So I ran an Analyze in VS 2017 with my C++ code. It gives me a buffer overrun with the following:
TCHAR *sTemp = new TCHAR[5]();
if (sTemp)
StringCchCopy(sTemp, 5, L"0123456789");
When I step through the code, sTemp is "0123", with the 4th position of course being \0.
When I run Analyze on the code, I get the C6386 error:
Warning C6386 Buffer overrun while writing to 'sTemp': the writable size is 'unsigned int' bytes, but '10' bytes might be written.
Why? I have also tried changing the array to 10 and the StringCchCopy to 5 and still the same error.
The warning refers to the fact, that the source string will not ever fit inside the destination. The source string has a length of 10, the destination a size of 5 code units. It's not relevant at all, that the static analyzer cannot determine the size of the dynamically allocated destination array.
If it were, and it would discover a mismatch between the actual size and the size you claimed, it would raise an error, not a warning.
The docs for StringCchCopy say that the second parameter must be the size of the destination buffer and that the destination buffer must be big enough to hold the source string. You're not checking the return code from the function but I suspect it will be STRSAFE_E_INSUFFICIENT_BUFFER, which means "The copy operation failed due to insufficient buffer space. The destination buffer contains a truncated, null-terminated version of the intended result. In situations where truncation is acceptable, this may not necessarily be seen as a failure condition."
https://learn.microsoft.com/en-us/windows/win32/api/strsafe/nf-strsafe-stringcchcopyw
I guess you're happy with, and expecting, the truncation, but the static analysis tool is seeing that your source string is longer than your destination buffer and triggering the warning.
Related
Suppose you have a char buffer that you want to copy an std::string into. Are there consequences of copying extra data into the buffer, outside of the strings scope, even though the buffer has adequate size?
Example
std::string my_string = "hello";
char my_buffer[128];
memset(my_buffer, 0, 128);
strncpy(my_buffer, my_string.c_str(), 128);
So "hello" gets copied into my_buffer, but so will 123 other bytes of data that comes after my_string. Are there any consequences of this? Is it harmful for the buffer to hold this other data?
but so will 123 other bytes of data that comes after my_string
This assumption is incorrect: strncpy pays attention to null termination of the source string, never reading past null terminator. The remaining data will be set to '\0' characters:
destination is padded with zeros until a total of num characters have been written to it. [reference]
This is in contrast to memcpy, which requires both the source and the destination to be of sufficient size in order to avoid undefined behavior.
OK, let's assume what you wanted is:
strncpy(my_buffer, my_string.c_str(), 128);
Thsi is always a 0-terminated string by definition, so considering:
Copies at most count characters of the character array pointed to by src (including the terminating null character, but not any of the characters that follow the null character) to character array pointed to by dest.
You won't get anything copied after "hello" from the original string, the rest will be 0s:
If, after copying the terminating null character from src, count is not reached, additional null characters are written to dest until the total of count characters have been written.
According to strncpy() description here 1, the copy is done up to the length you provided, for null terminated string, so that when end of the string come before, like in this case, copy is done up to it and no more copy is done, so rest of the "123 bytes" are not copied, and the copy loop terminates
The other answers to this question have addressed what happens with strncpy() (i.e. it will copy your string correctly because it stops at the 0-terminator byte), so perhaps what gets more to the intent of the question would be, what if you had this instead?
memcpy(my_buffer, my_string.c_str(), 128);
In this case, the function (memcpy()) doesn't know about 0-terminated-string semantics, and will always just blindly copy 128 bytes (starting at the address returned by my_string.c_str()) to the address my_buffer. The first 6 (or so) of those bytes will be from my_string's internal buffer, and the remaining bytes will be from whatever happens to be in memory after that.
So the question is, what happens then? Well, this memcpy() call reads from "mystery memory" whose purpose you're not aware of, so you're invoking undefined behavior by doing that, and therefore in principle anything could happen. In practice, the likely result is that your buffer will contain a copy of whatever bytes were read (although you probably won't notice them, since you'll be using string functions that don't look past the 0/terminator byte in your array anyway).
There is a small chance, though, that the "extra" memory bytes that memcpy() read could be part of a memory-page that is marked as off-limits, in which case trying to read from that page would likely cause a segmentation fault.
And finally, there's the real bugaboo of undefined behavior, which is that your C++ compiler's optimizer is allowed to do all kinds of crazy modifications to your code's logic, in the name of making your program more efficient -- and (assuming the optimizer isn't buggy) all of those optimizations will still result in the program running as intended -- as long as the program follows the rules and doesn't invoke undefined behavior. Which is to say, if your program invokes undefined behavior in any way, the optimizations may be applied in ways that are very difficult to predict or understand, resulting in bizarre/unexpected behavior in your program. So the general rule is, avoid undefined behavior like the plague, because even if you think it "should be harmless", there's a very real possibility that it will end up doing things you wouldn't expect it to do, and then you're in for a long, painful debugging session as you try to figure out what's going on.
Assuming you mean definitely copying that data, as your current code would end at a null terminator.
Essentially, no. Whatever data is in there will be used only as string data so unless you then tried to do something weird with that spare data (try to put a function pointer to it and execute it etc.) then it's basically safe.
The issue is with initially copying random data past the end of your original string, that could overflow into protected data that you don't have access to and throw a segfault(Inaccessible memory exception)
Function strncpy will copy everything up to NUL character or specified size. That's it.
In addition to other answers please remember that strncpy does not NUL terminate the dst buffer when the src string length (excluding NUL) is equal to buffer's size given. You should always do something like my_buffer[127]=0 to prevent buffer overruns when handling the string later on. In libbsd there is strlcpy defined that always NUL terminate the buffer.
Check out this snippet:
#include <string.h>
int main(int argc, char ** argv)
{
char buf[8];
char *str = "deadbeef";
strlcpy(buf, str, sizeof(buf));
printf("%s\n", buf);
strncpy(buf, str, sizeof(buf));
printf("%s\n", buf);
return 0;
}
To see the problem compile it with flag -fsanitize=address. The second buffer is not NUL terminated!
Using Borland C++ Builder 2009
I use vswprintf per RAD Studio's Help (F1) :
int vswprintf(wchar_t *buffer, const wchar_t *format, va_list arglist);
Up to now I always provided a big buffer wchar_t OutputStr[1000] and never had any issues. As a test and wanting to do an improvement action, I tried a small buffer wchar_t OutputStr[12] and noticed that the program crashes entirely. Even try{}catch(...){} doesn't catch it. Codeguard reports that a memcpy() fails, which seems to be the internal implementation. I had expected an EOF as return value instead.
When searching online for vswprintf I find the c++ variant takes a buffer size as input, but I can't seem to convince my compiler to use that variant ?
Any idea how to force it using BCB2009 ?
The whole point of the exercise was to implement a fall back scenario for when the buffer is too small in possibly one or two freak situations, so that I can allocate more memory for the function and try again. But this mechanism doesn't seem to work at all.
Not sure how to best test for the exact amount of bytes/characters needed either ?
You can use vswprintf_s. It returns negative value on failure
I have looked at this question as well as these PDFs' 1 and 2, this page and pretty much understand what happens if I do this printf(SOME_TEST_STRING). But what I do not understand is why exactly by ensuring the size of buffer vsnprintf becomes safe as compared to vsprintf?
What happens in these 2 cases ?
Case 1
char buf[3];
vsprint(buf, "%s", args);
Case 2
char buf[3];
vsnprint(buf, sizeof buf, "%s", args);
In case 1, if the string you're formatting has a length of 3 or greater, you have a buffer overrun, vsprintf might write to memory past the storage of the buf array, which is undefined behavior, possibly causing havoc/security concerns/crashes/etc.
In case 2. vsnprintf knows how big the buffer that will contain the result is, and it will make sure not to go past that(instead truncating the result to fit within buf ).
It's because vsnprintf has an additional size_t count parameter that vsprintf (and other non-n *sprintf methods) does not have. The implementation uses this to ensure that the data it writes to your buffer will not run off the end.
Data that runs off the end of a buffer can result in data corruption, or when maliciously exploited can be used as a buffer overrun attack.
The "n" in vsnprintf() means it takes the max size of the output string to avoid a buffer overflow. This makes it safe from buffer overflow, but does not make it safe if the format string comes from unsanitized user input. If your user gives you a giant format string, you'll avoid overflowing the target string, but if the user gives you %s and you don't pass a C string in the argument list at compile time, you are still left with undefined behavior.
I'm not sure what the problem is, since your question basically contains the answer already.
By passing your buffer size to vsnprintf you provide that function with information about your buffer size. The function now knows where the buffer ends and can make sure that it does not write past the end of the buffer.
vsprintf does not have information about buffer size, which is why it does not know where the buffer ends and cannot prevent buffer overflow.
I am using systemC with visual C++ 2008. I wrote a simple hello world program. However I am getting this error repeatedly:
warning C4996: 'sprintf': This function or variable may be unsafe.
Why this is happening? I would appreciate any help.
The compiler warns against sprintf() use because it may cause buffer overflow since it doesn't check buffer's limit. Instead, use snprintf() which never fills the buffer beyond the passed-in limit.
This advice is also given by the manpage:
Because sprintf() and vsprintf() assume an arbitrarily long string, callers must be
careful not to overflow the actual space; this is often impossible to assure. Note that
the length of the strings produced is locale-dependent and difficult to predict. Use
snprintf() and vsnprintf() instead (or asprintf(3) and vasprintf(3)).
It's insecure because - From MSDN
There is no way to limit the number of characters written, which means that code using sprintf is susceptible to buffer overruns. Consider using the related function _snprintf, which specifies a maximum number of characters to be written to buffer, or use _scprintf to determine how large a buffer is required. Also, ensure that format is not a user-defined string.
I assume this is a common way to use sprintf:
char pText[x];
sprintf(pText, "helloworld %d", Count );
but what exactly happens, if the char pointer has less memory allocated, than it will be print to?
i.e. what if x is smaller than the length of the second parameter of sprintf?
i am asking, since i get some strange behaviour in the code that follows the sprintf statement.
It's not possible to answer in general "exactly" what will happen. Doing this invokes what is called Undefined behavior, which basically means that anything might happen.
It's a good idea to simply avoid such cases, and use safe functions where available:
char pText[12];
snprintf(pText, sizeof pText, "helloworld %d", count);
Note how snprintf() takes an additional argument that is the buffer size, and won't write more than there is room for.
This is a common error and leads to memory after the char array being overwritten. So, for example, there could be some ints or another array in the memory after the char array and those would get overwritten with the text.
See a nice detailed description about the whole problem (buffer overflows) here. There's also a comment that some architectures provide a snprintf routine that has a fourth parameter that defines the maximum length (in your case x). If your compiler doesn't know it, you can also write it yourself to make sure you can't get such errors (or just check that you always have enough space allocated).
Note that the behaviour after such an error is undefined and can lead to very strange errors. Variables are usually aligned at memory locations divisible by 4, so you sometimes won't notice the error in most cases where you have written one or two bytes too much (i.e. forget to make place for a NUL), but get strange errors in other cases. These errors are hard to debug because other variables get changed and errors will often occur in a completely different part of the code.
This is called a buffer overrun.
sprintf will overwrite the memory that happens to follow pText address-wise. Since pText is on the stack, sprintf can overwrite local variables, function arguments and the return address, leading to all sorts of bugs. Many security vulnerabilities result from this kind of code — e.g. an attacker uses the buffer overrun to write a new return address pointing to his own code.
The behaviour in this situation is undefined. Normally, you will crash, but you might also see no ill effects, strange values appearing in unrelated variables and that kind of thing. Your code might also call into the wrong functions, format your hard-drive and kill other running programs. It is best to resolve this by allocating more memory for your buffer.
I have done this many times, you will receive memory corruption error. AFAIK, I remember i have done some thing like this:-
vector<char> vecMyObj(10);
vecMyObj.resize(10);
sprintf(&vecMyObj[0],"helloworld %d", count);
But when destructor of vector is called, my program receive memory corruption error, if size is less then 10, it will work successfully.
Can you spell Buffer Overflow ? One possible result will be stack corruption, and make your app vulnerable to Stack-based exploitation.