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.
so steps i need to do
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!
Syntax: char* strcpy (char* destination, const char* source); The strcpy() function is used to copy strings. It copies string pointed to by source into the destination . This function accepts two arguments of type pointer to char or array of characters and returns a pointer to the first string i.e destination .
Just call either the data() or c_str() member functions of the std::string class, to get the char* pointer of the string object. The strcpy() function doesn't have overload to accept two std::string objects as parameters.
strcpy() is a standard library function in C/C++ and is used to copy one string to another. In C it is present in string. h header file and in C++ it is present in cstring header file.
string at() in C++std::string::at can be used to extract characters by characters from a given string. Syntax 2: const char& string::at (size_type idx) const idx : index number Both forms return the character that has the index idx (the first character has index 0).
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)
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