I wanna do something like:
string result;
char* a[100];
a[0]=result;
it seems that result.c_str() has to be const char*. Is there any way to do this?
6 Answers
You can take the address of the first character in the string.
a[0] = &result[0];
This is guaranteed to work in C++11. (The internal string representation must be contiguous and null-terminated like a C-style string)
In C++03 these guarantees do not exist, but all common implementations will work.
2 Comments
Jeremy Friesner
I don't think it will compile unless a is declared const though (or unless you cast away const-ness)
Drew Dormann
@JeremyFriesner It's not const. en.cppreference.com/w/cpp/string/basic_string/operator_at
string result;
char a[100] = {0};
strncpy(a, result.c_str(), sizeof(a) - 1);
Comments
There is a member function (method) called "copy" to have this done.
but you need create the buffer first.
like this
string result;
char* a[100];
a[0] = new char[result.length() + 1];
result.copy(a[0], result.length(), 0);
a[0][result.length()] = '\0';
(references: http://www.cplusplus.com/reference/string/basic_string/copy/ )
by the way, I wonder if you means
string result;
char a[100];
Comments
You can do:
char a[100];
::strncpy(a, result.c_str(), 100);
Be careful of null termination.
Comments
The old fashioned way:
#include <string.h>
a[0] = strdup(result.c_str()); // allocates memory for a new string and copies it over
[...]
free(a[0]); // don't forget this or you leak memory!
Comments
If you really, truly can't avoid doing this, you shouldn't throw away all that C++ offers, and descend to using raw arrays and horrible functions like strncpy.
One reasonable possibility would be to copy the data from the string to a vector:
char const *temp = result.c_str();
std::vector<char> a(temp, temp+result.size()+1);
You can usually leave the data in the string though -- if you need a non-const pointer to the string's data, you can use &result[0].
const char*and then do astrcpy