I am trying to convert wchar_t to char using wcstombs. And it works fine if only 1 value is converted but when more than 1 value is converted it gives unexpected results.
These are two wchar_t values which I want to convert:
wchar_t szBuf[BUFF_LEN];
wchar_t szBuf1[BUFF_LEN];
and converting using wcstombs:
char user[]="";
int length = wcstombs(user,szBuf,250);
char pass[]="";
int length1 = wcstombs(pass,szBuf1,250);
say if I have alice in szBuf and alice123 in szBuf1 then pass will give the correct value but user will have a value like aalice123. What is the error?
You're invoking Undefined Behaviour, because you're writing outside of the buffer bounds. Your user and pass buffers are each of length 1 (they're arrays initialised by a copy of ""). So you're happily writing to random memory and anything could happen.
Your destination buffers are too small.
char user[] = ""; equals char user[1] = "";. This means that you will write to unallocated space and it is undefined behaviour. This is very bad, because there are no guarantees what will happen.
Specify size for your destionation buffers: char user[250] = "";
Related
New to C++, I was goofing around trying to get my head around complex systems like pointers and solving questions. I made this simple program which will concatenate char strings. But it always throws the same error: L'buffer is too small &&0.
Here is my code -
const char* str = "larry";
const char* yo{"is one of the stooges" };
size_t buffer{ std::strlen(str) + std::strlen(yo) + 1 };
std::cout << buffer;
char* buff = new char[buffer];
strcpy_s(buff, strlen(str), str);
strcat(buff, yo);
std::cout << *buff << std::endl;
delete[] buff;
and here is my program output as requested:
27 // the size of buffer
the program then crashes.
Please point out whats wrong.
Thanks
According to the C Standard (K.3.7.1.3 The strcpy_s function)
2 Neither s1 nor s2 shall be a null pointer. s1max shall not be
greater than RSIZE_MAX. s1max shall not equal zero. s1max shall be
greater than strnlen_s(s2, s1max). Copying shall not take place
between objects that overlap.
That is you have to write
strcpy_s(buff, strlen(str) + 1, str);
Otherwise the terminating zero of the string pointed to by str will not be copied. In this case the next call
strcat(buff, yo);
invokes undefined behaviour.
The second parameter to strcpy_s is the size of the destination array, including the 0 byte. However, you pass in the size of the source string, excluding the 0 byte. strcpy_s attempts to copy the 0 byte, which it can't, as the destination array is supposedly (as indicated by the parameter) too small. Pass buffer (not a good name, btw) as the 2nd parameter to strcpy_s to indicate the destination size correctly.
Given below is my sample code :
int function1(unsigned char *out, int length){
unsigned long crypto_out_len = 16;
unsigned char crypto_out[16] = {0};
.......
//produces 16 bytes output & stores in crypto_out
crypto_function(crypto_out, crypto_out_len);
//lets say crypto_output contents after are : "abcdefghijklmnop"
.......
memcpy(out, crypto_out,length);
return 0;
}
function2(){
unsigned char out[10] = {0};
function1(out, 10);
std::pair<unsigned char *,int> map_entry;
map_entry.first = out;
map_entry.second = 10;
}
Now, map_entry.first should contain : "abcdefghij", right?
But it contains "abcdefghij#$%f1^", some garbage associated with it. How should I avoid such unexpected behaviour so that map_entry.first should contain exactly "abcdefghij".
Since you haven't pasted the whole code, I can't be 100% sure but I think I know what's wrong. memcpy() is behaving correctly here, and everything is 100% defined behavior.
In this case, out is a 10-character string without a null terminator. You assign it to unsigned char* that contains no length information, and I suspect you simply don't use the number ten when you are referring to map_entry.first.
If you print it as unsigned char* or construct a std::string with it, C++ expects it to be a null-terminated string. Therefore, it reads it up until the first null character. Now, since out didn't have one it just runs over and starts reading characters on the stack after out which happen to be what you see as garbage.
What you need to do, is make sure that either the string is null-terminated or make sure that you always refer to it specifying the correct length. For the former, you'd want to make out 11-byte long, and leave the last byte as 0:
function2(){
unsigned char out[11] = {0};
function1(out, 10);
std::pair<unsigned char *,int> map_entry;
map_entry.first = out;
map_entry.second = 10;
}
Please also note that C++ will actually stop at the first null character it encounters. If your crypto_function() may output zero bytes in the middle of the string, you should be aware that the string will be truncated at the point.
For the latter, you'd have to use functions that actually allow you to specify the string length, and always pass the length of 10 to those. If you always work with it like this, you don't have to worry about zero bytes from crypto_function().
You are confusing char[] with strings. out does contain your expected data, but its not 0 terminated, so if you try to display it as a string it may look like it contains extra data. If the data is actually strings, you need to correctly 0 terminate them.
I've been trying to convert a const char to a char for the past 30 minutes.
Here's what I've got.
string atr;
getline(cin,atr); // Start off with a string because getline takes nothing else.
const char *buffA = atr.c_str(); // Create a const char of the string converted to a const char.
char *buff = ""; // Create a new char to hold the converted result.
strcat(buff,buffA); // Do the conversion.
parseargs(buff); // Pass the data on.
However, I get an unhandled exception. I have no idea why. I literally just typed 'try' into the console as my only argument.
Try using C++ instead of C idioms:
std::vector<char> data(atr.begin(), atr.end());
data.push_back('\0');
parseargs(&data[0]);
There are two things wrong with your code. First, you
initialize a char* with a string literal. This uses
a deprecated convertion; the type of a string literal is char
const[] (which converts to char const*, not to char*),
because any attempt to modify the literal is undefined behavior.
The second is that your string literal is only one char long,
so even if you could write to it, unless atr was empty, you're
writing beyond the end of the buffer.
You don't tell us anything about parseargs. If it doesn't
modify it's argument, you can just pass it atr.c_str(), and be
done with it. (If it's a legacy function which ignores const,
you may have to use a const_cast here.) If it does modify its
argument (say because it uses strtok), then you'll have to
explicitly push a '\0' onto the end of atr, and then pass it
&atr[0]. (Not a particularly clean solution, but if you're
not using atr afterwards, it should work.)
Both your contents of buff and buffA are in read-only memory of the process.
You will actually need to new your buff like
char* buff = new char[32];
This provides memory from the free-store and you can then strcat the string from buffA to buff.
You should prefer strncat, though to avoid buffer-overruns and delete your buff eventually.
This
char *buff = ""; // Create a new char to hold the converted result.
creates a char * that points to (probably read-only) memory of about 1 byte in extent. This:
strcat(buff,buffA); // Do the conversion.
attempts to overwrite that (probably read-only) memory of 1 or so bytes with an arbitrary string.
The chances are this will promptly crash. If the memory is read only, it will crash immediately. If the memory is not read only it will stomp over random data, resulting in very undefined behaviour.
Why on earth do you want to do that? Does parseArgs actually need a modifiable string? It's parsing arguments, it shouldn't need to change them. If it's really necessary, use a std::vector<char> and pass the address of the first element and hope that all it does is poke the contents of the array, rather than (say) running over the end.
I have got a const char which is made by concatenation like this:
const char *fileName = "background1";
std::stringstream sstm;
sstm << fileName << "-hd.png";
fileName = sstm.str().c_str();
My problem is that the following instruction:
printf("const char = %s size = %d", fileName, sizeof(fileName));
returns:
"const char = background1-hd.png size = 4"
whereas I would expect that it returns:
"const char = background1-hd.png size = 19"
For example, the following gives the convenient result (as there is no concatenation):
const char *fileName2 = "background1-hd";
printf("const char = %s size = %d", fileName2, sizeof(fileName2));
returns:
"const char = background1-hd.png size = 19"
How to avoid this issue and guarantee that the characters will be correctly counted in my concatenated char ?
Thanks !!
sizeof() returns the number of bytes the variable occupies in memory (in this case returns the size of the pointer fileName).
strlen() returns the length of the string (which is what you need).
You could as well try something like:
#include <iostream>
#include <cstdio>
int main()
{
std::string fileName("background1");
fileName.append("-hd.png");
printf("const char = %s size = %d", fileName.c_str(), fileName.length());
return 0;
}
sizeof returns the size of the variable you give to it; it's evaluated at compile time. The "4" is the size of a pointer on your system. You want to use strlen() to determine the length of a string.
The result of sizeof(fileName) is related to fileName being a pointer, not an array. It literally returns the size of a pointer to a constant character string, and on a 32-bit system, all pointers are 32 bits (so sizeof == 4).
What you should use instead is strlen or similar, which will count the characters in the string, up to the trailing null, and return that. The results with strlen in place of sizeof will be about what you expect.
Side-related, with const char strings there is only ever one character per "cell" (actually byte). There are character sets which make for multiple bytes per character, but packing multiple characters into a single byte is quite rare, at least in C-family languages.
sizeof calculates the size of the data type in bytes and not the size of its contents (what it points to). In your example you are calculating the sizeof char* which is 4 bytes on your system. To get the length of a C string use strlen.
There is a distinction in the language between arrays and pointers, even if this distinction seems diluted both by implicit conversions (arrays tend to decay into pointers quite easily), and common statements that they are the same.
How does this even relate to your code?
Well, a string literal is actually an array of constant characters, not a pointer to character(s). In the initialization const char *fileName = "background1"; you are creating a pointer variable that points to the first element of the array ("background1" is decaying into a pointer to the first element), and from there on the variable you are managing is pointer and not the literal.
If you mix this with the fact that sizeof will tell you the size of the variable, you get that in a platform with 32bit pointers and 8 bit chars, sizeof( const char* ) is always 4, regardless of the object that is pointed by that pointer (if there is even one).
Now, if you were treating the literal as what it actually is you would be having a bit more luck there:
const char filename[] = "background1";
assert( sizeof filename == 12 ); // note: NUL character is counted!
const char *fname = filename;
assert( sizeof filename == sizeof( void* ) );
In real code, you are not a so lucky and in many cases the literals have decayed into pointers well before you get a chance of getting the compile time size of the literal, so you cannot ask the compiler to tell you the size. In that case you need to calculate the length of the C style string, which can be done by calling strlen.
strlen has been suggested a number of times already, and for this case it's probably perfectly reasonable.
There is an alternative that will let you use sizeof though:
char fileName[] = "background1";
std::cout << sizeof(fileName) << "\n";
Since you're making fileName an array, it has all the characteristics of an array -- including the fact that your later attempt at assigning to it:
fileName = sstm.str().c_str();
...would fail (won't even compile when fileName is defined as an array). I should add, however, that it seems to me that you'd be better off just using std::string throughout:
std::string fileName("background1");
std::stringstream sstm;
sstm << fileName << "-hd.png";
fileName = sstm.str();
In this case, you can use string's size() or length() member.
I was working with a program that uses a function to set a new value in the registry, I used a const char * to get the value. However, the size of the value is only four bytes. I've tried to use std::string as a parameter instead, it didn't work.
I have a small example to show you what I'm talking about, and rather than solving my problem with the function I'd like to know the reason it does this.
#include <iostream>
void test(const char * input)
{
std::cout << input;
std::cout << "\n" << sizeof("THIS IS A TEST") << "\n" << sizeof(input) << "\n";
/* The code above prints out the size of an explicit string (THIS IS A TEST), which is 15. */
/* It then prints out the size of input, which is 4.*/
int sum = 0;
for(int i = 0; i < 15; i++) //Printed out each character, added the size of each to sum and printed it out.
//The result was 15.
{
sum += sizeof(input[i]);
std::cout << input[i];
}
std::cout << "\n" << sum;
}
int main(int argc, char * argv[])
{
test("THIS IS A TEST");
std::cin.get();
return 0;
}
Output:
THIS IS A TEST
15
4
THIS IS A TEST
15
What's the correct way to get string parameters? Do I have to loop through the whole array of characters and print each to a string (the value in the registry was only the first four bytes of the char)? Or can I use std::string as a parameter instead?
I wasn't sure if this was SO material, but I decided to post here as I consider this to be one of my best sources for programming related information.
sizeof(input) is the size of a const char* What you want is strlen(input) + 1
sizeof("THIS IS A TEST") is size of a const char[]. sizeof gives the size of the array when passed an array type which is why it is 15 .
For std::string use length()
sizeof gives a size based on the type you give it as a parameter. If you use the name of a variable, sizeof still only bases its result on the type of that variable. In the case of char *whatever, it's telling you the size of a pointer to char, not the size of the zero-terminated buffer it's point at. If you want the latter, you can use strlen instead. Note that strlen tells you the length of the content of the string, not including the terminating '\0'. As such, if (for example) you want to allocate space to duplicate a string, you need to add 1 to the result to tell you the total space occupied by the string.
Yes, as a rule in C++ you normally want to use std::string instead of pointers to char. In this case, you can use your_string.size() (or, equivalently, your_string.length()).
std::string is a C++ object, which cannot be passed to most APIs. Most API's take char* as you noticed, which is very different from a std::string. However, since this is a common need, std::string has a function for that: c_str.
std::string input;
const char* ptr = input.c_str(); //note, is const
In C++11, it is now also safe-ish to do this:
char* ptr = &input[0]; //nonconst
and you can alter the characters, but the size is fixed, and the pointer is invalidated if you call any mutating member of the std::string.
As for the code you posted, "THIS IS A TEST" has the type of const char[15], which has a size of 15 bytes. The char* input however, has a type char* (obviously), which has a size of 4 on your system. (Might be other sizes on other systems)
To find the size of a c-string pointed at by a char* pointer, you can call strlen(...) if it is NULL-terminated. It will return the number of characters before the first NULL character.
If the registry you speak of is the Windows registry, it may be an issue of Unicode vs. ASCII.
Modern Windows stores almost all strings as Unicode, which uses 2 bytes per character.
If you try to put a Unicode string into an std::string, it may be getting a 0 (null), which some implementations of string classes treat as "end of string."
You may try using a std::wstring (wide string) or vector< wchar_t > (wide character type). These can store strings of two-byte characters.
sizeof() is also not giving you the value you may think it is giving you. Your system probably runs 32-bit Windows -- that "4" value is the size of the pointer to the first character of that string.
If this doesn't help, please post the specific results that occur when you use std::string or std::wstring (more than saying that it doesn't work).
To put it simply, the size of a const char * != the size of a const char[] (if they are equal, it's by coincidence). The former is a pointer. A pointer, in the case of your system, is 4 bytes REGARDLESS of the datatype. It could be int, char, float, whatever. This is because a pointer is always a memory address, and is numeric. Print out the value of your pointer and you'll see it's actually 4 bytes. const char[] now, is the array itself and will return the length of the array when requested.