C signed-integer-based attacks - c++

I was reading this question and one of the comments mentioned C signed-integer-based attacks.
I know what is an int overflow is, but I don't understand how can this be used to attack
a program.
what exactly is meant by attacking a program ? and if you know the program has this bug, how can you use it ?
Is this only limited to signed int.
If yes then why?
and what is the case in C++ ?
my apologies if the question is trivial

For example, there was a bug in the getpeername function from FreeBSD.
To illustrate it, let's take a function void copyFromKernel(char* dest, int size) that copies from a restricted memory area size bytes.
As you might already know, the memcpy function is declared like that:
void * memcpy ( void * destination, const void * source, size_t num );
Where size_t is an unsigned type. If in our function, we do something like:
void copy_from_kernel(void *user_dest, int maxlen) {
int len = KSIZE < maxlen ? KSIZE : maxlen;
memcpy(user_dest, kbuf, len);
}
, where KSIZE is the maximum number of bytes we want to allow for the user to copy. If the caller sends a positive value for maxlen, the function works as expected. But if the caller sends a negative value for maxlen, then the comparison would pass and memcpy's third parameter would be that negative value. As it is converted to unsigned, the number of bytes copied would be huge, thus the caller may get restricted data.

A very simple case could be an overflow on id in the following example. Imagine that id is the id of a user. And you create a ton of fake users or I don't know to create an overflow. And 0 is the id of the administrator.
if (id > 0) {
// you don't have many privileges
} else {
// you are basically root and can do evil stuff.
}

Most "anti-overflow" code is a combination of a range check followed by using the index to access memory. So if you can use int wrap-around to pass the range check (e.g. "if (index < max)" with index being negative) then you can access memory outside the intended target (e.g. "array[index] = value"). This coding mistake is less likely using unsigned.

read about bit representations of signed and unsigned integers.
basically if the int is signed... and if it is accessible by the user (for example, it is loaded from user input)... if the user puts a bigger number then the integer can contain, the int will be in result negative...
if this int is the size of something in the program, it will turn out negative.
there is no such problem with unsigned ints

You can compare it with sql injection.
You supply the input with a very large no which the integer can't store and the program may show undefined behavior.
Whenever we store a bigger no that the capacity of int it wraps around and can turn out to be a negative no, which can be used by attacker. for eg:-
int j ;
cin>>j ;
for(i=0;i<j;++i)
{
// Do some stuff
}
Now if the attacker enters a bigger no j becomes negative beacuse of wrapping around, and the for loop code is skipped by the program.

Related

What does size_t mean in c++? [duplicate]

I'm just wondering should I use std::size_t for loops and stuff instead of int?
For instance:
#include <cstdint>
int main()
{
for (std::size_t i = 0; i < 10; ++i) {
// std::size_t OK here? Or should I use, say, unsigned int instead?
}
}
In general, what is the best practice regarding when to use std::size_t?
A good rule of thumb is for anything that you need to compare in the loop condition against something that is naturally a std::size_t itself.
std::size_t is the type of any sizeof expression and as is guaranteed to be able to express the maximum size of any object (including any array) in C++. By extension it is also guaranteed to be big enough for any array index so it is a natural type for a loop by index over an array.
If you are just counting up to a number then it may be more natural to use either the type of the variable that holds that number or an int or unsigned int (if large enough) as these should be a natural size for the machine.
size_t is the result type of the sizeof operator.
Use size_t for variables that model size or index in an array. size_t conveys semantics: you immediately know it represents a size in bytes or an index, rather than just another integer.
Also, using size_t to represent a size in bytes helps making the code portable.
The size_t type is meant to specify the size of something so it's natural to use it, for example, getting the length of a string and then processing each character:
for (size_t i = 0, max = strlen (str); i < max; i++)
doSomethingWith (str[i]);
You do have to watch out for boundary conditions of course, since it's an unsigned type. The boundary at the top end is not usually that important since the maximum is usually large (though it is possible to get there). Most people just use an int for that sort of thing because they rarely have structures or arrays that get big enough to exceed the capacity of that int.
But watch out for things like:
for (size_t i = strlen (str) - 1; i >= 0; i--)
which will cause an infinite loop due to the wrapping behaviour of unsigned values (although I've seen compilers warn against this). This can also be alleviated by the (slightly harder to understand but at least immune to wrapping problems):
for (size_t i = strlen (str); i-- > 0; )
By shifting the decrement into a post-check side-effect of the continuation condition, this does the check for continuation on the value before decrement, but still uses the decremented value inside the loop (which is why the loop runs from len .. 1 rather than len-1 .. 0).
By definition, size_t is the result of the sizeof operator. size_t was created to refer to sizes.
The number of times you do something (10, in your example) is not about sizes, so why use size_t? int, or unsigned int, should be ok.
Of course it is also relevant what you do with i inside the loop. If you pass it to a function which takes an unsigned int, for example, pick unsigned int.
In any case, I recommend to avoid implicit type conversions. Make all type conversions explicit.
short answer:
Almost never. Use signed version ptrdiff_t or non-standard ssize_t. Use function std::ssize instead of std::size.
long answer:
Whenever you need to have a vector of char bigger that 2gb on a 32 bit system. In every other use case, using a signed type is much safer than using an unsigned type.
example:
std::vector<A> data;
[...]
// calculate the index that should be used;
size_t i = calc_index(param1, param2);
// doing calculations close to the underflow of an integer is already dangerous
// do some bounds checking
if( i - 1 < 0 ) {
// always false, because 0-1 on unsigned creates an underflow
return LEFT_BORDER;
} else if( i >= data.size() - 1 ) {
// if i already had an underflow, this becomes true
return RIGHT_BORDER;
}
// now you have a bug that is very hard to track, because you never
// get an exception or anything anymore, to detect that you actually
// return the false border case.
return calc_something(data[i-1], data[i], data[i+1]);
The signed equivalent of size_t is ptrdiff_t, not int. But using int is still much better in most cases than size_t. ptrdiff_t is long on 32 and 64 bit systems.
This means that you always have to convert to and from size_t whenever you interact with a std::containers, which not very beautiful. But on a going native conference the authors of c++ mentioned that designing std::vector with an unsigned size_t was a mistake.
If your compiler gives you warnings on implicit conversions from ptrdiff_t to size_t, you can make it explicit with constructor syntax:
calc_something(data[size_t(i-1)], data[size_t(i)], data[size_t(i+1)]);
if just want to iterate a collection, without bounds cheking, use range based for:
for(const auto& d : data) {
[...]
}
here some words from Bjarne Stroustrup (C++ author) at going native
For some people this signed/unsigned design error in the STL is reason enough, to not use the std::vector, but instead an own implementation.
size_t is a very readable way to specify the size dimension of an item - length of a string, amount of bytes a pointer takes, etc.
It's also portable across platforms - you'll find that 64bit and 32bit both behave nicely with system functions and size_t - something that unsigned int might not do (e.g. when should you use unsigned long
Use std::size_t for indexing/counting C-style arrays.
For STL containers, you'll have (for example) vector<int>::size_type, which should be used for indexing and counting vector elements.
In practice, they are usually both unsigned ints, but it isn't guaranteed, especially when using custom allocators.
Soon most computers will be 64-bit architectures with 64-bit OS:es running programs operating on containers of billions of elements. Then you must use size_t instead of int as loop index, otherwise your index will wrap around at the 2^32:th element, on both 32- and 64-bit systems.
Prepare for the future!
size_t is returned by various libraries to indicate that the size of that container is non-zero. You use it when you get once back :0
However, in the your example above looping on a size_t is a potential bug. Consider the following:
for (size_t i = thing.size(); i >= 0; --i) {
// this will never terminate because size_t is a typedef for
// unsigned int which can not be negative by definition
// therefore i will always be >= 0
printf("the never ending story. la la la la");
}
the use of unsigned integers has the potential to create these types of subtle issues. Therefore imho I prefer to use size_t only when I interact with containers/types that require it.
When using size_t be careful with the following expression
size_t i = containner.find("mytoken");
size_t x = 99;
if (i-x>-1 && i+x < containner.size()) {
cout << containner[i-x] << " " << containner[i+x] << endl;
}
You will get false in the if expression regardless of what value you have for x.
It took me several days to realize this (the code is so simple that I did not do unit test), although it only take a few minutes to figure the source of the problem. Not sure it is better to do a cast or use zero.
if ((int)(i-x) > -1 or (i-x) >= 0)
Both ways should work. Here is my test run
size_t i = 5;
cerr << "i-7=" << i-7 << " (int)(i-7)=" << (int)(i-7) << endl;
The output: i-7=18446744073709551614 (int)(i-7)=-2
I would like other's comments.
It is often better not to use size_t in a loop. For example,
vector<int> a = {1,2,3,4};
for (size_t i=0; i<a.size(); i++) {
std::cout << a[i] << std::endl;
}
size_t n = a.size();
for (size_t i=n-1; i>=0; i--) {
std::cout << a[i] << std::endl;
}
The first loop is ok. But for the second loop:
When i=0, the result of i-- will be ULLONG_MAX (assuming size_t = unsigned long long), which is not what you want in a loop.
Moreover, if a is empty then n=0 and n-1=ULLONG_MAX which is not good either.
size_t is an unsigned type that can hold maximum integer value for your architecture, so it is protected from integer overflows due to sign (signed int 0x7FFFFFFF incremented by 1 will give you -1) or short size (unsigned short int 0xFFFF incremented by 1 will give you 0).
It is mainly used in array indexing/loops/address arithmetic and so on. Functions like memset() and alike accept size_t only, because theoretically you may have a block of memory of size 2^32-1 (on 32bit platform).
For such simple loops don't bother and use just int.
I have been struggling myself with understanding what and when to use it. But size_t is just an unsigned integral data type which is defined in various header files such as <stddef.h>, <stdio.h>, <stdlib.h>, <string.h>, <time.h>, <wchar.h> etc.
It is used to represent the size of objects in bytes hence it's used as the return type by the sizeof operator. The maximum permissible size is dependent on the compiler; if the compiler is 32 bit then it is simply a typedef (alias) for unsigned int but if the compiler is 64 bit then it would be a typedef for unsigned long long. The size_t data type is never negative(excluding ssize_t)
Therefore many C library functions like malloc, memcpy and strlen declare their arguments and return type as size_t.
/ Declaration of various standard library functions.
// Here argument of 'n' refers to maximum blocks that can be
// allocated which is guaranteed to be non-negative.
void *malloc(size_t n);
// While copying 'n' bytes from 's2' to 's1'
// n must be non-negative integer.
void *memcpy(void *s1, void const *s2, size_t n);
// the size of any string or `std::vector<char> st;` will always be at least 0.
size_t strlen(char const *s);
size_t or any unsigned type might be seen used as loop variable as loop variables are typically greater than or equal to 0.
size_t is an unsigned integral type, that can represent the largest integer on you system.
Only use it if you need very large arrays,matrices etc.
Some functions return an size_t and your compiler will warn you if you try to do comparisons.
Avoid that by using a the appropriate signed/unsigned datatype or simply typecast for a fast hack.
size_t is unsigned int. so whenever you want unsigned int you can use it.
I use it when i want to specify size of the array , counter ect...
void * operator new (size_t size); is a good use of it.

C++ string.length() Strange Behavior

I just came across an extremely strange problem. The function I have is simply:
int strStr(string haystack, string needle) {
for(int i=0; i<=(haystack.length()-needle.length()); i++){
cout<<"i "<<i<<endl;
}
return 0;
}
Then if I call strStr("", "a"), although haystack.length()-needle.length()=-1, this will not return 0, you can try it yourself...
This is because .length() (and .size()) return size_t, which is an unsigned int. You think you get a negative number, when in fact it underflows back to the maximum value for size_t (On my machine, this is 18446744073709551615). This means your for loop will loop through all the possible values of size_t, instead of just exiting immediately like you expect.
To get the result you want, you can explicitly convert the sizes to ints, rather than unsigned ints (See aslgs answer), although this may fail for strings with sufficient length (Enough to over/under flow a standard int)
Edit:
Two solutions from the comments below:
(Nir Friedman) Instead of using int as in aslg's answer, include the header and use an int64_t, which will avoid the problem mentioned above.
(rici) Turn your for loop into for(int i = 0;needle.length() + i <= haystack.length();i ++){, which avoid the problem all together by rearranging the equation to avoid the subtraction all together.
(haystack.length()-needle.length())
length returns a size_t, in other words an unsigned int. Given the size of your strings, 0 and 1 respectively, when you calculate the difference it underflows and becomes the maximum possible value for an unsigned int. (Which is approximately 4.2 billions for a storage of 4 bytes, but could be a different value)
i<=(haystack.length()-needle.length())
The indexer i is converted by the compiler into an unsigned int to match the type. So you're gonna have to wait until i is greater than the max possible value for an unsigned int. It's not going to stop.
Solution:
You have to convert the result of each method to int, like so,
i <= ( (int)haystack.length() - (int)needle.length() )

C++ for-loop condition

I want to know why this loops runs even when result.bad_matches.size()=0
for (int i = 1; i <= result.badmatches.size() - 1; i++)
{
...
}
Also, is there any other way I could stop it from running when badmatches size is 0 without using an if condition?
This depends on the type size() returns. It is probably a standard container and thus will be an unsigned type and those types wrap around on overflow. That means it the result of subtracting one will be the maximum value of that type.
Either use a comparison that doesn't require you to subtract from the size (<, !=) or just use iterators or a for-auto loop. Under any circumstance you should at least use the same type for iterating as the nested size_type of the container and not int.
for(auto& x : result.badmatches) {
// ...
}
use while(result.badmatches.size()) to NOT execute it.
result.badmatches.size()-1 this will be converted to -1. If its an unsigned integer, then -1 is interpreted as 0xFFFFFFFF(on a 32 bit machine). This will make the loop run for 2^32 or 2^64 times. To avoid this, use while() as before IF you're certain that result.badmatches.size() will return 0.
size must be returning an unsigned so 0-1 is getting upgraded to unsigned and so is the left value.
So for int size of 4 bytes, -1 will be represented as 2^32 -1 in unsigned int.
If you don't want this behavior then just cast it like this : static_cast <signed int > (result.badmatches.size());
PS: I've not touched C++ for past 4 years pl. excuse little mistakes.
The right way is:
for (int i=0;i< result.badmatches.size() ;++i)
{
}
If you specifically don't want this loop to enter when the sise of the collection is zero then you could check for ! badmatches.empty() assuming that badmatches is an STL container. However, if you structure your code slightly differently, you'll probably overcome this issue without having to do that:
for (size_t i=0; i < result.badmatches.size(); i++)
{
}
I've changed the int to size_t which is the same type that size() returns (an unsigned integer), changed the initial value to 0 and the comparison so that it will exit if i >= result.badmatches.size() Generally, I'd say that this is the clearest way of presenting an indexed approach as it matches the natural indexing of collections and if you need 1, 2, 3 ... rather than 0, 1, 2 in your loop, then you can address that within it.
If you're still having problems, two questions:
Is there anything in your loop that might alter the value of result.badmatches.size()?
Is your code multithreaded with a possibility that result.badmatches.size() could change by actions on another thread?
After understanding the problem explained by #Prototype Stark #Aga , i came to a more simpler solution , using which i can keep my initial index to 1 .
for(int i=1;i+1<=result.badmatches.size();i++)
Thanks for all the help , it's much clearer now .

What's an efficient way to avoid integer overflow converting an unsigned int to int in C++?

Is the following an efficient and problem free way to convert an unsigned int to an int in C++:
#include <limits.h>
void safeConvert(unsigned int passed)
{
int variable = static_cast<int>(passed % (INT_MAX+1));
...
}
Or is there a better way?
UPDATE
As pointed out by James McNellis it is not undefined to assign an unsigned int > INT_MAX to an integer - rather this is implementation defined. As such the context here is now specifically on my preference is to ensure this integer resets to zero when the unsigned int exceeds INT_MAX.
Original Context
I have a number of unsigned int's used as counters, but want to pass them around as integers in a specific case.
Under normal operation these counts will remain within the bounds of INT_MAX. However to avoid running into undefined implementation specific behaviour should the abnormal (but valid) case occur I want some efficient conversion here.
This should also work:
int variable = passed & INT_MAX;
Under normal operation these counts will remain within the bounds of INT_MAX. However to avoid running into undefined behaviour should the abnormal (but valid) case occur I want some efficient conversion here.
Efficient conversion to what? If all the shared values for int and unsigned int correspond, and you want other unsigned values such as INT_MAX + 1 to each have distinct values, then you can only map them onto the negative integer values. This is done by default, and can be explicitly requested with static_cast<int>(my_unsigned). Otherwise, you could map them all to 0, or -1, or INT_MIN, or throw away the high bit... easiest way is simply: if (my_unsigned > INT_MAX) my_unsigned = XXX, or ...my_unsigned &= INT_MAX to clear the high bit. But will the called functions work properly if the int overflows? Perhaps a better solution would be to use 64-bit ints to begin with?

A warning - comparison between signed and unsigned integer expressions

I am currently working through Accelerated C++ and have come across an issue in exercise 2-3.
A quick overview of the program - the program basically takes a name, then displays a greeting within a frame of asterisks - i.e. Hello ! surrounded framed by *'s.
The exercise - In the example program, the authors use const int to determine the padding (blank spaces) between the greeting and the asterisks. They then ask the reader, as part of the exercise, to ask the user for input as to how big they want the padding to be.
All this seems easy enough, I go ahead ask the user for two integers (int) and store them and change the program to use these integers, removing the ones used by the author, when compiling though I get the following warning;
Exercise2-3.cpp:46: warning: comparison between signed and unsigned integer expressions
After some research it appears to be because the code attempts to compare one of the above integers (int) to a string::size_type, which is fine. But I was wondering - does this mean I should change one of the integers to unsigned int? Is it important to explicitly state whether my integers are signed or unsigned?
cout << "Please enter the size of the frame between top and bottom you would like ";
int padtopbottom;
cin >> padtopbottom;
cout << "Please enter size of the frame from each side you would like: ";
unsigned int padsides;
cin >> padsides;
string::size_type c = 0; // definition of c in the program
if (r == padtopbottom + 1 && c == padsides + 1) { // where the error occurs
Above are the relevant bits of code, the c is of type string::size_type because we do not know how long the greeting might be - but why do I get this problem now, when the author's code didn't get the problem when using const int? In addition - to anyone who may have completed Accelerated C++ - will this be explained later in the book?
I am on Linux Mint using g++ via Geany, if that helps or makes a difference (as I read that it could when determining what string::size_type is).
It is usually a good idea to declare variables as unsigned or size_t if they will be compared to sizes, to avoid this issue. Whenever possible, use the exact type you will be comparing against (for example, use std::string::size_type when comparing with a std::string's length).
Compilers give warnings about comparing signed and unsigned types because the ranges of signed and unsigned ints are different, and when they are compared to one another, the results can be surprising. If you have to make such a comparison, you should explicitly convert one of the values to a type compatible with the other, perhaps after checking to ensure that the conversion is valid. For example:
unsigned u = GetSomeUnsignedValue();
int i = GetSomeSignedValue();
if (i >= 0)
{
// i is nonnegative, so it is safe to cast to unsigned value
if ((unsigned)i >= u)
iIsGreaterThanOrEqualToU();
else
iIsLessThanU();
}
else
{
iIsNegative();
}
I had the exact same problem yesterday working through problem 2-3 in Accelerated C++. The key is to change all variables you will be comparing (using Boolean operators) to compatible types. In this case, that means string::size_type (or unsigned int, but since this example is using the former, I will just stick with that even though the two are technically compatible).
Notice that in their original code they did exactly this for the c counter (page 30 in Section 2.5 of the book), as you rightly pointed out.
What makes this example more complicated is that the different padding variables (padsides and padtopbottom), as well as all counters, must also be changed to string::size_type.
Getting to your example, the code that you posted would end up looking like this:
cout << "Please enter the size of the frame between top and bottom";
string::size_type padtopbottom;
cin >> padtopbottom;
cout << "Please enter size of the frame from each side you would like: ";
string::size_type padsides;
cin >> padsides;
string::size_type c = 0; // definition of c in the program
if (r == padtopbottom + 1 && c == padsides + 1) { // where the error no longer occurs
Notice that in the previous conditional, you would get the error if you didn't initialize variable r as a string::size_type in the for loop. So you need to initialize the for loop using something like:
for (string::size_type r=0; r!=rows; ++r) //If r and rows are string::size_type, no error!
So, basically, once you introduce a string::size_type variable into the mix, any time you want to perform a boolean operation on that item, all operands must have a compatible type for it to compile without warnings.
The important difference between signed and unsigned ints
is the interpretation of the last bit. The last bit
in signed types represent the sign of the number, meaning:
e.g:
0001 is 1 signed and unsigned
1001 is -1 signed and 9 unsigned
(I avoided the whole complement issue for clarity of explanation!
This is not exactly how ints are represented in memory!)
You can imagine that it makes a difference to know if you compare
with -1 or with +9. In many cases, programmers are just too lazy
to declare counting ints as unsigned (bloating the for loop head f.i.)
It is usually not an issue because with ints you have to count to 2^31
until your sign bit bites you. That's why it is only a warning.
Because we are too lazy to write 'unsigned' instead of 'int'.
At the extreme ranges, an unsigned int can become larger than an int.
Therefore, the compiler generates a warning. If you are sure that this is not a problem, feel free to cast the types to the same type so the warning disappears (use C++ cast so that they are easy to spot).
Alternatively, make the variables the same type to stop the compiler from complaining.
I mean, is it possible to have a negative padding? If so then keep it as an int. Otherwise you should probably use unsigned int and let the stream catch the situations where the user types in a negative number.
The primary issue is that underlying hardware, the CPU, only has instructions to compare two signed values or compare two unsigned values. If you pass the unsigned comparison instruction a signed, negative value, it will treat it as a large positive number. So, -1, the bit pattern with all bits on (twos complement), becomes the maximum unsigned value for the same number of bits.
8-bits: -1 signed is the same bits as 255 unsigned
16-bits: -1 signed is the same bits as 65535 unsigned
etc.
So, if you have the following code:
int fd;
fd = open( .... );
int cnt;
SomeType buf;
cnt = read( fd, &buf, sizeof(buf) );
if( cnt < sizeof(buf) ) {
perror("read error");
}
you will find that if the read(2) call fails due to the file descriptor becoming invalid (or some other error), that cnt will be set to -1. When comparing to sizeof(buf), an unsigned value, the if() statement will be false because 0xffffffff is not less than sizeof() some (reasonable, not concocted to be max size) data structure.
Thus, you have to write the above if, to remove the signed/unsigned warning as:
if( cnt < 0 || (size_t)cnt < sizeof(buf) ) {
perror("read error");
}
This just speaks loudly to the problems.
1. Introduction of size_t and other datatypes was crafted to mostly work,
not engineered, with language changes, to be explicitly robust and
fool proof.
2. Overall, C/C++ data types should just be signed, as Java correctly
implemented.
If you have values so large that you can't find a signed value type that works, you are using too small of a processor or too large of a magnitude of values in your language of choice. If, like with money, every digit counts, there are systems to use in most languages which provide you infinite digits of precision. C/C++ just doesn't do this well, and you have to be very explicit about everything around types as mentioned in many of the other answers here.
or use this header library and write:
// |notEqaul|less|lessEqual|greater|greaterEqual
if(sweet::equal(valueA,valueB))
and don't care about signed/unsigned or different sizes