most cases not detect logical errors, such as this, where the programmer allocates memory but never frees it.
This is what we call a "memory leak".
Here temp is a pointer not an identifier.
So temp POINTS to the memory allocated my malloc.
So when you allocate new memory and use the SAME pointer to point to the newly allocated memory, it NO LONGER points the the old allocated memory block and it is lost forever. So there you have a memory leak.
If you no longer need the memory pointed to by temp after the end of the current iteration of the loop, use following code instead.
#include<stdio.h>
#include<stdlib.h>
int main()
{
char *temp;
while(1)
{
temp= malloc(sizeof(char)*10);
/*some code*/
/* free the memory pointed by temp */
free(temp);
}
}
This way you do not have a memory leak as the memory pointed to my temp is freed at the end of the iteration before a new block is allocated at the start of the next loop.
Update:
Also you can use the same memory block pointed to by temp throughout your loop. No need to reallocate in each iteration.
And if you do not want to retain any old memory you can simply use memset, like this
#include<stdio.h>
#include<stdlib.h>
int main()
{
/* allocate memopry only once */
char *temp = malloc(sizeof(char)*10);
while(1){
/* zero out the memory so no old data is left */
memset((void*)temp, 0, sizeof(char)*10);
/* some code that uses temp */
}
/* free the allocated memory */
free(temp);
}
Related
This question already has answers here:
Why do I get different results when I dereference a pointer after freeing it?
(7 answers)
Closed 8 years ago.
I'm trying to understand the how the memory is managed in C. In the sample code below, in the comments I have added my understanding on how the memory is managed.
I would like to know if I'm correct
In the final part of the code, I'm trying to access a part of memory which is already freed. I'm surprised why the compiler did not through any warnings
struct node
{
int keys[2];
int *pkeys;
};
int main()
{
int temp[2] = {10,20}; // creates a temp array in stack
struct node* node = malloc(sizeof(struct node)); //allocates a block of memory on heap pointed by node
node->keys[0]=1;node->keys[1]=2; // assigns values for keys present in heap
node->pkeys = malloc(2*sizeof(int)); // creates a pointer pkeys in heap
node->pkeys = temp; // the pointer pkeys in heap points to the temp array in stack
free(node); // the block of memory on heap pointed by node is freed
node->keys[0] = 9; //**Why does the compiler ignore that I'm accessing freed memory?**
}
The compiler in C does not do that sort of checking. If gives enough rope to hang yourself.
It is up to you to do the checks. That is also true for bounds checking of arrays.
Also you need to note that malloc/free does not always have to beg/give to the OS. That means it may still be accessible by the process without seg fault.
The compiler does not check for illegal memory access and your program will cause undefined behaviour or may even crash (segmentation fault). The behaviour is unpredictable and next time you run your program, it may crash.
Few things about your code:
The signature of main should either be int main(void) or int main(int argc, char *argv[]).
The comment for this statement is wrong.
node->pkeys = malloc(2*sizeof(int)); // creates a pointer pkeys in heap
You created the pointer pkeys when you allocated memory and made node point to it. This statement is dynamically allocating memory for array of 2 ints and making pkeys point to it. So when you do node->pkeys = temp, you lose handle on the dynamically allocated array and will cause memory leak. So before reassigning pkeys, you must free the memory it points to.
free(node->pkeys);
// now reassign node->pkeys
The space allocated to node is freed but node still points to the same location. So it is still accessible. But if you are unlucky (that is the OS reused the memory for its own purposes) you will get a segmentation fault.
int a = 0;
int *b = malloc (sizeof(int));
b = malloc (sizeof(int));
The above code is bad because it allocates memory on the heap and then doesn't free it, meaning you lose access to it. But you also created 'a' and never used it, so you also allocated memory on the stack, which isn't freed until the scope ends.
So why is it bad practice to not free memory on the heap but okay for memory on the stack to not be freed (until the scope ends)?
Note: I know that memory on the stack can't be freed, I want to know why its not considered bad.
The stack memory will get released automatically when the scope ends. The memory allocated on the heap will remain occupied unless you release it explicitly. As an example:
void foo(void) {
int a = 0;
void *b = malloc(1000);
}
for (int i=0; i<1000; i++) {
foo();
}
Running this code will decrease the available memory by 1000*1000 bytes required by b, whereas the memory required by a will always get released automatically when you return from the foo call.
Simple: Because you'll leak memory. And memory leaks are bad. Leaks: bad, free: good.
When calling malloc or calloc, or indeed any *alloc function, you're claiming a chunk of memory (the size of which is defined by the arguments passed to the allocating function).
Unlike stack variables, which reside in a portion of memory the program has, sort of, free reign over, the same rules don't apply to heap memory. You may need to allocate heap memory for any number of reasons: the stack isn't big enough, you need an array of pointers, but have no way of knowing how big this array will need to be at compile time, you need to share some chunk of memory (threading nightmares), a struct that requires the members to be set at various places (functions) in your program...
Some of these reasons, by their very nature, imply that the memory can't be freed as soon as pointer to that memory goes out of scope. Another pointer might still be around, in another scope, that points to the same block of memory.
There is, though, as mentioned in one of the comments, a slight drawback to this: heap memory requires not just more awareness on the programmers part, but it's also more expensive, and slower than working on the stack.
So some rules of thumb are:
You claimed the memory, so you take care of it... you make sure it's freed when you're done playing around with it.
Don't use heap memory without a valid reason. Avoiding stack overflow, for example, is a valid reason.
Anyway,
Some examples:
Stack overflow:
#include <stdio.h>
int main()
{
int foo[2000000000];//stack overflow, array is too large!
return 0;
}
So, here we've depleted the stack, we need to allocate the memory on the heap:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int *foo= malloc(2000000000*sizeof(int));//heap is bigger
if (foo == NULL)
{
fprintf(stderr, "But not big enough\n");
}
free(foo);//free claimed memory
return 0;
}
Or, an example of an array, whose length depends on user input:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int *arr = NULL;//null pointer
int arrLen;
scanf("%d", &arrLen);
arr = malloc(arrLen * sizeof(int));
if (arr == NULL)
{
fprintf(stderr, "Not enough heap-mem for %d ints\n", arrLen);
exit ( EXIT_FAILURE);
}
//do stuff
free(arr);
return 0;
}
And so the list goes on... Another case where malloc or calloc is useful: An array of strings, that all might vary in size. Compare:
char str_array[20][100];
In this case str_array is an array of 20 char arrays (or strings), each 100 chars long. But what if 100 chars is the maximum you'll ever need, and on average, you'll only ever use 25 chars, or less?
You're writing in C, because it's fast and your program won't use any more resources than it actually needs? Then this isn't what you actually want to be doing. More likely, you want:
char *str_array[20];
for (int i=0;i<20;++i) str_array[i] = malloc((someInt+i)*sizeof(int));
Now each element in the str_array has exactly the amount of memory I need allocated too it. That's just way more clean. However, in this case calling free(str_array) won't cut it. Another rule of thumb is: Each alloc call has to have a free call to match it, so deallocating this memory looks like this:
for (i=0;i<20;++i) free(str_array[i]);
Note:
Dynamically allocated memory isn't the only cause for mem-leaks. It has to be said. If you read a file, opening a file pointer using fopen, but failing to close that file (fclose) will cause a leak, too:
int main()
{//LEAK!!
FILE *fp = fopen("some_file.txt", "w");
if (fp == NULL) exit(EXIT_FAILURE);
fwritef(fp, "%s\n", "I was written in a buggy program");
return 0;
}
Will compile and run just fine, but it will contain a leak, that is easily plugged (and it should be plugged) by adding just one line:
int main()
{//OK
FILE *fp = fopen("some_file.txt", "w");
if (fp == NULL) exit(EXIT_FAILURE);
fwritef(fp, "%s\n", "I was written in a bug-free(?) program");
fclose(fp);
return 0;
}
As an asside: if the scope is really long, chances are you're trying to cram too much into a single function. Even so, if you're not: you can free up claimed memory at any point, it needn't be the end of the current scope:
_Bool some_long_f()
{
int *foo = malloc(2000000000*sizeof(int));
if (foo == NULL) exit(EXIT_FAILURE);
//do stuff with foo
free(foo);
//do more stuff
//and some more
//...
//and more
return true;
}
Because stack and heap, mentioned many times in the other answers, are sometimes misunderstood terms, even amongst C programmers, Here is a great conversation discussing that topic....
So why is it bad practice to not free memory on the heap but okay for memory on the stack to not be freed (until the scope ends)?
Memory on the stack, such as memory allocated to automatic variables, will be automatically freed upon exiting the scope in which they were created.
whether scope means global file, or function, or within a block ( {...} ) within a function.
But memory on the heap, such as that created using malloc(), calloc(), or even fopen() allocate memory resources that will not be made available to any other purpose until you explicity free them using free(), or fclose()
To illustrate why it is bad practice to allocate memory without freeing it, consider what would happen if an application were designed to run autonomously for very long time, say that application was used in the PID loop controlling the cruise control on your car. And, in that application there was un-freed memory, and that after 3 hours of running, the memory available in the microprocessor is exhausted, causing the PID to suddenly rail. "Ah!", you say, "This will never happen!" Yes, it does. (look here). (not exactly the same problem, but you get the idea)
If that word picture doesn't do the trick, then observe what happens when you run this application (with memory leaks) on your own PC. (at least view the graphic below to see what it did on mine)
Your computer will exhibit increasingly sluggish behavior until it eventually stops working. Likely, you will be required to re-boot to restore normal behavior.
(I would not recommend running it)
#include <ansi_c.h>
char *buf=0;
int main(void)
{
long long i;
char text[]="a;lskdddddddd;js;'";
buf = malloc(1000000);
strcat(buf, "a;lskdddddddd;js;dlkag;lkjsda;gkl;sdfja;klagj;aglkjaf;d");
i=1;
while(strlen(buf) < i*1000000)
{
strcat(buf,text);
if(strlen(buf) > (i*10000) -10)
{
i++;
buf = realloc(buf, 10000000*i);
}
}
return 0;
}
Memory usage after just 30 seconds of running this memory pig:
I guess that has to do with scope 'ending' really often (at the end of a function) meaning if you return from that function creating a and allocating b, you will have freed in a sense the memory taken by a, and lost for the remainder of the execution memory used by b
Try calling that function a a handful of times, and you'll soon exhaust all of your memory. This never happens with stack variables (except in the case of a defectuous recursion)
Memory for local variables automatically is reclaimed when the function is left (by resetting the frame pointer).
The problem is that memory you allocate on the heap never gets freed until your program ends, unless you explicitly free it. That means every time you allocate more heap memory, you reduce available memory more and more, until eventually your program runs out (in theory).
Stack memory is different because it's laid-out and used in a predictable pattern, as determined by the compiler. It expands as needed for a given block, then contracts when the block ends.
So why is it bad practice to not free memory on the heap but okay for memory on the stack to not be freed (until the scope ends)?
Imagine the following:
while ( some_condition() )
{
int x;
char *foo = malloc( sizeof *foo * N );
// do something interesting with x and foo
}
Both x and foo are auto ("stack") variables. Logically speaking, a new instance for each is created and destroyed in each loop iteration1; no matter how many times this loop runs, the program will only allocate enough memory for a single instance of each.
However, each time through the loop, N bytes are allocated from the heap, and the address of those bytes is written to foo. Even though the variable foo ceases to exist at the end of the loop, that heap memory remains allocated, and now you can't free it because you've lost the reference to it. So each time the loop runs, another N bytes of heap memory is allocated. Over time, you run out of heap memory, which may cause your code to crash, or even cause a kernel panic depending on the platform. Even before then, you may see degraded performance in your code or other processes running on the same machine.
For long-running processes like Web servers, this is deadly. You always want to make sure you clean up after yourself. Stack-based variables are cleaned up for you, but you're responsible for cleaning up the heap after you're done.
1. In practice, this (usually) isn't the case; if you look at the generated machine code, you'll (usually) see the stack space allocated for x and foo at function entry. Usually, space for all local variables (regardless of their scope within the function) is allocated at once.
Suppose I have the following piece of code in my program:
char *ptr;
ptr=malloc(sizeof(char)*10);
ptr=malloc(sizeof(char)*10);
ptr=malloc(sizeof(char)*10);
ptr=malloc(sizeof(char)*10);
Will a pointer to the same memory block be assigned to ptr each time,or a separate piece of memory be reserved each time and its pointer assigned to ptr,resulting in memory leak each time malloc() is called?
I am still learning C so bear with me if it's too basic.I tried googling but found no answer.
EDIT::
Thanks for your answers.Please tell me if this approach of me deals with the memory leak risk.My program simply asks for names of 5 people and displays it,without using static arrays.After reading your answers,I put the free(ptr) inside the loop,else before I had planned to use it only once outside the loop,after the loop.Am I correct now?
#include <stdio.h>
#include<stdlib.h>
#include<string.h>
int main ()
{
char *names[5],*ptr;
int i;
for(i=0;i<=4;i++)
{
ptr=malloc(sizeof(char)*10);
printf("Enter name no.%d : \n",i+1);
scanf("%s",ptr);
names[i]=malloc(strlen(ptr)+1);
strcpy(names[i],ptr);
free(ptr);
}
for(i=0;i<=4;i++)
printf("%s\n",names[i]);
}
malloc() will never return the same memory block multiple times, unless (of course) it has been free()'d since the last time it was returned. This is guaranteed by the C standard. Therefore, your code also leaks memory. Any memory allocator that handed out the same piece of memory twice would be worse than useless: Applications would step on their own toes, using the same piece of memory for different purposes, likely concurrently.
Edit: Barring buffer overflow issues, your code is correct in that it frees the 10-char buffers referenced via ptr. Calling free(ptr) just once, outside the loop, would indeed be incorrect. However, your code (as shown here) does not free the memory allocated later in the loop body stored in names[i].
A separate block of memory will be allocated each time, since C does not perform any garbage collection and no free calls are done in between the malloc calls. The first three blocks are leaked since you're not keeping pointers to them.
You can find out for yourself that the pointers are all distinct using a simple printf call:
for (int i = 0; i < 4; i++) {
ptr = malloc(10); // no need for sizeof(char), it's 1 by definition
printf("%p\n", ptr);
}
This should print four distinct numbers.
This will cause a memory leak assuming each call is successful, all but the memory allocated to the last malloc will be inaccessible. You either need to keep a pointer to each block allocated by malloc or you need to free between successive calls. so yes, malloc will attempt to allocate distinct block of memory for each call. A quick test could be as follows:
#include <stdio.h>
#include <stdlib.h>
int main()
{
char *ptr;
ptr=malloc(sizeof(char)*10);
printf("%p\n", ptr);
ptr=malloc(sizeof(char)*10);
printf("%p\n", ptr);
ptr=malloc(sizeof(char)*10);
printf("%p\n", ptr);
ptr=malloc(sizeof(char)*10);
printf("%p\n", ptr);
}
This will print out the address allocated each time. If you added a free between each call to malloc you may end up with the same pointer, a quick test demonstrated this behavior.
Yes, each malloc call will allocate a new block, which means that the previous one will be leaked.
EDIT: answer to the edited part of the question
You are now correctly releasing the blocks to which ptr points, however you are not releasing those allocated for names. You would also need:
for(i=0;i<=4;i++) {
printf("%s\n",names[i]);
free(names[i]);
}
In your case, you could skip using ptr and work on names directly. Here's an equally [un]safe version:
char *names[5];
int i;
for(i=0; i < 5; i++)
{
names[i] = malloc(10);
printf("Enter name no.%d : \n",i+1);
scanf("%s",names[i]); // what if name is longer than 10 characters?
}
for(i=0; i < 5; i++)
{
printf("%s\n",names[i]);
free(names[i]);
}
However, operating system will reclaim the memory consumed by a process once the process exits, so in this simple example you don't have to worry, but I guess that this is only an illustration. I also assume that you don't care about user entering names longer than 10 characters which would write over the boundaries of allocated buffers.
Your second part is right: it's a memory leak.
A new block of memory is allocated each time you call malloc. This will create a memory leak as you have no way to reference the first three allocations to free them.
You can check memory leaks by using valgrind,its very useful in detecting memory leaks.
valgrind ./progr
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.
When I alloc memory outside a while loop for example, is it okay to free it inside it ?
Are these two codes equivalent ?
int* memory = NULL;
memory = malloc(sizeof(int));
if (memory != NULL)
{
memory=10;
free(memory);
}
int* memory = NULL;
memory = malloc(sizeof(int));
if (memory != NULL)
{
memory=10;
}
free(memory);
Yes, they are equivalent. You do not have to call free() if allocation did not succeed.Pay attention, that memory is pointer to int and you have to dereference it to assign something to its memory block;
int* memory = NULL;
memory = malloc(sizeof(int));
if (memory)
*memory=10;
free(memory);
memory = NULL;
int* memory = NULL;
memory = malloc(sizeof(int));
if (memory != NULL)
{
memory=10;
free(memory);
}
This will crash. You're setting the pointer to memory location 10 and then asking the system to release the memory. It's extremely unlikely that you previously allocated some memory that happened to start at 10 (0xA in hexadecimal), even in the crazy world of virtual address spaces. Furthermore, IF malloc failed, no memory has been allocated, so you do not need to free it.
int* memory = NULL;
memory = malloc(sizeof(int));
if (memory != NULL)
{
memory=10;
}
free(memory);
This is also a bug. If malloc fails, then you are setting the pointer to 10 and freeing that memory. (as before.) If malloc succeeds, then you're immediately freeing the memory, which means it was pointless to allocate it! Now, I imagine this is just example code simplified to get the point across, and that this isn't present in your real program? :)
Doesn't matter IF you are sure you only free it once.
A good idea is to always set a pointer =NULL when you free it, then if it gets free'ed again it won't matter
Yes. The important thing is too free it once. (After you are finished using the memory)
A general rule of thumb that I like to follow is to free memory in the same scope that it was allocated in.
First of all you can only free it once. So if you want to free it inside a loop (as your question suggest), the second solution is the only correct way of doing.
If you want to free it inside an if statement, both solution are technically correct. However it is a good practice to always free the memory you use and its then easier to always free the memory at the end outside any if statement.
So my advice is : whatever you are trying to do : always free the memory outside any loop/if statement (assuming you don't need it anymore of course). I usually free it at the end of the function where it's last used but it really depends on the function itself, its length, ....
Try finding your way of doing things and stick to it : it will save you a lot of useless brain time (which is really precious)