char *buffer1 = "abc";
const char *buffer2 = (const char*) buffer;
std :: string str (buffer2);
This works, but I want to declare the std::string object i.e. str, once and use it many times to store different const char*.
What's the way out?
You can just re-assign:
const char *buf1 = "abc";
const char *buf2 = "def";
std::string str(buf1);
str = buf2; // Calls str.operator=(const char *)
Ah well, as I commented above, found the answer soon after posting the question :doh:
const char* g;
g = (const char*)buffer;
std :: string str;
str.append (g);
So, I can call append() function as many times (after using the clear()) as I want on the same object with "const char *".
Though the "push_back" function won't work in place of "append".
str is actually copying the characters from buffer2, so it is not connected in any way.
If you want it to have another value, you just assign a new one
str = "Hello";
Make a Class say MyString which compose String buffer.
Have a constant of that class.
and then u can reassign the value of the composed string buffer, while using the same constant.
Related
string b="some string another string";
std::regex r("string");
std::sregex_iterator m(b.begin(),b.end(),r);
for (std::sregex_iterator end; m!=end; m++)
{
// want a char* to m->str() in here
}
I'm gettng totally lost trying to determine what's what because of the templates. I've tried
const char *c = m->str().c_str() // compiles but points to ""
Answer
Pointed out by lightness below
const char *c = &b[m->position()]; // length of str is m->length()
const char *c = m->str().c_str()
m->str() is a temporary value, not a reference to data inside m, so c is immediately dangling.
Just store the string first:
const std::string str = m->str();
const char* c = str.c_str();
Yes, the copy is unfortunate.
A cheaper way than the copy would be:
const std::string_view sv{
b.data() + m->position(),
m->length()
};
This is great if you can use string views. But it is not null-terminated! So, if you really do need a const char*, this won't work for you and you're pretty much stuck with a fresh buffer.
Following are two legacy routines. I cannot change the routine declarations.
static bool GetString(char * str); //str is output parameter
static bool IsStringValid(const char * str); //str is input parameter
With call as follows
char inputString[1000];
GetString(inputString);
IsStringValid(inputString);
Instead of using fixed char array, I want to use std::string as the input. I am not able get the semantics right (string::c_str).
With IsEmpty it should not be a problem:
std::string str = "Some text here";
IsEmpty(str.c_str());
Though it's pretty useless if you have a std::string as then you would normally just call str.empty().
The other function though, that's harder. The reason is that it's argument is not const, and std::string doesn't allow you to modify the string using a pointer.
It can be solved, by writing a wrapper-function which takes a string reference, and have an internal array used for the actual GetString call, and uses that array to initialize the passed string reference.
Wrapper examples:
// Function which "creates" a string from scratch
void GetString(std::string& str)
{
char tempstr[4096];
GetString(tempstr);
str = tempstr;
}
// Function which modifies an existing string
void ModifyString(std::string& str)
{
const size_t length = str.size() + 1;
char* tempstr = new char[length];
std::copy_n(str.c_str(), tempstr, length);
ModifyString(tempstr);
str = tempstr;
delete[] tempstr;
}
You can't use c_str for the first function, because it returns a const char*. You can pass a std::string by reference and assign to it. As for is empty, you can call c_str on your string, but you'd be better of calling the member empty().
I think you can use the string container of STL ( Standard template Library ) .
#include <string>
bool isempty ( int x ) {
return ( x == 0 ) ? true : false ;
}
// inside main()
string s ;
cin >> s ; // or getline ( cin , s) ;
bool empty = isEmpty (s.length()) ;
std::string has c_str() which you can use for IsEmpty. There ist no function which gives you a non const pointer. Since std::string's allocation is not guaranteed to be contiguous you cannot do something like &s[0] either. The only thing you can do is to use a temporary char buffer as you do in your example.
std::string s;
char inputString[1000];
std::vector<char> v(1000);
GetString(inputString);
GetString(&v[0]);
s = &v[0];
IsEmpty(s.c_str());
If i define something like below,
char *s1 = "Hello";
why I can't do something like below,
*s1 = 'w'; // gives segmentation fault ...why???
What if I do something like below,
string s1 = "hello";
Can I do something like below,
*s1 = 'w';
Because "Hello" creates a const char[]. This decays to a const char* not a char*. In C++ string literals are read-only. You've created a pointer to such a literal and are trying to write to it.
But when you do
string s1 = "hello";
You copy the const char* "hello" into s1. The difference being in the first example s1 points to read-only "hello" and in the second example read-only "hello" is copied into non-const s1, allowing you to access the elements in the copied string to do what you wish with them.
If you want to do the same with a char* you need to allocate space for char data and copy hello into it
char hello[] = "hello"; // creates a char array big enough to hold "hello"
hello[0] = 'w'; // writes to the 0th char in the array
string literals are usually allocated in read-only data segment.
Because Hello resides in read only memory. Your signature should actually be
const char* s1 = "Hello";
If you want a mutable buffer then declare s1 as a char[]. std::string overloads operator [], so you can index into it, i.e., s1[index] = 'w'.
Time to confuse matters:
char s0[] = "Hello";
s0[0] = 'w';
This is perfectly valid! Of course, this doesn't answer the original question so here we go: string literals are created in read-only memory. That is, their type is char const[n] where n is the size of the string (including the terminating null character, i.e. n == 6 for the string literal "Hello". But why, oh, why can this type be used to initialize a char const*? The answer is simply backward compatibility, respectively compatibility to [old] C code: by the time const made it into the language, lots of places already initialized char* with string literals. Any decent compiler should warn about this abuse, however.
I am not able to concat two const char*.
I do the following:
const char* p = new char[strlen(metadata.getRoot())+strlen(metadata.getPath())];
strcat(const_cast<char*>(p),metadata.getRoot());
strcat(const_cast<char*>(p),metadata.getPath());
strcpy(const_cast<char*>(args2->fileOrFolderPath),p);
function(args2->fileOrFolderPath);
Now when I print the variable args2->fileOrFolderPath on the console then the correct output appears... But when I call a method with the variable as parameter, and work with the variable then I got a segmentation fault. What is the problem?
I did not declare them like this but i know they have this information
So, I have this:
const char* ruta1 = "C:\\Users\\Deivid\\Desktop\\";
const char* ruta2 = "lenaGris.xls";
Then I used this for concatenation:
char * RutaFinal = new char[strlen(ruta1) + strlen(ruta2) + 1];
strcpy(RutaFinal, ruta1);
strcat(RutaFinal, ruta2);
printf(RutaFinal);
This worked for me.
I would prefer using std::string for this, but if you like char* and the str... functions, at least initialize p before using strcat:
*p = 0;
BTW:
using std::string, this would be:
std::string p = std::string(metadata.getRoot()) + metadata.getPath();
strcpy(const_cast<char*>(args2->fileOrFolderPath), p.c_str());
function(args2->fileOrFolderPath);
And you don't have to deallocate p somewhere.
1.
const char* p=new char[strlen(metadata.getRoot())+strlen(metadata.getPath())+1];
the length plus 1 to store '\0'.
2.
strcpy(const_cast<char*>(args2->fileOrFolderPath),p);
You can not guarantee args2->fileOrFolderPath 's length is longger than strlen(p).
This works well
#include <iostream>
using namespace std;
void foo(const char*s){
cout<<s<<endl;
}
int main(int argc,char*argv[]){
const char* s1 = "hello ";
const char* s2 = "world!";
const char* p = new char [strlen(s1)+strlen(s2)+1];
const char* s = new char [strlen(s1)+strlen(s2)+1];
strcat(const_cast<char*>(p),s1);
strcat(const_cast<char*>(p),s2);
strcpy(const_cast<char*>(s),p);
cout<<s<<endl;
foo(s);
return 0;
}
You have char pointers, pointing to char constants which can't be modified . What you can do is to copy your const char array to some char array and do like this to concate const strings :
char result[MAX];
strcpy(result,some_char_array); // copy to result array
strcat(result,another_char_array); // concat to result array
I believe you need to include space for the null terminator, and the first parameter to strcat shouldn't be const as you're trying to modify the memory being pointed to.
You want to do something like this:
char * str1 = "Hello, ";
char * str2 = "World!\n";
char * buffer = malloc(strlen(str1) + strlen(str2) + 1);
strcpy(buffer, str1);
strcat(buffer, str2);
printf(buffer);
Which prints out "Hello, World!" as expected.
As for the error you're seeing when using a parameter, I've wrote some tests to see why it doesn't break when using a const local variable. While compiling using a const char * for the pointer I'm using as the target I get this warning:
./strings.c:10: warning: passing argument 1 of ‘strcat’ discards qualifiers from pointer target type
As it states, const is discarded and it works as expected. However, if I pass a parameter which is a const char * pointer, then I get a bus error when trying to modify the buffer it writes to. I suspect what is happening is that it ignores the const on the argument, but it can't then modify the buffer because it's defined as const elsewhere in the code.
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());