If i pass a char * into a function. I want to then take that char * convert it to a std::string and once I get my result convert it back to char * from a std::string to show the result.
I don't know how to do this for conversion ( I am not talking const char * but just char *)
I am not sure how to manipulate the value of the pointer I send in.
so steps i need to do
take in a char *
convert it into a string.
take the result of that string and put it back in the form of a char *
return the result such that the value should be available outside the function and not get destroyed.
If possible can i see how it could be done via reference vs a pointer (whose address I pass in by value however I can still modify the value that pointer is pointing to. so even though the copy of the pointer address in the function gets destroyed i still see the changed value outside.
thanks!
Converting a char* to a std::string:
char* c = "Hello, world";
std::string s(c);
Converting a std::string to a char*:
std::string s = "Hello, world";
char* c = new char[s.length() + 1];
strcpy(c, s.c_str());
// and then later on, when you are done with the `char*`:
delete[] c;
I prefer to use a std::vector<char> instead of an actual char*; then you don't have to manage your own memory:
std::string s = "Hello, world";
std::vector<char> v(s.begin(), s.end());
v.push_back('\0'); // Make sure we are null-terminated
char* c = &v[0];
You need to watch how you handle the memory from the pointer you return, for example the code below will not work because the memory allocated in the std::string will be released when fn() exits.
const char* fn(const char*psz) {
std::string s(psz);
// do something with s
return s.c_str(); //BAD
}
One solution is to allocate the memory in the function and make sure the caller of the function releases it:
const char* fn(const char*psz) {
std::string s(psz);
// do something with s
char *ret = new char[s.size()]; //memory allocated
strcpy(ret, s.c_str());
return ret;
}
....
const char* p = fn("some text");
//do something with p
delete[] p;// release the array of chars
Alternatively, if you know an upper bound on the size of the string you can create it on the stack yourself and pass in a pointer, e.g.
void fn(const char*in size_t bufsize, char* out) {
std::string s(psz);
// do something with s
strcpy_s(out, bufsize, s.c_str()); //strcpy_s is a microsoft specific safe str copy
}
....
const int BUFSIZE = 100;
char str[BUFSIZE];
fn("some text", BUFSIZE, str);
//ok to use str (memory gets deleted when it goes out of scope)
You can maintain a garbage collector for your library implemented as
std::vector<char*> g_gc; which is accessible in your library 'lib'. Later, you can release all pointers in g_gc at your convenience by calling lib::release_garbage();
char* lib::func(char*pStr)
{
std::string str(pStr);
char *outStr = new char[str.size()+1];
strcpy(outStr, str.c_str());
g_gc.push_back(outStr); // collect garbage
return outStr;
}
release_garbage function will look like:
void lib::release_garbage()
{
for(int i=0;i<g_gc.size();i++)
{
delete g_gc[i];
}
g_gc.clear();
}
In a single threaded model, you can keep this g_gc static. Multi-threaded model would involve locking/unlocking it.
Related
I have a char pointer:
char* s = new char[150];
Now how do i fill it? This:
s="abcdef";
Gives warning about deprecation of conversion between string literal and char*, but generally works.
This:
char* s = new[150]("abcdef");
Does not work, gives an error.
How to do this properly? Note that I want the memory allocation to have 150*sizeof(char) bytes and contain "abcdef". I know about malloc, but is it possible to do with new?
Its for an assignment where i cant use the standard library.
This sequence of statements
char* s = new char[150];
s="abcdef";
results in a memory leak because at first a memory was allocated and its address was assigned to the pointer s and then the pointer was reassigned with the address of the string literal "abcdef". And moreover string literals in C++ (opposite to C) have types of constant character arrays.
If you allocated a memory for a string then you should copy a string in the memory either by using the C standard function strcpy or C standard function strncpy.
For example
char* s = new char[150];
std::strcpy( s, "abcdef" );
Or
const size_t N = 150;
char* s = new char[N];
std::strncpy( s, "abcdef", N );
s[N-1] = '\0';
Or even the following way
#include <iostream>
#include <cstring>
int main()
{
const size_t N = 150;
char *s = new char[N]{ '\0' };
std::strncpy( s, "abcdef", N - 1 );
std::cout << s << '\n';
delete []s;
}
In any case it is better just to use the standard class std::string.
std::string s( "abcdef" );
or for example
std::string s;
s.assign( "abcdef" );
The basic procedure for creating a memory area for a string and then filling it without using the Standard Library in C++ is as follows:
create the appropriate sized memory area with new
use a loop to copy characters from a string into the new area
So the source code would look like:
// function to copy a zero terminated char string to a new char string.
// loop requires a zero terminated char string as the source.
char *strcpyX (char *dest, const char *source)
{
char *destSave = dest; // save copy of the destination address to return
while (*dest++ = *source++); // copy characters up to and including zero terminator.
return destSave; // return destination pointer per standard library strcpy()
}
// somewhere in your code
char *s1 = new char [150];
strcpyX (s1, "abcdef");
Given a character array:
char * s = new char [256];
Here's how to fill the pointer:
std::fill(&s, &s + sizeof(s), 0);
Here's how to fill the array:
std::fill(s, s+256, '\0');
Here's how to assign or copy text into the array:
std::strcpy(s, "Hello");
You could also use std::copy:
static const char text[] = "World";
std::copy(text, text + sizeof(text), s);
Remember that a pointer, array and C-Style string are different concepts and objects.
Edit 1: Prefer std::string
In C++, prefer to use std::string for text rather than character arrays.
std::string s;
s = "abcdef";
std::cout << s << "\n";
Once you've allocated the memory for this string, you could use strcpy to populate it:
strcpy(s, "abcdef");
I have a number that I need to convert to a const char * (an API I'm using them requires const char * as input to many of its functions). The following works:
int num = 5;
std::string s = std::to_string(5);
const char * p = s.c_str();
as suggested by answers like those in how to convert from int to char*?, but it involves creating the seemingly unnecessary variable s, so I tried the following, but it doesn't work (p points to an empty string afterwards):
int num = 5;
const char * p = std::to_string(num).c_str();
Is there a clean way I can accomplish this? Why doesn't the second example work? The behavior is very similar to what happens if I made this obvious mistake:
const char * p;
{
std::string tempStr( "hi" );
p = tempStr.c_str( );
// p points to "hi" string.
}
// now p points to "" string.
Which makes me suspect that the issue std::to_string(num) immediately goes out of scope or something similar because it's not used to directly initialize anything.
std::string encapsulates managing dynamic memory (created with new[] and delete[]). Let's break it down.
const char * p = std::to_string(num).c_str();
Create a std::string (with a human-readable representation of num).
Get the new[]ly allocated const char* to the string.
Assign that value to p.
Destroy the std::string → delete[] the allocated const char*.
p points to... deallocated data
If you are using a pointer, the data that the pointer points to must exist throughout the lifetime of that pointer.
So, no, there is no way around this other than new[]ing a copy of the string, which you will have to explicitly delete[] later. And at that point, you've thrown the baby out with the bath and have no need to use std::string.
Create a string that lives at least as long as you want to refer to its internal data.
Just use std::string it does everything you want and everything that you would have to do manually if you don't use it.
When you need to pass a const char* to a const char* function simply use std::string::c_str() like this:
some_api_function(mystring.c_str()); // passes a const char*
What you need is a function which returns a char* which holds your value and can be used to manage its lifetime. The problematic version is broken because the char* points to memory which it does not manage.
For example:
std::unique_ptr<char[]> str(int32_t x)
{
std::unique_ptr<char[]> res(new char[12]);
snprintf(res.get(), 12, "%d", x);
return res;
}
Usestd::string everywhere and don't use const char* when not nessecary. They are basically the same thing. I use const char* only when I'm using a file-path.
Use std::string everywhere and your program should work.
So, i have a lib file where i have function which converts string to char* :
void Additional::str2Char(string s,char** cstr)
{
*cstr = new char[s.length() + 1];
*cstr = (char*) s.c_str();
}
Then, i create console app, and do this:
int main()
{
Additional *a = new Additional();
string b = "fdfd";
char *test;
a->str2Char(b, &test);
cout << test << endl;
delete a;
}
The ouput is really bad..
Help me, i dont know how to get the pointer from lib right.
First you allocate some memory.
Then you reassign the pointer to that memory, to point to the contents of the local string s, leaking the memory you allocated.
Then you return from the function, destroying s, leaving the pointer dangling. It no longer points to valid data, and dereferencing it gives undefined behaviour.
The best solution is to stop messing around with pointers and new, and use std::string for all your strings. If you really want to do this, then you'll need to copy the string contents to the new memory:
*cstr = new char[s.length() + 1]; // no change
std::strcpy(*cstr, s.c_str()); // copy the data, including the terminator
If the library is not going to modify the text then, the c_str() call is enough. No need to create an char pointer/array.
If it is going to modify the text: https://stackoverflow.com/a/13294114/47351
strncpy(tab2, tmp.c_str(), sizeof(tab2));
tab2[sizeof(tab2) - 1] = 0;
Since you want to handle the memory allocation inside the function, a different API could be used:
// Creates a const char* and copies the contents of s to it. Caller
// takes ownership of the returned data.
const char* str2cstr(const string& s) {
char* cstr = new char[s.size()];
strncopy(cstr, s.c_str(), s.size());
return cstr;
}
Now you can do const char* cstr = str2cstr(str);. Notice also that I'm using a const string& param to avoid the extra string copy.
I am trying to write a function to convert a std::string to char* .
The first one I have written was this:
char* $ (string str)
{
char* cstr;
const unsigned int length=str.size();
cstr=new char[1000];
for(int i=0;i<length;i++)
cstr[i]=str[i];
cstr[length]=0;
return cstr;
}
But the problem was the memory leak: let's suppose that I do this:
char* cstr;
string str1("hello"),str2("hello2");
cstr=$(str1);
cstr=$(str2);
There is a memory leak in this case.The first allocated string is not reachable but it's reference is lost.
So I made the same using static:
char* $ (string str)
{
static char cstr[1000];
const unsigned int length=str.size();
for(int i=0;i<length;i++)
cstr[i]=str[i];
cstr[length]=0;
return cstr;
}
But the problem now is that the static char fields are accessible:
char* cstr;
string str("hello");
cstr=$(str);
$(str)[5]='!';
This is possibile, the 6th character is modified and so also the C-style string pointed by cstr is modified.
Using const:
const char* $ (string str)
{
static char cstr[1000];
const unsigned int length=str.size();
for(int i=0;i<length;i++)
cstr[i]=str[i];
cstr[length]=0;
return cstr;
}
The problem is that a char pointer is not compatible with a const char pointer, so I can't do this:
string str("hello");
char* cstr;
cstr=$(str);
But I can only use a const char pointer.
What I would do is to have a function which the return value could be placed only as right operand, but not as left operand of an assignment.How could this be done?
I tried to do this:
char* toCharArray(string& str)
{
std::unique_ptr<char>p(new char[1000]);
char* temp=p.get();
for(unsigned int i=0;i<str.size();i++)
{
*temp=str[i];
temp++;
}
return p.get();
}
But the problem is still there, I don't see the difference between this and the other solution I posted using static.Since a code like this:
char* cstr;
string str("hello");
cstr=toCharArray(str);
toCharArray(str)[0]='o';
cout << cstr;
Modifies the string (prints "oello").
Problem still not solved.
You can prevent memory leaks by returning the allocated array as either std::unique_ptr<char[]> or std::vector<char>; both will release the memory if they are reassigned or go out of scope.
You can get a char* pointer to the contents as ptr.get() or &vec[0] respectively.
By the way, since the length is known, the array length really should be length+1, not 1000. Fixed-sized buffers are an overrun waiting to happen. Also, $ is not a portable name for a function.
http://www.cplusplus.com/reference/string/string/c_str/
Look at the example, specifically:
char * cstr, *p;
string str ("Please split this phrase into tokens");
cstr = new char [str.size()+1];
strcpy (cstr, str.c_str());
You need to create a new char * and copy the contents of the std::string over it.
You can use strcpy.
I don't see why you're surprised by the memory leak. If you're making a copy of something, you've got to put it somewhere.
You can either put all the copies in the same place (running a risk of overflow, and putting the onus of taking a proper copy on the client, or they run the risk of unexpected corruption), or you do the allocation (they still have to do the freeing, but at least they don't have to copy), or you get them to pass a buffer and a size into your function.
And seriously, $ as a function name?
What about str.c_str()? Or strdup(str.c_str()) if you want a copy.
I want to convert a std::string into a char* or char[] data type.
std::string str = "string";
char* chr = str;
Results in: “error: cannot convert ‘std::string’ to ‘char’ ...”.
What methods are there available to do this?
It won't automatically convert (thank god). You'll have to use the method c_str() to get the C string version.
std::string str = "string";
const char *cstr = str.c_str();
Note that it returns a const char *; you aren't allowed to change the C-style string returned by c_str(). If you want to process it you'll have to copy it first:
std::string str = "string";
char *cstr = new char[str.length() + 1];
strcpy(cstr, str.c_str());
// do stuff
delete [] cstr;
Or in modern C++:
std::vector<char> cstr(str.c_str(), str.c_str() + str.size() + 1);
More details here, and here but you can use
string str = "some string" ;
char *cstr = &str[0];
As of C++11, you can also use the str.data() member function, which returns char *
string str = "some string" ;
char *cstr = str.data();
If I'd need a mutable raw copy of a c++'s string contents, then I'd do this:
std::string str = "string";
char* chr = strdup(str.c_str());
and later:
free(chr);
So why don't I fiddle with std::vector or new[] like anyone else? Because when I need a mutable C-style raw char* string, then because I want to call C code which changes the string and C code deallocates stuff with free() and allocates with malloc() (strdup uses malloc). So if I pass my raw string to some function X written in C it might have a constraint on it's argument that it has to allocated on the heap (for example if the function might want to call realloc on the parameter). But it is highly unlikely that it would expect an argument allocated with (some user-redefined) new[]!
(This answer applies to C++98 only.)
Please, don't use a raw char*.
std::string str = "string";
std::vector<char> chars(str.c_str(), str.c_str() + str.size() + 1u);
// use &chars[0] as a char*
If you just want a C-style string representing the same content:
char const* ca = str.c_str();
If you want a C-style string with new contents, one way (given that you don't know the string size at compile-time) is dynamic allocation:
char* ca = new char[str.size()+1];
std::copy(str.begin(), str.end(), ca);
ca[str.size()] = '\0';
Don't forget to delete[] it later.
If you want a statically-allocated, limited-length array instead:
size_t const MAX = 80; // maximum number of chars
char ca[MAX] = {};
std::copy(str.begin(), (str.size() >= MAX ? str.begin() + MAX : str.end()), ca);
std::string doesn't implicitly convert to these types for the simple reason that needing to do this is usually a design smell. Make sure that you really need it.
If you definitely need a char*, the best way is probably:
vector<char> v(str.begin(), str.end());
char* ca = &v[0]; // pointer to start of vector
This would be better as a comment on bobobobo's answer, but I don't have the rep for that. It accomplishes the same thing but with better practices.
Although the other answers are useful, if you ever need to convert std::string to char* explicitly without const, const_cast is your friend.
std::string str = "string";
char* chr = const_cast<char*>(str.c_str());
Note that this will not give you a copy of the data; it will give you a pointer to the string. Thus, if you modify an element of chr, you'll modify str.
Assuming you just need a C-style string to pass as input:
std::string str = "string";
const char* chr = str.c_str();
To obtain a const char * from an std::string use the c_str() member function :
std::string str = "string";
const char* chr = str.c_str();
To obtain a non-const char * from an std::string you can use the data() member function which returns a non-const pointer since C++17 :
std::string str = "string";
char* chr = str.data();
For older versions of the language, you can use range construction to copy the string into a vector from which a non-const pointer can be obtained :
std::string str = "string";
std::vector<char> str_copy(str.c_str(), str.c_str() + str.size() + 1);
char* chr = str_copy.data();
But beware that this won't let you modify the string contained in str, only the copy's data can be changed this way. Note that it's specially important in older versions of the language to use c_str() here because back then std::string wasn't guaranteed to be null terminated until c_str() was called.
To be strictly pedantic, you cannot "convert a std::string into a char* or char[] data type."
As the other answers have shown, you can copy the content of the std::string to a char array, or make a const char* to the content of the std::string so that you can access it in a "C style".
If you're trying to change the content of the std::string, the std::string type has all of the methods to do anything you could possibly need to do to it.
If you're trying to pass it to some function which takes a char*, there's std::string::c_str().
Here is one more robust version from Protocol Buffer
char* string_as_array(string* str)
{
return str->empty() ? NULL : &*str->begin();
}
// test codes
std::string mystr("you are here");
char* pstr = string_as_array(&mystr);
cout << pstr << endl; // you are here
Conversion in OOP style
converter.hpp
class StringConverter {
public: static char * strToChar(std::string str);
};
converter.cpp
char * StringConverter::strToChar(std::string str)
{
return (char*)str.c_str();
}
usage
StringConverter::strToChar("converted string")
For completeness' sake, don't forget std::string::copy().
std::string str = "string";
const size_t MAX = 80;
char chrs[MAX];
str.copy(chrs, MAX);
std::string::copy() doesn't NUL terminate. If you need to ensure a NUL terminator for use in C string functions:
std::string str = "string";
const size_t MAX = 80;
char chrs[MAX];
memset(chrs, '\0', MAX);
str.copy(chrs, MAX-1);
You can make it using iterator.
std::string str = "string";
std::string::iterator p=str.begin();
char* chr = &(*p);
Good luck.
A safe version of orlp's char* answer using unique_ptr:
std::string str = "string";
auto cstr = std::make_unique<char[]>(str.length() + 1);
strcpy(cstr.get(), str.c_str());
char* result = strcpy((char*)malloc(str.length()+1), str.c_str());
Alternatively , you can use vectors to get a writable char* as demonstrated below;
//this handles memory manipulations and is more convenient
string str;
vector <char> writable (str.begin (), str.end) ;
writable .push_back ('\0');
char* cstring = &writable[0] //or &*writable.begin ()
//Goodluck
This will also work
std::string s;
std::cout<<"Enter the String";
std::getline(std::cin, s);
char *a=new char[s.size()+1];
a[s.size()]=0;
memcpy(a,s.c_str(),s.size());
std::cout<<a;
No body ever mentioned sprintf?
std::string s;
char * c;
sprintf(c, "%s", s.c_str());