This question already has answers here:
How dangerous is it to access an array out of bounds?
(12 answers)
Closed 8 years ago.
In c program we can initialize an array like int array[10]. So it can store 10 integer value.But when I give input using loop it takes input more than 10 and doesn't show any error.
actually what is happening??
#include<stdio.H>
main()
{
int array[10],i;
for(i=0;i<=11;i++)
scanf("%d",&array[i]);
for(i=0;i<10;i++)
printf("%d",array[i]);
}
Because C doesn't do any array bounds checking. You as a programmer are responsible for making sure that you don't index out of bounds.
Depending on the used compiler and the system the code is running on, you might read random data from memory or get a SIGSEGV eventually when reading/writing out of bounds.
The C compiler and the runtime are not required to perform any array bounds checking.
What you describe is an example of a whole class of programming errors that result in undefined behavior. From Wikipedia:
In computer programming, undefined behavior refers to computer code whose behavior is specified to be arbitrary.
What this means is that the program is allowed to misbehave (or not) in any way it pleases.
In practice, any of the following are reasonably likely to happen when you write past the end of an array:
The program crashes, either immediately or at a later point.
Other, unrelated, data gets overwritten. This could result in arbitrary misbehaviour and/or in serious security vulnerabilities.
Internal data structures that are used to keep track of allocated memory get corrupted by the out-of-bounds write.
The program works exactly as if more memory had been allocated in the first place (memory is often allocated in block, and by luck there might happen to be some spare capacity after the end of the array).
(This is not an exhaustive list.)
There exist tools, such as Valgrid, that can help discover and diagnose this type of errors.
The C-language standard does not dictate how variables should be allocated in memory.
So the theoretical answer is that you are performing an unsafe memory access operation, which will lead to undefined behavior (anything could happen).
Technically, however, all compilers allocate local variables in the stack and global variables in the data-section, so the practical answer is:
In the case of a local array, you will either override some other local variable or perform an illegal memory access operation.
In the case of a global array, you will either override some other global variable or perform an illegal memory access operation.
Related
This question already has answers here:
How dangerous is it to access an array out of bounds?
(12 answers)
Closed 3 years ago.
My understanding is that if char *my_word is allocated ONE byte of memory malloc(1), then technically, then the following code would produce an out-of-bounds error
char *my_word = malloc(1);
my_word[0] = 'y';
my_word[1] = 'e';
my_word[2] = 's';
and yet, the code runs just fine and doesn't produce any error. In fact, printf("%s", my_word) prints the word just fine.
Why is this not producing an out-of-bounds error if I specifically only allocated 1 byte of memory?
C doesn't have explicit bounds checking. That's part of what makes it fast. But when you write past the bounds of allocated memory, you invoke undefined behavior.
Once you invoke undefined behavior, you can't reliable predict what the program will do. It may crash, it may output strange results, or (as in this case) it may appear to work properly. Additionally, making a seemingly unrelated change such as adding a printf call for debugging or adding an unused local variable can change how undefined behavior manifests itself.
Just because the program could crash doesn't mean it will.
This comes down to the system it is running on. Generally a malloc will allocate in multiples of a certain block size. Eg the block size maybe 16 bytes on your system and malloc will allocate 16 even though you only asked for 1. So in this case you are getting away with overflowing the buffer because you are not writing on memory that is used by anything else.
However you should never rely on this. Always assume that when you write outside the amount requested that bad things will happen.
C does not provide any built-in mechanism to protect you from buffer overflowing, it is up to you to know the size of your buffers and ensure that you never read/write outside of them.
For example if you allocated a buffer a multiple of the block size then writing to the next byte will probably start overwriting critical memory control blocks which may show up as bizarre errors later when you try to free or allocate more memory.
C does not performs bound check. It is just undefined behavior when you access out of bounds which means it can work as normal.
This question already has answers here:
How dangerous is it to access an array out of bounds?
(12 answers)
Closed 5 years ago.
I have written this code and I am not getting any error, so I came to know that it's entirely programmer's botheration to keep a check bounds check. but I am confused by the fact that in static memory allocation the int a[10] has allocated i.e. 40 bytes during compile time so how come it can allocate more memory to it during runtime, because it's also printing the same.
#include<stdio.h>
main()
{
int a[10];
for(i=0;i<=100;i++)
{
a[i]=i;
}
}
Nope, this is undefined behavior. And well that's why it may work and it may not. (Depends on the moon). You should avoid it. Don't think that this is making your program work by allocating more memory - it doesn't.
If you write this, you are writing in some part of your program's memory which will in turn result in errorneous result or if it is trying to write to some memory which is non-modifiable, you will get error. In either way you are invoking undefined behavior - anything may happen from crashing your code to a correct execution. Don't rely on that.
so if it's not allocating memory so where he is storing the elements. I just want to know what is happening internally in the system??
Well it is writing to the memory out of the bound of the array where it is not supposed to write. That is what is happening. That is why this is not what you should be doing. You are basically changing part of memory that program is being assigned and moreover which you are not supposed to change.
This question already has answers here:
How dangerous is it to access an array out of bounds?
(12 answers)
Closed 6 years ago.
I wrote a short program in C just to see what happens when you index past the end of an array.
I found that it mostly produces random values (I know they are not actually random) up until a point (52 indexes past the end in this case) where it produced 0 every single time. Every value past this point and the program crashes. Why is this? is it the end of the programs allocated memory space?
main()
{
int ar[4];
ar[0] = 99;
ar[1] = 45;
printf("array: %d, %d random value: %d", ar[0], ar[1], ar[55]);
}
Edit: I also found that if I alter this value that always ends up being 0 (i.e. ar[55] = 1000) then the return code for the program goes up.
... just to see what happens when you index past the end of an array
Trying to access out of bound memory, invokes undefined behavior. Anything can happen, just anything.
In your case, for some reason, the memory address for upto index 52 is accessible from the process, so it allows the access. Index past 52 points to a memory region not allocated to your process address space and thus, raises the access violation leading to segfault. This is not a deterministic behaviour, at all and there's no way you can rely on the output of a program invoking UB.
Accessing array elements beyond array boundaries (before 0 or from its size up) is undefined behavior. It may or may not produce values, it may cause the program to end abruptly, it may cause your system to stop, restart or catch fire...
Modern systems try to confine undefined behavior within reasonable limits via memory protection, user space limitations etc. but even user space code errors can have dire consequences:
pacemaker messing with its timing values can cause premature death ;
banking software overflowing array boundaries can overwrite account balance information crediting some random account with untold amounts of dollars.
your self driving car could behave worse than drunk drivers...
think of nuclear power-plant control software, airplane instruments, military stuff...
There is no question undefined behavior should be avoided.
Regarding the exit status, your program uses an obsolete syntax for the definition of main(), implicit return type, which is no longer supported in C99 and later, but does not return anything, which means its return value can be any random value, including a different value for every execution. C99 specified a kludge for the main() function and forces an implicit return 0; at the end of main(), but relying on it is bad style.
Similarly, invoking printf() without a proper prototype is undefined behavior. You should include <stdio.h> before the definition of function main().
Lastly, ar[0] and ar[1] are initialized in main(), but ar[2] and ar[3] are not. Be aware that accessing uninitialized values also has undefined behavior. The values can be anything at all, what you describe as random values, but on some systems, they could be trap values, causing undefined behavior by just reading them.
Some very handy tools are available to track this kind of problems in simple and complex programs, most notably valgrind. If you are curious about this subject, You should definitely look at it.
This question already has answers here:
How dangerous is it to access an array out of bounds?
(12 answers)
Closed 7 years ago.
I have seen several posts on the dangers of WRITING outside of array boundaries. I was wondering though, is there any problem with READING outside of them? My reason for this is as follows:
I have commands and data in a randomly generated array, but sometimes the commands require uncertain amounts of data. Do I need to put checks in each command's subroutine so that data is not read from outside the string, or can I temporarily read from outside the array, and realloc later?
According to the (draft) C standard (ISO 9899:201x) Annex J, undefined behaviour includes :
— Addition or subtraction of a pointer into, or just beyond, an array
object and an integer type produces a result that points just beyond
the array object and is used as the operand of a unary * operator that
is evaluated (6.5.6).
— An array subscript is out of range, even if an object is apparently
accessible with the given subscript
In C this expression a[2] is equivalent to *(a+2)
sometimes the commands require uncertain amounts of data
Initially, an amount of memory needs to be allocated based on the best guess (using malloc or calloc).
can I temporarily read junk from outside the array, and realloc later
As the program proceeds, the previous uncertainty about the amount of data presumably gives way to certainty as to how much new memory is needed to store new data.
Then, realloc needs to be called with the newly available information.
But the program always needs to check the return values of alloc routines and keep proper accounting of the valid range(s) of memory blocks that it was given and should keep within those bounds.
Otherwise it would lead to undefined behavior
Better yet, you can use c++ standard library to do the memory management for you. The facilities provided by standard library allocate and expand memory as needed so you don't have to do it.
Why can i do this?
int array[4]; // i assume this creates an array with 4 indexes
array[25]=1; // i have assigned an index larger than the int declaration
NSLog(#"result: %i", array[25]); // this prints "1" to the screen
Why does this work, if the index exceeds the declaration? what is the significance of the number in the declaration if it has no effect on what you can actually do with the array?
Thanks.
You are getting undefined behavior. It could print anything, it could crash, it could burst into singing (okay that isn't likely but you get the idea).
If it happens to write to a location that is mapped with the adequate permissions it will work. Until one day when it won't because of a different layout.
it is undefined. some OS will give you segmentation fault, while some tolerate this. anyhow, exceeding the array's size should be avoided.
An array is really just a pointer to the start of a contiguous, allocated block of memory.
In this case, you have allocated 4 ints worth of memory.
So if you went array[2] it would think "the memory at array + sizeof(int) * 2"
Change the 2 to 25, and you're just looking at 25 int's worth of memory past the start. Since there's no checks to verify you're in bounds (either when assigning or printing) it works.
There's nothing to say somehting else might be allocated there though!
The number in the decleration determines how many memory should be reserved, in that case 4 * sizeof(int).
If you write to memory out of the bounds, this is possible but not recommended. In C you can access any point in memory available to your program, but if you write to that if it's not reserved for that thing, you can cause memory corruption. Arrays are pointers (but not the other way around).
The behavior depends on the compiler, the platform and some randomness. Don't do it.
It's doing very bad things. If the array is declared locally to a function, you're probably writing on stack locations above the current function's stack frame. If it is declared globally, you're probably writing on memory allocated to adjacent variables. Either way, it is "working" by pure luck.
It is possible your C compiler had padded your array for memory alignment purposes, and by luck your array overrun just happens to still be within the rounded-up allocation. Don't rely on it though.
This is unsafe programming. It really should be avoided because it may not crash your program. Which is really the best thing you could hope for. It could give you garbage results. These are unpredictable and could really screw up your program. However since you don't know that is wrong because it not crashing it will ruin the integrity of your data. Since there is no try/catch with C you really should check inputs. Remember scanf returns an int.
C by design does not perform array bounds checking. It was designed as a systems level language, and the overhead of explicit run-time bounds checking could be prohibitive in a great many cases. Consequently C will allow 'dangerous' code and must be used carefully. If you need something 'safer' then C# or Java may be more appropriate, but there is a definite performance hit.
Automated static analysis tools may help, and there are run-time bounds checking tools for use in development and debugging.
In C an array is a contiguous block of memory. By accessing the array out-of-bounds, you are simply accessing memory beyond the end of the array. What accessing such memory will do is non-deterministic, it may be junk, it may belong to an adjacent variable, or it may belong to the variables of the calling function or above. It maybe a return address for the current function or a calling function. In a memory protected OS such as Windows or Linux, if you access so far out of bounds as to be no longer within the address range assigned to the process, a fault exception will occur.