What is a segmentation fault?

Asked 27/2, 2010 at 9:23 Answered 9/1, 2023 at 19:5

874

What is a segmentation fault? Is it different in C and C++? How are segmentation faults and dangling pointers related?

Ellett answered 27/2, 2010 at 9:23 Comment(3)

Just a memory dump when something goes wrong! – Antipyretic 25/4, 2015 at 1:18

Typically called by attempting to dereference a null pointer, so a segmentation fault is often analogous to a Java NullPointerException. – Twelve 18/12, 2017 at 9:8

Segmentation comes from Memory Segmentation. You're accessing a segment of memory that does not belong to you. – Brassie 13/5, 2020 at 10:33

963

Segmentation fault is a specific kind of error caused by accessing memory that “does not belong to you.” It’s a helper mechanism that keeps you from corrupting the memory and introducing hard-to-debug memory bugs. Whenever you get a segfault you know you are doing something wrong with memory – accessing a variable that has already been freed, writing to a read-only portion of the memory, etc. Segmentation fault is essentially the same in most languages that let you mess with memory management, there is no principal difference between segfaults in C and C++.

There are many ways to get a segfault, at least in the lower-level languages such as C(++). A common way to get a segfault is to dereference a null pointer:

int *p = NULL;
*p = 1;

Another segfault happens when you try to write to a portion of memory that was marked as read-only:

char *str = "Foo"; // Compiler marks the constant string as read-only
*str = 'b'; // Which means this is illegal and results in a segfault

Dangling pointer points to a thing that does not exist anymore, like here:

char *p = NULL;
{
    char c;
    p = &c;
}
// Now p is dangling

The pointer p dangles because it points to the character variable c that ceased to exist after the block ended. And when you try to dereference dangling pointer (like *p='A'), you would probably get a segfault.

Smyrna answered 27/2, 2010 at 9:36 Comment(9)

The last example is particularly nasty, when I build: int main() { char *p = 0; { char c = 'x'; p = &c; } printf( "%c\n",*p); return 0; } With either gcc or several other compilers, it 'appears' to work. No warnings on compile. No segfault. This is because the '}' out of scope, doesn't actually delete the data, just marks it as free to be used again. The code can run fine on a production system for years, you alter another part of the code, change compiler or something else and BOOOOOM! – Sectionalism 13/4, 2010 at 9:6

Sorry for the bump but just a side note... none of your examples necessarily cause a segfault, in fact it's just undefined behavior ;-) – Rumpus 15/9, 2012 at 3:1

@oldrinb: It is impossible to write code that necessarily causes a segfault. Not least because there are systems out there that operate without memory protection, thus cannot tell whether a piece of memory actually "belongs to you", and thus don't know segfaults, only undefined behaviour... (classic AmigaOS, for example) – Candelabrum 29/5, 2014 at 18:3

@ChrisHuang-Leaver, you need to understand that c is local, it means that it have been pushed on the stack after { and pop-ed out of it after }. the dangling pointer is just a reference to an offset which is now out of the stack. that's why modifying it in a simple program will never trigger any segfault. on the other hand it may lead to segfault in a more complex use case, where other function calls might lead the stack to grow and contain the data pointed to by the dangling pointer. writing to that data (local vars) would lead to undefined behavior (segfault &Co) – Cassock 19/1, 2016 at 21:23

@ChrisHuang-Leaver, what if the block occurs outside the method, causing a new stack frame, e.g.,

int main() {    char* p = 0;   hello(p);   printf( "%c\n",*p); return 0;  }   void hello(char* d) {   char c = 'x';   d = &c; }

– Carden 28/1, 2016 at 22:55

@ChrisHuang-Leaver I had the same problem, but you can discover it by having a function that creates a lot of objects on stack/heap memory after you've done some stuff in your program, then continue using your program. This is a very generic way of checking if you have these kind of problems in your application. The only thing you need to do is to use every possible feature before you run the function, and then again try every possible feature after the memory-intensive function. – Diathermic 8/2, 2016 at 5:28

@ChrisHuang-Leaver, normally when you get out of scope, the compiler has to recover some stack space to free the unused stack space, but this doesn't happen always (with gcc being one of this compilers). Also, the allocated stack space is normally reused again, so I have heard of no operating systems that return unused stack pages to the system, making that space subject for a SIGSEGV, so I won't expect such a signal from mangling with the stack. – Gibberish 22/7, 2016 at 11:59

For further reading, adding static to char c will solve the problem. Because adding static inside a function tells the variable to go in the .data section, not on the stack. – Embark 19/11, 2016 at 14:2

The dangling pointer is even more dangerous as it may not cause a (direct) segmentation fault but instead display all kinds of undefined behavior. For instance giving the address of a temporary variable to a system function that will later come back and change the data at that address (asynchronous I/O mechanisms for instance). This disrupts and possibly crashes whatever function is currently using that memory as the new owner - so it happens even within "innocent" code. – Santinasantini 25/5, 2020 at 7:46

131

It would be worth noting that segmentation fault isn't caused by directly accessing another process memory (this is what I'm hearing sometimes), as it is simply not possible. With virtual memory every process has its own virtual address space and there is no way to access another one using any value of pointer. Exception to this can be shared libraries which are same physical address space mapped to (possibly) different virtual addresses and kernel memory which is even mapped in the same way in every process (to avoid TLB flushing on syscall, I think). And things like shmat ;) - these are what I count as 'indirect' access. One can, however, check that they are usually located long way from process code and we are usually able to access them (this is why they are there, nevertheless accessing them in a improper way will produce segmentation fault).

Still, segmentation fault can occur in case of accessing our own (process) memory in improper way (for instance trying to write to non-writable space). But the most common reason for it is the access to the part of the virtual address space that is not mapped to physical one at all.

And all of this with respect to virtual memory systems.

Welty answered 3/7, 2011 at 23:22 Comment(3)

With shared memory/memory mapped files it is possible for someone else to mess with your memory. In WIN32 there are nasty API's like 'WriteProcessMemory' too! – Oidium 17/2, 2014 at 23:46

@paulm: Yes, I know. This is what I had on mind in "And things like shmat ;) - these are what I count as 'indirect' access." – Welty 18/2, 2014 at 10:8

In a virtual memory operating system there's no way (normally, so please, operating system implementors, don't flame me for this) for a process to access another process virtual memory, not being some kind of memory attach system call that allows you to access. Virtual memory addresses normally mean different things depending on the process being considered. – Gibberish 22/7, 2016 at 12:2

A segmentation fault is caused by a request for a page that the process does not have listed in its descriptor table, or an invalid request for a page that it does have listed (e.g. a write request on a read-only page).

A dangling pointer is a pointer that may or may not point to a valid page, but does point to an "unexpected" segment of memory.

Futilitarian answered 27/2, 2010 at 9:27 Comment(1)

This is true, but would it really help you if you already didn’t know what a segmentation fault is? – Smyrna 27/2, 2010 at 9:37

To be honest, as other posters have mentioned, Wikipedia has a very good article on this so have a look there. This type of error is very common and often called other things such as Access Violation or General Protection Fault.

They are no different in C, C++ or any other language that allows pointers. These kinds of errors are usually caused by pointers that are

Used before being properly initialised
Used after the memory they point to has been realloced or deleted.
Used in an indexed array where the index is outside of the array bounds. This is generally only when you're doing pointer math on traditional arrays or c-strings, not STL / Boost based collections (in C++.)

Fervor answered 27/2, 2010 at 20:35 Comment(0)

According to Wikipedia:

A segmentation fault occurs when a program attempts to access a memory location that it is not allowed to access, or attempts to access a memory location in a way that is not allowed (for example, attempting to write to a read-only location, or to overwrite part of the operating system).

Suannesuarez answered 27/2, 2010 at 9:30 Comment(0)

Segmentation fault is also caused by hardware failures, in this case the RAM memories. This is the less common cause, but if you don't find an error in your code, maybe a memtest could help you.

The solution in this case, change the RAM.

edit:

Here there is a reference: Segmentation fault by hardware

Ameliaamelie answered 24/6, 2014 at 16:59 Comment(1)

A quick-and-dirty test for faulty RAM is to run your crashing program over and over again in a loop. If the program has no internal nondeterminism—that is, it always produces the same output for the same input, or at least it's supposed to—but, for some particular input, it crashes sometimes, not always but not never either: then you should start worrying about bad RAM. – Leaseholder 26/10, 2017 at 21:39

Wikipedia's Segmentation_fault page has a very nice description about it, just pointing out the causes and reasons. Have a look into the wiki for a detailed description.

In computing, a segmentation fault (often shortened to segfault) or access violation is a fault raised by hardware with memory protection, notifying an operating system (OS) about a memory access violation.

The following are some typical causes of a segmentation fault:

Dereferencing NULL pointers – this is special-cased by memory management hardware
Attempting to access a nonexistent memory address (outside process's address space)
Attempting to access memory the program does not have rights to (such as kernel structures in process context)
Attempting to write read-only memory (such as code segment)

These in turn are often caused by programming errors that result in invalid memory access:

Dereferencing or assigning to an uninitialized pointer (wild pointer, which points to a random memory address)
Dereferencing or assigning to a freed pointer (dangling pointer, which points to memory that has been freed/deallocated/deleted)
A buffer overflow.
A stack overflow.
Attempting to execute a program that does not compile correctly. (Some compilers will output an executable file despite the presence of compile-time errors.)

Lemuel answered 14/10, 2014 at 10:5 Comment(0)

Segmentation fault occurs when a process (running instance of a program) is trying to access read-only memory address or memory range which is being used by other process or access the non-existent (invalid) memory address. Dangling Reference (pointer) problem means that trying to access an object or variable whose contents have already been deleted from memory, e.g:

int *arr = new int[20];
delete arr;
cout<<arr[1];  //dangling problem occurs here

Charpentier answered 10/12, 2013 at 22:34 Comment(1)

The correct way to delete an array is delete [] arr; – Accountant 28/3, 2016 at 15:48

In simple words: segmentation fault is the operating system sending a signal to the program saying that it has detected an illegal memory access and is prematurely terminating the program to prevent memory from being corrupted.

Senary answered 19/7, 2017 at 13:43 Comment(0)

There are several good explanations of "Segmentation fault" in the answers, but since with segmentation fault often there's a dump of the memory content, I wanted to share where the relationship between the "core dumped" part in Segmentation fault (core dumped) and memory comes from:

From about 1955 to 1975 - before semiconductor memory - the dominant technology in computer memory used tiny magnetic doughnuts strung on copper wires. The doughnuts were known as "ferrite cores" and main memory thus known as "core memory" or "core".

Taken from here.

Fenestrated answered 13/10, 2018 at 18:39 Comment(0)

In computing, a segmentation fault or access violation is a fault, or failure condition, raised by hardware with memory protection, notifying an operating system the software has attempted to access a restricted area of memory. -WIKIPEDIA

You might be accessing the computer memory with the wrong data type. Your case might be like the code below:

#include <stdio.h>

int main(int argc, char *argv[]) {
    
    char A = 'asd';
    puts(A);
    
    return 0;
    
}

'asd' -> is a character chain rather than a single character char data type. So, storing it as a char causes the segmentation fault. Stocking some data at the wrong position.

Storing this string or character chain as a single char is trying to fit a square peg in a round hole.

Terminated due to signal: SEGMENTATION FAULT (11)

Segm. Fault is the same as trying to breath in under water, your lungs were not made for that. Reserving memory for an integer and then trying to operate it as another data type won't work at all.

Gloam answered 23/4, 2021 at 3:29 Comment(0)

"Segmentation fault" means that you tried to access memory that you do not have access to.

The first problem is with your arguments of main. The main function should be int main(int argc, char *argv[]), and you should check that argc is at least 2 before accessing argv[1].

Also, since you're passing in a float to printf (which, by the way, gets converted to a double when passing to printf), you should use the %f format specifier. The %s format specifier is for strings ('\0'-terminated character arrays).

Korten answered 1/3, 2019 at 11:44 Comment(0)

There are enough definitions of segmentation fault, I would like to quote few examples which I came across while programming, which might seem like silly mistakes, but will waste a lot of time.

You can get a segmentation fault in below case while argument type mismatch in printf:

#include <stdio.h>
int main(){   
  int a = 5;
  printf("%s",a);
  return 0;
}

output : Segmentation Fault (SIGSEGV)

When you forgot to allocate memory to a pointer, but try to use it.

#include <stdio.h> 
typedef struct{
  int a;
} myStruct;   
int main(){
  myStruct *s;
  /* few lines of code */
  s->a = 5;
  return 0;
}

output : Segmentation Fault (SIGSEGV)

Re answered 10/11, 2019 at 15:44 Comment(0)

Simple meaning of Segmentation fault is that you are trying to access some memory which doesn't belong to you. Segmentation fault occurs when we attempt to read and/or write tasks in a read only memory location or try to freed memory. In other words, we can explain this as some sort of memory corruption.

Below I mention common mistakes done by programmers that lead to Segmentation fault.

Use scanf() in wrong way(forgot to put &).

int num;
scanf("%d", num);// must use &num instead of num

Use pointers in wrong way.

int *num; 
printf("%d",*num); //*num should be correct as num only
//Unless You can use *num but you have to point this pointer to valid memory address before accessing it.

Modifying a string literal(pointer try to write or modify a read only memory.)

char *str;  

//Stored in read only part of data segment
str = "GfG";      

//Problem:  trying to modify read only memory
*(str+1) = 'n';

Try to reach through an address which is already freed.

// allocating memory to num 
int* num = malloc(8); 
*num = 100; 

// de-allocated the space allocated to num 
free(num); 

// num is already freed there for it cause segmentation fault
*num = 110;

Stack Overflow -: Running out of memory on the stack
Accessing an array out of bounds'
Use wrong format specifiers when using printf() and scanf()'

Pep answered 3/1, 2020 at 17:50 Comment(0)

Consider the following snippets of Code,

SNIPPET 1

int *number = NULL;
*number = 1;

SNIPPET 2

int *number = malloc(sizeof(int));
*number = 1;

I'd assume you know the meaning of the functions: malloc() and sizeof() if you are asking this question.

Now that that is settled, SNIPPET 1 would throw a Segmentation Fault Error. while SNIPPET 2 would not.

Here's why.

The first line of snippet one is creating a variable(*number) to store the address of some other variable but in this case it is initialized to NULL. on the other hand, The second line of snippet two is creating the same variable(*number) to store the address of some other and in this case it is given a memory address(because malloc() is a function in C/C++ that returns a memory address of the computer)

The point is you cannot put water inside a bowl that has not been bought OR a bowl that has been bought but has not been authorized for use by you. When you try to do that, the computer is alerted and it throws a SegFault error.

You should only face this errors with languages that are close to low-level like C/C++. There is an abstraction in other High Level Languages that ensure you do not make this error.

It is also paramount to understand that Segmentation Fault is not language-specific.

Impoverish answered 15/10, 2020 at 10:17 Comment(0)

Segmentation fault occurs when a process (running instance of a program) is trying to access a read-only memory address or memory range which is being used by another process or access the non-existent memory address.

seg fault,when type gets mismatched

Evy answered 9/1, 2023 at 19:5 Comment(2)

Is this segfault due to type mismatch, or because it's trying to read memory address 116 (decimal form of t)? There's multiple things wrong with the code, and my compiler just warns of the formatting mismatch. – Prader 30/11, 2023 at 20:56

@Prader Well yes, but it is trying to read that memory address because of a type mismatch. – Autotruck 30/3 at 0:46

A segmentation fault or access violation occurs when a program attempts to access a memory location that is not exist, or attempts to access a memory location in a way that is not allowed.

 /* "Array out of bounds" error 
   valid indices for array foo
   are 0, 1, ... 999 */
   int foo[1000];
   for (int i = 0; i <= 1000 ; i++) 
   foo[i] = i;

Here i[1000] not exist, so segfault occurs.

Causes of segmentation fault:

it arise primarily due to errors in use of pointers for virtual memory addressing, particularly illegal access.

De-referencing NULL pointers – this is special-cased by memory management hardware.

Attempting to access a nonexistent memory address (outside process’s address space).

Attempting to access memory the program does not have rights to (such as kernel structures in process context).

Attempting to write read-only memory (such as code segment).

Lagrange answered 8/12, 2015 at 16:14 Comment(2)

First of all, seg fault has nothing to do with the address does or doesn't exist. It is about you are accessing it where you are not allowed to do so. And in your special example it is even guranteed by standard that that location exists. since the standard says in array case it must be given that there is a valid address for an pointer pointg on an well aligned array within its bounds AND 1 behind. – Imprison 8/12, 2015 at 16:25

it is also releated with address, if you don't have the address and if u try to access this address, also there is seg. fault. And in my example, it is only for understand point of view. – Lagrange 12/12, 2015 at 3:11

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