std::strlen doesn't handle c strings that are not \0 terminated. Is there a safe version of it?
PS I know that in c++ std::string should be used instead of c strings, but in this case my string is stored in a shared memory.
EDIT
Ok, I need to add some explanation.
My application is getting a string from a shared memory (which is of some length), therefore it could be represented as an array of characters. If there is a bug in the library writing this string, then the string would not be zero terminated, and the strlen could fail.
You've added that the string is in shared memory. That's guaranteed readable, and of fixed size. You can therefore use size_t MaxPossibleSize = startOfSharedMemory + sizeOfSharedMemory - input; strnlen(input, MaxPossibleSize) (mind the extra n in strnlen).
This will return MaxPossibleSize if there's no \0 in the shared memory following input, or the string length if there is. (The maximal possible string length is of course MaxPossibleSize-1, in case the last byte of shared memory is the first \0)
C strings that are not null-terminated are not C strings, they are simply arrays of characters, and there is no way of finding their length.
If you define a c-string as
char* cowSays = "moo";
then you autmagically get the '\0' at the end and strlen would return 3. If you define it like:
char iDoThis[1024] = {0};
you get an empty buffer (and array of characters, all of which are null characters). You can then fill it with what you like as long as you don't over-run the buffer length. At the start strlen would return 0, and once you have written something you would also get the correct number from strlen.
You could also do this:
char uhoh[100];
int len = strlen(uhoh);
but that would be bad, because you have no idea what is in that array. It could hit a null character you might not. The point is that the null character is the defined standard manner to declare that the string is finished.
Not having a null character means by definition that the string is not finished. Changing that will break the paradigm of how the string works. What you want to do is make up your own rules. C++ will let you do that, but you will have to write a lot of code yourself.
EDIT
From your newly added info, what you want to do is loop over the array and check for the null character by hand. You should also do some validation if you are expecting ASCII characters only (especially if you are expecting alpha-numeric characters). This assumes that you know the maximum size.
If you do not need to validate the content of the string then you could use one of the strnlen family of functions:
http://msdn.microsoft.com/en-us/library/z50ty2zh%28v=vs.80%29.aspx
http://linux.about.com/library/cmd/blcmdl3_strnlen.htm
size_t safe_strlen(const char *str, size_t max_len)
{
const char * end = (const char *)memchr(str, '\0', max_len);
if (end == NULL)
return max_len;
else
return end - str;
}
Yes, since C11:
size_t strnlen_s( const char *str, size_t strsz );
Located in <string.h>
Get a better library, or verify the one you have - if you can't trust you library to do what it says it will, then how the h%^&l do you expect your program to?
Thats said, Assuming you know the length of the buiffer the string resides, what about
buffer[-1+sizeof(buffer)]=0 ;
x = strlen(buffer) ;
make buffer bigger than needed and you can then test the lib.
assert(x<-1+sizeof(buffer));
C11 includes "safe" functions such as strnlen_s. strnlen_s takes an extra maximum length argument (a size_t). This argument is returned if a null character isn't found after checking that many characters. It also returns the second argument if a null pointer is provided.
size_t strnlen_s(const char *, size_t);
While part of C11, it is recommended that you check that your compiler supports these bounds-checking "safe" functions via its definition of __STDC_LIB_EXT1__. Furthermore, a user must also set another macro, __STDC_WANT_LIB_EXT1__, to 1, before including string.h, if they intend to use such functions. See here for some Stack Overflow commentary on the origins of these functions, and here for C++ documentation.
GCC and Clang also support the POSIX function strnlen, and provide it within string.h. Microsoft too provide strnlen which can also be found within string.h.
You will need to encode your string. For example:
struct string
{
size_t len;
char *data;
} __attribute__(packed);
You can then accept any array of characters if you know the first sizeof(size_t) bytes of the shared memory location is the size of the char array. It gets tricky when you want to chain arrays this way.
It's better to trust your other end to terminate it's strings or roll your own strlen that does not go outside the bounderies of the shared memory segment (providing you know at least the size of that segment).
If you need to get the size of shared memory, try to use
// get memory size
struct shmid_ds shm_info;
size_t shm_size;
int shm_rc;
if((shm_rc = shmctl(shmid, IPC_STAT, &shm_info)) < 0)
exit(101);
shm_size = shm_info.shm_segsz;
Instead of using strlen you can use shm_size - 1 if you are sure that it is null terminated. Otherwise you can null terminate it by data[shm_size - 1] = '\0'; then use strlen(data);
a simple solution:
buff[BUFF_SIZE -1] = '\0'
ofc this will not tell you if the string originally was exactly BUFF_SIZE-1 long or it was just not terminated... so you need xtra logic for that.
How about this portable nugget:
int safeStrlen(char *buf, int max)
{
int i;
for(i=0;buf[i] && i<max; i++){};
return i;
}
As Neil Butterworth already said in his answer above: C-Strings which are not terminated by a \0 character, are no C-Strings!
The only chance you do have is to write an immutable Adaptor or something which creates a valid copy of the C-String with a \0 terminating character. Of course, if the input is wrong and there is an C-String defined like:
char cstring[3] = {'1','2','3'};
will indeed result in unexpected behavior, because there can be something like 123#4x\0 in the memory now. So the result of of strlen() for example is now 6 and not 3 as expected.
The following approach shows how to create a safe C-String in any case:
char *createSafeCString(char cStringToCheck[]) {
//Cast size_t to integer
int size = static_cast<int>(strlen(cStringToCheck)) ;
//Initialize new array out of the stack of the method
char *pszCString = new char[size + 1];
//Copy data from one char array to the new
strncpy(pszCString, cStringToCheck, size);
//set last character to the \0 termination character
pszCString[size] = '\0';
return pszCString;
}
This ensures that if you manipulate the C-String to not write on the memory of something else.
But this is not what you wanted. I know, but there is no other way to achieve the length of a char array without termination. This isn't even an approach. It just ensures that even if the User (or Dev) is inserting ***** to work fine.
Related
According to this StackOverflow comment strncpy should never be used with a non-fixed length array.
strncpy should never be used unless you're working with fixed-width, not-necessarily-terminated string fields in structures/binary files. – R.. Jan 11 '12 at 16:22
I understand that it is redundant if you are dynamically allocating memory for the string but is there a reason why it would be bad to use strncpy over strcpy
strncpy will copy data up to the limit you specify--but if it reaches that limit before the end of the string, it'll leave the destination unterminated.
In other words, there are two possibilities with strncpy. One is that you get behavior precisely like strcpy would have produced anyway (except slower, since it fills the remainder of the destination buffer with NULs, which you virtually never actually want or care about). The other is that it produces a result you generally can't put to any real use.
If you want to copy a string up to a maximum length into a fixed-length buffer, you can (for example) use sprintf to do the job:
char buffer[256];
sprintf(buffer, "%255s", source);
Unlike strncpy, this always zero-terminates the result, so the result is always usable as a string.
If you don't want to use sprintf (or similar), I'd advise just writing a function that actually does what you want, something on this general order:
void copy_string(char const *dest, char const *source, size_t max_len) {
size_t i;
for (i=0; i<max_len-1 && source[i]; i++)
dest[i] = source[i];
dest[i] = '\0';
}
Since you've tagged this as C++ (in addition to C): my advice would be to generally avoid this whole mess in C++ by just using std::string.
If you really have to work with NUL-terminated sequences in C++, you might consider another possibility:
template <size_t N>
void copy_string(char const (&dest)[N], char const *source) {
size_t i;
for (i=0; i<N-1 && source[i]; i++)
dest[i] = source[i];
dest[i] = '\0';
}
This only works when the destination is an actual array (not a pointer), but for that case, it gets the compiler to deduce the size of the array, instead of requiring the user to pass it explicitly. This will generally make the code a tiny bit faster (less overhead in the function call) and much harder to screw up and pass the wrong size.
The argument against using strncpy is that it does not guarentee that your string will be null terminated.
The less error prone way to copy a string in C when using non-fixed length arrays is to use snprintf which does guarentee null termination of your string.
A good Blog Post Commenting on *n* functions.
These functions let you specify the size of the buffer but – and this is really important – they do not guarantee null-termination. If you ask these functions to write more characters than will fill the buffer then they will stop – thus avoiding the buffer overrun – but they will not null-terminate the buffer.
Which means that the use of strncpy and other such functions when not dealing with fixed arrays introduces unnessisary risk of non-null terminated strings which can be time-bombs in your code.
char * strncpy ( char * destination, const char * source, size_t num );
Limitations of strncpy():
It doesn't put a null-terminator on the destination string if it is completely filled. And, no null-character is implicitly appended at the end of destination if source is longer than num.
If num is greater than the length of source string, the destination string is padded with null characters up to num length.
Like strcpy, it is not a memory-safe operation. Because it does not check for sufficient space in destination before it copies source, it is a potential cause of buffer overruns.
Refer: Why should you use strncpy instead of strcpy?
We have 2 versions for copy string from one to another
1> strcpy
2> strncpy
These two versions is used for fixed and non-fixed length array. The strcpy don't check the upper bound for destination string when copy string, strncpy will check it. When the destination string is reached to this upper bound, the function strncpy will return error code, in the meantime the function strcpy will cause some effect in memory of the current process and terminate the process immediately. So that the strncpy is more secure than strcpy
i am learning c/c++ recently.but i don't understand the difference between
int a(chat* str,int len)
{
cout<<str<<len;
}
and
int a(char* str)
{
cout<<str<<strlen(str);
}
When you pass char* without a length, how would you know how many elements to process? char* means a pointer to a character. When you pass a pointer, you have no idea (and cannot find out) how much memory (if any) was allocated for the pointer.
That's why C-strings use are null-terminated (they end with a '\0' character), so you can detect their length by iterating the pointer. Hence, if you want to use a pointer without giving the length of its allocated memory, you need to obey some conventions. But in general, e.g. when passing a buffer, you shouldn't expect any end-signalling character, so in this case you need to pass the length, otherwise may end up reading/writing out of bounds.
For your particular example, you're fine with passing only a pointer provided you use your function only on C-strings, since strlen(str) uses this convention of counting until encountering a '\0'.
Buffer overflows are one the most messy and nightmarish programming errors, which can result in serious security issues. That's why you should try (whenever possible) to use std::string from the C++ standard library instead of C-style char* strings.
A C-String should always contain a termination character, we call it null character. It's technically 0 (not the number 0, but ASCII 0)
When we create a char* and initialize it with some text, it automatically adds the '\0' to the end.
char* c = "Hello";
This will create an array of char with six elements. Yes, six elements.
c = {'H', 'e', 'l', 'l', 'o', '\0'}
When you print c, it will search till it finds that '\0'. What if someone replaces it.
c[5] = '!';
Then the system can't determine the end of the text. Then it will keep on reading the memory (which does not belong to that variable, or may be even the program) until it hits a null char.
That is the main reason to pass the size (or number or chars to read) to a function.
On the other hand, if you need to read some data from a stream, you can use a buffer. In that case, you should specify how many bytes to read, in that way you will not cause buffer overflows.
Above answers are to the point. So I'm going to discuss other perspective behind of practise of passing length along with char *.
As others said, not always, the string pointed by char * end up with \0. Only when the string ends with \0 strlen() would actually work. There are certain use-cases for example binary coding, where data is represented as string. In such case, char * would not end with \0. Besides, there can be certain use-cases to read / write only up to certain length / size. In such case, it is always necessary to test whether the input length is within the range of length of total string. So as a common case, length has been passed explicitly, which can be used in any way as desired by the caller.
There is a function which sends data to the server:
int send(
_In_ SOCKET s,
_In_ const char *buf,
_In_ int len,
_In_ int flags
);
Providing length seems to me a little bit weird. I need to write a function, sending a line to the server and wrapping this one such that we don't have to provide length explicitly. I'm a Java-developer and in Java we could just invoke String::length() method, but now we're not in Java. How can I do that, unless providing length as a template parameter? For instance:
void sendLine(SOCKET s, const char *buf)
{
}
Is it possible to implement such a function?
Use std string:
void sendLine(SOCKET s, const std::string& buf) {
send (s, buf.c_str(), buf.size()+1, 0); //+1 will also transmit terminating \0.
}
On a side note: your wrapper function ignores the return value and doesn't take any flags.
you can retrieve the length of C-string by using strlen(const char*) function.
make sure all the strings are null terminated and keep in mind that null-termination (the length grows by 1)
Edit: My answer originally only mentioned std::string. I've now also added std::vector<char> to account for situations where send is not used for strictly textual data.
First of all, you absolutely need a C++ book. You are looking for either the std::string class or for std::vector<char>, both of which are fundamental elements of the language.
Your question is a bit like asking, in Java, how to avoid char[] because you never heard of java.lang.String, or how to avoid arrays in general because you never heard of java.util.ArrayList.
For the first part of this answer, let's assume you are dealing with just text output here, i.e. with output where a char is really meant to be a text character. That's the std::string use case.
Providing lenght seems to me a little bit wierd.
That's the way strings work in C. A C string is really a pointer to a memory location where characters are stored. Normally, C strings are null-terminated. This means that the last character stored for the string is '\0'. It means "the string stops here, and if you move further, you enter illegal territory".
Here is a C-style example:
#include <string.h>
#include <stdio.h>
void f(char const* s)
{
int l = strlen(s); // l = 3
printf(s); // prints "foo"
}
int main()
{
char* test = new char[4]; // avoid new[] in real programs
test[0] = 'f';
test[1] = 'o';
test[2] = 'o';
test[3] = '\0';
f(test);
delete[] test;
}
strlen just counts all characters at the specified position in memory until it finds '\0'. printf just writes all characters at the specified position in memory until it finds '\0'.
So far, so good. Now what happens if someone forgets about the null terminator?
char* test = new char[3]; // don't do this at home, please
test[0] = 'f';
test[1] = 'o';
test[2] = 'o';
f(test); // uh-oh, there is no null terminator...
The result will be undefined behaviour. strlen will keep looking for '\0'. So will printf. The functions will try to read memory they are not supposed to. The program is allowed to do anything, including crashing. The evil thing is that most likely, nothing will happen for a while because a '\0' just happens to be stored there in memory, until one day you are not so lucky anymore.
That's why C functions are sometimes made safer by requiring you to explicitly specify the number of characters. Your send is such a function. It works fine even without null-terminated strings.
So much for C strings. And now please don't use them in your C++ code. Use std::string. It is designed to be compatible with C functions by providing the c_str() member function, which returns a null-terminated char const * pointing to the contents of the string, and it of course has a size() member function to tell you the number of characters without the null-terminated character (e.g. for a std::string representing the word "foo", size() would be 3, not 4, and 3 is also what a C function like yours would probably expect, but you have to look at the documentation of the function to find out whether it needs the number of visible characters or number of elements in memory).
In fact, with std::string you can just forget about the whole null-termination business. Everything is nicely automated. std::string is exactly as easy and safe to use as java.lang.String.
Your sendLine should thus become:
void sendLine(SOCKET s, std::string const& line)
{
send(s, line.c_str(), line.size());
}
(Passing a std::string by const& is the normal way of passing big objects in C++. It's just for performance, but it's such a widely-used convention that your code would look strange if you just passed std::string.)
How can I do that, unless providing lenght as a template parameter?
This is a misunderstanding of how templates work. With a template, the length would have to be known at compile time. That's certainly not what you intended.
Now, for the second part of the answer, perhaps you aren't really dealing with text here. It's unlikely, as the name "sendLine" in your example sounds very much like text, but perhaps you are dealing with raw data, and a char in your output does not represent a text character but just a value to be interpreted as something completely different, such as the contents of an image file.
In that case, std::string is a poor choice. Your output could contain '\0' characters that do not have the meaning of "data ends here", but which are part of the normal contents. In other words, you don't really have strings anymore, you have a range of char elements in which '\0' has no special meaning.
For this situation, C++ offers the std::vector template, which you can use as std::vector<char>. It is also designed to be usable with C functions by providing a member function that returns a char pointer. Here's an example:
void sendLine(SOCKET s, std::vector<char> const& data)
{
send(s, &data[0], data.size());
}
(The unusual &data[0] syntax means "pointer to the first element of the encapsulated data. C++11 has nicer-to-read ways of doing this, but &data[0] also works in older versions of C++.)
Things to keep in mind:
std::string is like String in Java.
std::vector is like ArrayList in Java.
std::string is for a range of char with the meaning of text, std::vector<char> is for a range of char with the meaning of raw data.
std::string and std::vector are designed to work together with C APIs.
Do not use new[] in C++.
Understand the null termination of C strings.
I have to write a function that fills a char* buffer for an assigned length with the content of a string. If the string is too long, I just have to cut it. The buffer is not allocated by me but by the user of my function. I tried something like this:
int writebuff(char* buffer, int length){
string text="123456789012345";
memcpy(buffer, text.c_str(),length);
//buffer[length]='\0';
return 1;
}
int main(){
char* buffer = new char[10];
writebuff(buffer,10);
cout << "After: "<<buffer<<endl;
}
my question is about the terminator: should it be there or not? This function is used in a much wider code and sometimes it seems I get problems with strange characters when the string needs to be cut.
Any hints on the correct procedure to follow?
A C-style string must be terminated with a zero character '\0'.
In addition you have another problem with your code - it may try to copy from beyond the end of your source string. This is classic undefined behavior. It may look like it works, until the one time that the string is allocated at the end of a heap memory block and the copy goes off into a protected area of memory and fails spectacularly. You should copy only until the minimum of the length of the buffer or the length of the string.
P.S. For completeness here's a good version of your function. Thanks to Naveen for pointing out the off-by-one error in your terminating null. I've taken the liberty of using your return value to indicate the length of the returned string, or the number of characters required if the length passed in was <= 0.
int writebuff(char* buffer, int length)
{
string text="123456789012345";
if (length <= 0)
return text.size();
if (text.size() < length)
{
memcpy(buffer, text.c_str(), text.size()+1);
return text.size();
}
memcpy(buffer, text.c_str(), length-1);
buffer[length-1]='\0';
return length-1;
}
If you want to treat the buffer as a string you should NULL terminate it. For this you need to copy length-1 characters using memcpy and set the length-1 character as \0.
it seems you are using C++ - given that, the simplest approach is (assuming that NUL termination is required by the interface spec)
int writebuff(char* buffer, int length)
{
string text = "123456789012345";
std::fill_n(buffer, length, 0); // reset the entire buffer
// use the built-in copy method from std::string, it will decide what's best.
text.copy(buffer, length);
// only over-write the last character if source is greater than length
if (length < text.size())
buffer[length-1] = 0;
return 1; // eh?
}
char * Buffers must be null terminated unless you are explicitly passing out the length with it everywhere and saying so that the buffer is not null terminated.
Whether or not you should terminate the string with a \0 depends on the specification of your writebuff function. If what you have in buffer should be a valid C-style string after calling your function, you should terminate it with a \0.
Note, though, that c_str() will terminate with a \0 for you, so you could use text.size() + 1 as the size of the source string. Also note that if length is larger than the size of the string, you will copy further than what text provides with your current code (you can use min(length - 2, text.size() + 1/*trailing \0*/) to prevent that, and set buffer[length - 1] = 0 to cap it off).
The buffer allocated in main is leaked, btw
my question is about the terminator: should it be there or not?
Yes. It should be there. Otherwise how would you later know where the string ends? And how would cout would know? It would keep printing garbage till it encounters a garbage whose value happens to be \0. Your program might even crash.
As a sidenote, your program is leaking memory. It doesn't free the memory it allocates. But since you're exiting from the main(), it doesn't matter much; after all once the program ends, all the memory would go back to the OS, whether you deallocate it or not. But its good practice in general, if you don't forget deallocating memory (or any other resource ) yourself.
I agree with Necrolis that strncpy is the way to go, but it will not get the null terminator if the string is too long. You had the right idea in putting an explicit terminator, but as written your code puts it one past the end. (This is in C, since you seemed to be doing more C than C++?)
int writebuff(char* buffer, int length){
char* text="123456789012345";
strncpy(buffer, text, length);
buffer[length-1]='\0';
return 1;
}
It should most defiantly be there*, this prevents strings that are too long for the buffer from filling it completely and causing an overflow later on when its accessed. though imo, strncpy should be used instead of memcpy, but you'll still have to null terminate it. (also your example leaks memory).
*if you're ever in doubt, go the safest route!
First, I don't know whether writerbuff should terminate the string or not. That is a design question, to be answered by the person who decided that writebuff should exist at all.
Second, taking your specific example as a whole, there are two problems. One is that you pass an unterminated string to operator<<(ostream, char*). Second is the commented-out line writes beyond the end of the indicated buffer. Both of these invoke undefined behavior.
(Third is a design flaw -- can you know that length is always less than the length of text?)
Try this:
int writebuff(char* buffer, int length){
string text="123456789012345";
memcpy(buffer, text.c_str(),length);
buffer[length-1]='\0';
return 1;
}
int main(){
char* buffer = new char[10];
writebuff(buffer,10);
cout << "After: "<<buffer<<endl;
}
In main(), you should delete the buffer you allocated with new., or allocate it statically (char buf[10]). Yes, it's only 10 bytes, and yes, it's a memory "pool," not a leak, since it's a one-time allocations, and yes, you need that memory around for the entire running time of the program. But it's still a good habit to be into.
In C/C++ the general contract with character buffers is that they be null-terminiated, so I would include it unless I had been explicitly told not to do it. And if I did, I would comment it, and maybe even use a typedef or name on the char * parameter indicating that the result is a string that is not null terminated.
I want to copy a string into a char array, and not overrun the buffer.
So if I have a char array of size 5, then I want to copy a maximum of 5 bytes from a string into it.
what's the code to do that?
This is exactly what std::string's copy function does.
#include <string>
#include <iostream>
int main()
{
char test[5];
std::string str( "Hello, world" );
str.copy(test, 5);
std::cout.write(test, 5);
std::cout.put('\n');
return 0;
}
If you need null termination you should do something like this:
str.copy(test, 4);
test[4] = '\0';
First of all, strncpy is almost certainly not what you want. strncpy was designed for a fairly specific purpose. It's in the standard library almost exclusively because it already exists, not because it's generally useful.
Probably the simplest way to do what you want is with something like:
sprintf(buffer, "%.4s", your_string.c_str());
Unlike strncpy, this guarantees that the result will be NUL terminated, but does not fill in extra data in the target if the source is shorter than specified (though the latter isn't a major issue when the target length is 5).
Use function strlcpybroken link, and material not found on destination site if your implementation provides one (the function is not in the standard C library), yet it is rather widely accepted as a de-facto standard name for a "safe" limited-length copying function for zero-terminated strings.
If your implementation does not provide strlcpy function, implement one yourself. For example, something like this might work for you
char *my_strlcpy(char *dst, const char *src, size_t n)
{
assert(dst != NULL && src != NULL);
if (n > 0)
{
char *pd;
const char *ps;
for (--n, pd = dst, ps = src; n > 0 && *ps != '\0'; --n, ++pd, ++ps)
*pd = *ps;
*pd = '\0';
}
return dst;
}
(Actually, the de-facto accepted strlcpy returns size_t, so you might prefer to implement the accepted specification instead of what I did above).
Beware of the answers that recommend using strncpy for that purpose. strncpy is not a safe limited-length string copying function and is not supposed to be used for that purpose. While you can force strncpy to "work" for that purpose, it is still akin to driving woodscrews with a hammer.
Update: Thought I would try to tie together some of the answers, answers which have convinced me that my own original knee-jerk strncpy response was poor.
First, as AndreyT noted in the comments to this question, truncation methods (snprintf, strlcpy, and strncpy) are often not a good solution. Its often better to check the size of the string string.size() against the buffer length and return/throw an error or resize the buffer.
If truncation is OK in your situation, IMHO, strlcpy is the best solution, being the fastest/least overhead method that ensures null termination. Unfortunately, its not in many/all standard distributions and so is not portable. If you are doing a lot of these, it maybe worth providing your own implementation, AndreyT gave an example. It runs in O(result length). Also the reference specification returns the number of bytes copied, which can assist in detecting if the source was truncated.
Other good solutions are sprintf and snprintf. They are standard, and so are portable and provide a safe null terminated result. They have more overhead than strlcpy (parsing the format string specifier and variable argument list), but unless you are doing a lot of these you probably won't notice the difference. It also runs in O(result length). snprintf is always safe and that sprintf may overflow if you get the format specifier wrong (as other have noted, format string should be "%.<N>s" not "%<N>s"). These methods also return the number of bytes copied.
A special case solution is strncpy. It runs in O(buffer length), because if it reaches the end of the src it zeros out the remainder of the buffer. Only useful if you need to zero the tail of the buffer or are confident that destination and source string lengths are the same. Also note that it is not safe in that it doesn't necessarily null terminate the string. If the source is truncated, then null will not be appended, so call in sequence with a null assignment to ensure null termination: strncpy(buffer, str.c_str(), BUFFER_LAST); buffer[BUFFER_LAST] = '\0';
Some nice libc versions provide non-standard but great replacement for strcpy(3)/strncpy(3) - strlcpy(3).
If yours doesn't, the source code is freely available here from the OpenBSD repository.
void stringChange(string var){
char strArray[100];
strcpy(strArray, var.c_str());
}
I guess this should work. it'll copy form string to an char array.
i think snprintf() is much safe and simlest
snprintf ( buffer, 100, "The half of %d is %d", 60, 60/2 );
null character is append it end automatically :)
The most popular answer is fine but the null-termination is not generic. The generic way to null-terminate the char-buffer is:
std::string aString = "foo";
const size_t BUF_LEN = 5;
char buf[BUF_LEN];
size_t len = aString.copy(buf, BUF_LEN-1); // leave one char for the null-termination
buf[len] = '\0';
len is the number of chars copied so it's between 0 and BUF_LEN-1.
std::string my_string("something");
char* my_char_array = new char[5];
strncpy(my_char_array, my_string.c_str(), 4);
my_char_array[4] = '\0'; // my_char_array contains "some"
With strncpy, you can copy at most n characters from the source to the destination. However, note that if the source string is at most n chars long, the destination will not be null terminated; you must put the terminating null character into it yourself.
A char array with a length of 5 can contain at most a string of 4 characters, since the 5th must be the terminating null character. Hence in the above code, n = 4.
std::string str = "Your string";
char buffer[5];
strncpy(buffer, str.c_str(), sizeof(buffer));
buffer[sizeof(buffer)-1] = '\0';
The last line is required because strncpy isn't guaranteed to NUL terminate the string (there has been a discussion about the motivation yesterday).
If you used wide strings, instead of sizeof(buffer) you'd use sizeof(buffer)/sizeof(*buffer), or, even better, a macro like
#define ARRSIZE(arr) (sizeof(arr)/sizeof(*(arr)))
/* ... */
buffer[ARRSIZE(buffer)-1]='\0';
char mystring[101]; // a 100 character string plus terminator
char *any_input;
any_input = "Example";
iterate = 0;
while ( any_input[iterate] != '\0' && iterate < 100) {
mystring[iterate] = any_input[iterate];
iterate++;
}
mystring[iterate] = '\0';
This is the basic efficient design.
If you always have a buffer of size 5, then you could do:
std::string s = "Your string";
char buffer[5]={s[0],s[1],s[2],s[3],'\0'};
Edit:
Of course, assuming that your std::string is large enough.