I have declare buffer const char* buf;
Later on I want to re assign size using memset
buffer_len = 1024;
memset(buf, '\0', buffer_len);
buf[strlen(buf)-1]='\0';
gives error:
client.cpp:73:30: error: invalid conversion from ‘const void*’ to ‘void*’ [-fpermissive]
In file included from client.cpp:2:0:
/usr/include/string.h:62:14: error: initializing argument 1 of ‘void* memset(void*, int, size_t)’ [-fpermissive]
client.cpp:75:21: error: assignment of read-only location ‘*(buf + (((sizetype)strlen(buf)) + -1u))’
I know it's due to const but is there any alternative or way to perform it event it is const?
The assignment buf[strlen(buf)-1]='\0'; is invalid because you defined buf as const: const char* buf; Read compiler's error message: error: assignment of read-only location.
One point: You set buf with nul \0 so length of buf is 0 (\0 at zero index) then if suppose you don't declare buf as const even then you would be assiging at negative index because strlen(buf) - 1 == 0 - 1 = -1 – Undefined behaviour
memset does not assign size. It fills a buffer with bytes. Filling a buffer declared as const char* makes no sense, since the reason you declare it const is for yourself not to write to it.
You could create a different array instead, since this const does not prevent you from changing the pointer itself.
Reassigning size should probably be called reallocating memory, you can use one of malloc, calloc or others to do it. Or since you've tagged this with c++, probably using the new operator would be the best idea.
Obvious the author is asking a solution for what he want to operate. And simply explanation of the comping error is satisfactory. I do not want to explain why the author's code is not compiling, cause the above answers explain very well about this.
I am try to give a solution if the author really want to do all the operations he meant in his code, though I do not recommend to do that in real world software project.
Use const_cast to remove the const property of the variable:
const char* buf;
...
buffer_len = 1024;
...
char *ptr = const_cast<char *>(buf);
memset(ptr, '\0', buffer_len);
ptr[strlen(buf)-1]='\0';
A way to achieve flexible strings in C is to use realloc:
//Get some malloc'ed strings.
char* myString = asprintf("my cool, malloc()-allocated string\n");
char* myOtherString = asprintf("Oh, and by the way: Hello World!\n");
//Make room in myString to append the other string.
size_t finalStringSize = strlen(myString) + strlen(myOtherString);
myString = realloc(myString, finalStringSize + 1); //+1 for termination
//Concatenate the strings
strcat(myString, myOtherString);
free(myOtherString), myOtherString = NULL;
But, of course, using C++ std::strings should be less of a hassle.
Related
I wan't to initialize a static char buffer with a string str in the code below but I am getting the following error :
error: cannot convert ‘std::string’ to >‘char’ in initialization
and if I use
static char buf[500] = str.c_str();
I get the following error:
error: invalid conversion from ‘const char*’ to ‘char*’
Below is my code :
std::string str = "<Version="+version+" Ret=\"false\"/>";
static char buf[500] = str;
int len=strlen(buf);
buf[len]='\0';
INFO("Static Buffer :: "<<buf);
First of all, you cannot directly initialise a char[] from an std::string. It's just not possible. Even if you could, you would write = str, not = { str }.
So, you need to create the array first then assign the std::string's contents to it manually. Sadly, arrays are not assignable, so you're going to have to use an "algorithm" to do it.
Here we go:
const std::string str = "Hello world";
static char buf[500] = {};
std::copy(
// from the start of the string
std::begin(str),
// to the end of the string, or to 499 chars in, whichever comes first
std::begin(str) + std::min(str.size(), sizeof(buf)),
// into buf
std::begin(buf)
);
Yuck.
If you can, and this is likely the case, avoid it.
If you really need a C-string with the std::string's contents, just access str.c_str() whenever you need to. There is, in general, no need to keep a raw char array lying about, especially when you already have the right tool for the job doing that job.
Besides, as you are not initialising buf with that data, if it's function-static, this code probably does not have the intended effect.
You can use std::string::copy().
std::string text = "Hello there!";
char* cStrText = new char[text.length()];
//Copy the string into the buffer
text.copy(cStrText, text.length(), 0);
Obviously, this performs a copy which may not be optimal. You might want to look into trying to move it.
https://www.cplusplus.com/reference/string/string/copy/
Consider the following function:
unique_ptr<char> f(const wstring key, const unsigned int length)
{
assert(length <= key.length());
const wstring suffix = key.substr(length, key.length() - length);
const size_t outputSize = suffix.length() + 1; // +1 for null terminator
char * output = new char[outputSize];
size_t charsConverted = 0;
const wchar_t * outputWide = suffix.c_str();
wcstombs_s(&charsConverted, output, outputSize, outputWide, suffix.length());
return unique_ptr<char>(output);
}
The intent here is to accept a wstring, select length characters from the end, and return them as a C-style string that's wrapped in a unique_ptr (as required by another library - I certainly didn't chose that type :)).
One of my peers said in passing that he thinks this leaks memory, but he didn't have time to elaborate, and I don't see it. Can anybody spot it, and if so explain how I ought to fix it? I probably have my blinders on.
It's not necessarily a leak, but it is undefined behavior. You created the char array using new[] but the unique_ptr<char> will call delete, and not delete[] to free the memory. Use unique_ptr<char[]> instead.
Also, your conversion may not always behave the way you want it to. You should make 2 calls to wcstombs_s, in the first one pass nullptr as the second argument. This will return the number of characters required in the output string.
wcstombs_s(&charsConverted, nullptr, 0, outputWide, suffix.length());
Check the return value, and then use the result stored in charsConverted to allocate the output buffer.
auto output = std::unique_ptr<char[]>(new char[charsConverted]);
// now use output.get() to get access to the raw pointer
I have been working with C++ strings and trying to load char * strings into std::string by using C functions such as strcpy(). Since strcpy() takes char * as a parameter, I have to cast it which goes something like this:
std::string destination;
unsigned char *source;
strcpy((char*)destination.c_str(), (char*)source);
The code works fine and when I run the program in a debugger, the value of *source is stored in destination, but for some odd reason it won't print out with the statement
std::cout << destination;
I noticed that if I use
std::cout << destination.c_str();
The value prints out correctly and all is well. Why does this happen? Is there a better method of copying an unsigned char* or char* into a std::string (stringstreams?) This seems to only happen when I specify the string as foo.c_str() in a copying operation.
Edit: To answer the question "why would you do this?", I am using strcpy() as a plain example. There are other times that it's more complex than assignment. For example, having to copy only X amount of string A into string B using strncpy() or passing a std::string to a function from a C library that takes a char * as a parameter for a buffer.
Here's what you want
std::string destination = source;
What you're doing is wrong on so many levels... you're writing over the inner representation of a std::string... I mean... not cool man... it's much more complex than that, arrays being resized, read-only memory... the works.
This is not a good idea at all for two reasons:
destination.c_str() is a const pointer and casting away it's const and writing to it is undefined behavior.
You haven't set the size of the string, meaning that it won't even necessealy have a large enough buffer to hold the string which is likely to cause an access violation.
std::string has a constructor which allows it to be constructed from a char* so simply write:
std::string destination = source
Well what you are doing is undefined behavior. Your c_str() returns a const char * and is not meant to be assigned to. Why not use the defined constructor or assignment operator.
std::string defines an implicit conversion from const char* to std::string... so use that.
You decided to cast away an error as c_str() returns a const char*, i.e., it does not allow for writing to its underlying buffer. You did everything you could to get around that and it didn't work (you shouldn't be surprised at this).
c_str() returns a const char* for good reason. You have no idea if this pointer points to the string's underlying buffer. You have no idea if this pointer points to a memory block large enough to hold your new string. The library is using its interface to tell you exactly how the return value of c_str() should be used and you're ignoring that completely.
Do not do what you are doing!!!
I repeat!
DO NOT DO WHAT YOU ARE DOING!!!
That it seems to sort of work when you do some weird things is a consequence of how the string class was implemented. You are almost certainly writing in memory you shouldn't be and a bunch of other bogus stuff.
When you need to interact with a C function that writes to a buffer there's two basic methods:
std::string read_from_sock(int sock) {
char buffer[1024] = "";
int recv = read(sock, buffer, 1024);
if (recv > 0) {
return std::string(buffer, buffer + recv);
}
return std::string();
}
Or you might try the peek method:
std::string read_from_sock(int sock) {
int recv = read(sock, 0, 0, MSG_PEEK);
if (recv > 0) {
std::vector<char> buf(recv);
recv = read(sock, &buf[0], recv, 0);
return std::string(buf.begin(), buf.end());
}
return std::string();
}
Of course, these are not very robust versions...but they illustrate the point.
First you should note that the value returned by c_str is a const char* and must not be modified. Actually it even does not have to point to the internal buffer of string.
In response to your edit:
having to copy only X amount of string A into string B using strncpy()
If string A is a char array, and string B is std::string, and strlen(A) >= X, then you can do this:
B.assign(A, A + X);
passing a std::string to a function from a C library that takes a char
* as a parameter for a buffer
If the parameter is actually const char *, you can use c_str() for that. But if it is just plain char *, and you are using a C++11 compliant compiler, then you can do the following:
c_function(&B[0]);
However, you need to ensure that there is room in the string for the data(same as if you were using a plain c-string), which you can do with a call to the resize() function. If the function writes an unspecified amount of characters to the string as a null-terminated c-string, then you will probably want to truncate the string afterward, like this:
B.resize(B.find('\0'));
The reason you can safely do this in a C++11 compiler and not a C++03 compiler is that in C++03, strings were not guaranteed by the standard to be contiguous, but in C++11, they are. If you want the guarantee in C++03, then you can use std::vector<char> instead.
What is the proper way to initialize unsigned char*? I am currently doing this:
unsigned char* tempBuffer;
tempBuffer = "";
Or should I be using memset(tempBuffer, 0, sizeof(tempBuffer)); ?
To "properly" initialize a pointer (unsigned char * as in your example), you need to do just a simple
unsigned char *tempBuffer = NULL;
If you want to initialize an array of unsigned chars, you can do either of following things:
unsigned char *tempBuffer = new unsigned char[1024]();
// and do not forget to delete it later
delete[] tempBuffer;
or
unsigned char tempBuffer[1024] = {};
I would also recommend to take a look at std::vector<unsigned char>, which you can initialize like this:
std::vector<unsigned char> tempBuffer(1024, 0);
The second method will leave you with a null pointer. Note that you aren't declaring any space for a buffer here, you're declaring a pointer to a buffer that must be created elsewhere. If you initialize it to "", that will make the pointer point to a static buffer with exactly one byte—the null terminator. If you want a buffer you can write characters into later, use Fred's array suggestion or something like malloc.
As it's a pointer, you either want to initialize it to NULL first like this:
unsigned char* tempBuffer = NULL;
unsigned char* tempBuffer = 0;
or assign an address of a variable, like so:
unsigned char c = 'c';
unsigned char* tempBuffer = &c;
EDIT:
If you wish to assign a string, this can be done as follows:
unsigned char myString [] = "This is my string";
unsigned char* tmpBuffer = &myString[0];
If you know the size of the buffer at compile time:
unsigned char buffer[SIZE] = {0};
For dynamically allocated buffers (buffers allocated during run-time or on the heap):
1.Prefer the new operator:
unsigned char * buffer = 0; // Pointer to a buffer, buffer not allocated.
buffer = new unsigned char [runtime_size];
2.Many solutions to "initialize" or fill with a simple value:
std::fill(buffer, buffer + runtime_size, 0); // Prefer to use STL
memset(buffer, 0, runtime_size);
for (i = 0; i < runtime_size; ++i) *buffer++ = 0; // Using a loop
3.The C language side provides allocation and initialization with one call.
However, the function does not call the object's constructors:
buffer = calloc(runtime_size, sizeof(unsigned char))
Note that this also sets all bits in the buffer to zero; you don't get a choice in the initial value.
It depends on what you want to achieve (e.g. do you ever want to modify the string). See e.g. http://c-faq.com/charstring/index.html for more details.
Note that if you declare a pointer to a string literal, it should be const, i.e.:
const unsigned char *tempBuffer = "";
If the plan is for it to be a buffer and you want to move it later to point to something, then initialise it to NULL until it really points somewhere to which you want to write, not an empty string.
unsigned char * tempBuffer = NULL;
std::vector< unsigned char > realBuffer( 1024 );
tempBuffer = &realBuffer[0]; // now it really points to writable memory
memcpy( tempBuffer, someStuff, someSizeThatFits );
The answer depends on what you inted to use the unsigned char for. A char is nothing else but a small integer, which is of size 8 bits on 99% of all implementations.
C happens to have some string support that fits well with char, but that doesn't limit the usage of char to strings.
The proper way to initialize a pointer depends on 1) its scope and 2) its intended use.
If the pointer is declared static, and/or declared at file scope, then ISO C/C++ guarantees that it is initialized to NULL. Programming style purists would still set it to NULL to keep their style consistent with local scope variables, but theoretically it is pointless to do so.
As for what to initialize it to... set it to NULL. Don't set it to point at "", because that will allocate a static dummy byte containing a null termination, which will become a tiny little static memory leak as soon as the pointer is assigned to something else.
One may question why you need to initialize it to anything at all in the first place. Just set it to something valid before using it. If you worry about using a pointer before giving it a valid value, you should get a proper static analyzer to find such simple bugs. Even most compilers will catch that bug and give you a warning.
Const-correctness in C++ is still giving me headaches. In working with some old C code, I find myself needing to assign turn a C++ string object into a C string and assign it to a variable. However, the variable is a char * and c_str() returns a const char []. Is there a good way to get around this without having to roll my own function to do it?
edit: I am also trying to avoid calling new. I will gladly trade slightly more complicated code for less memory leaks.
C++17 and newer:
foo(s.data(), s.size());
C++11, C++14:
foo(&s[0], s.size());
However this needs a note of caution: The result of &s[0]/s.data()/s.c_str() is only guaranteed to be valid until any member function is invoked that might change the string. So you should not store the result of these operations anywhere. The safest is to be done with them at the end of the full expression, as my examples do.
Pre C++-11 answer:
Since for to me inexplicable reasons nobody answered this the way I do now, and since other questions are now being closed pointing to this one, I'll add this here, even though coming a year too late will mean that it hangs at the very bottom of the pile...
With C++03, std::string isn't guaranteed to store its characters in a contiguous piece of memory, and the result of c_str() doesn't need to point to the string's internal buffer, so the only way guaranteed to work is this:
std::vector<char> buffer(s.begin(), s.end());
foo(&buffer[0], buffer.size());
s.assign(buffer.begin(), buffer.end());
This is no longer true in C++11.
There is an important distinction you need to make here: is the char* to which you wish to assign this "morally constant"? That is, is casting away const-ness just a technicality, and you really will still treat the string as a const? In that case, you can use a cast - either C-style or a C++-style const_cast. As long as you (and anyone else who ever maintains this code) have the discipline to treat that char* as a const char*, you'll be fine, but the compiler will no longer be watching your back, so if you ever treat it as a non-const you may be modifying a buffer that something else in your code relies upon.
If your char* is going to be treated as non-const, and you intend to modify what it points to, you must copy the returned string, not cast away its const-ness.
I guess there is always strcpy.
Or use char* strings in the parts of your C++ code that must interface with the old stuff.
Or refactor the existing code to compile with the C++ compiler and then to use std:string.
There's always const_cast...
std::string s("hello world");
char *p = const_cast<char *>(s.c_str());
Of course, that's basically subverting the type system, but sometimes it's necessary when integrating with older code.
If you can afford extra allocation, instead of a recommended strcpy I would consider using std::vector<char> like this:
// suppose you have your string:
std::string some_string("hello world");
// you can make a vector from it like this:
std::vector<char> some_buffer(some_string.begin(), some_string.end());
// suppose your C function is declared like this:
// some_c_function(char *buffer);
// you can just pass this vector to it like this:
some_c_function(&some_buffer[0]);
// if that function wants a buffer size as well,
// just give it some_buffer.size()
To me this is a bit more of a C++ way than strcpy. Take a look at Meyers' Effective STL Item 16 for a much nicer explanation than I could ever provide.
You can use the copy method:
len = myStr.copy(cStr, myStr.length());
cStr[len] = '\0';
Where myStr is your C++ string and cStr a char * with at least myStr.length()+1 size. Also, len is of type size_t and is needed, because copy doesn't null-terminate cStr.
Just use const_cast<char*>(str.data())
Do not feel bad or weird about it, it's perfectly good style to do this.
It's guaranteed to work in C++11. The fact that it's const qualified at all is arguably a mistake by the original standard before it; in C++03 it was possible to implement string as a discontinuous list of memory, but no one ever did it. There is not a compiler on earth that implements string as anything other than a contiguous block of memory, so feel free to treat it as such with complete confidence.
If you know that the std::string is not going to change, a C-style cast will work.
std::string s("hello");
char *p = (char *)s.c_str();
Of course, p is pointing to some buffer managed by the std::string. If the std::string goes out of scope or the buffer is changed (i.e., written to), p will probably be invalid.
The safest thing to do would be to copy the string if refactoring the code is out of the question.
std::string vString;
vString.resize(256); // allocate some space, up to you
char* vStringPtr(&vString.front());
// assign the value to the string (by using a function that copies the value).
// don't exceed vString.size() here!
// now make sure you erase the extra capacity after the first encountered \0.
vString.erase(std::find(vString.begin(), vString.end(), 0), vString.end());
// and here you have the C++ string with the proper value and bounds.
This is how you turn a C++ string to a C string. But make sure you know what you're doing, as it's really easy to step out of bounds using raw string functions. There are moments when this is necessary.
If c_str() is returning to you a copy of the string object internal buffer, you can just use const_cast<>.
However, if c_str() is giving you direct access tot he string object internal buffer, make an explicit copy, instead of removing the const.
Since c_str() gives you direct const access to the data structure, you probably shouldn't cast it. The simplest way to do it without having to preallocate a buffer is to just use strdup.
char* tmpptr;
tmpptr = strdup(myStringVar.c_str();
oldfunction(tmpptr);
free tmpptr;
It's quick, easy, and correct.
In CPP, if you want a char * from a string.c_str()
(to give it for example to a function that only takes a char *),
you can cast it to char * directly to lose the const from .c_str()
Example:
launchGame((char *) string.c_str());
C++17 adds a char* string::data() noexcept overload. So if your string object isn't const, the pointer returned by data() isn't either and you can use that.
Is it really that difficult to do yourself?
#include <string>
#include <cstring>
char *convert(std::string str)
{
size_t len = str.length();
char *buf = new char[len + 1];
memcpy(buf, str.data(), len);
buf[len] = '\0';
return buf;
}
char *convert(std::string str, char *buf, size_t len)
{
memcpy(buf, str.data(), len - 1);
buf[len - 1] = '\0';
return buf;
}
// A crazy template solution to avoid passing in the array length
// but loses the ability to pass in a dynamically allocated buffer
template <size_t len>
char *convert(std::string str, char (&buf)[len])
{
memcpy(buf, str.data(), len - 1);
buf[len - 1] = '\0';
return buf;
}
Usage:
std::string str = "Hello";
// Use buffer we've allocated
char buf[10];
convert(str, buf);
// Use buffer allocated for us
char *buf = convert(str);
delete [] buf;
// Use dynamic buffer of known length
buf = new char[10];
convert(str, buf, 10);
delete [] buf;