Thread: string vs dynamic char array

Ah, of course. Using resize() & size() are correct.

Was I at least right about needing to manually add a '\0' to the vector?

Cpjust: vec.push_back( '\0' ); // I can't remember if this is required, but it doesn't hurt.

Apparrently its working without the '\0'.
At least i dont have garbage at the end and nothing missing either.

And by the way why would you need to allocate space in a vector?

Arent vectors like strings?

"Actually you could say that a string is a vector<char> A container of chars, or an advanced array of chars. "

Yes, but only if you use the vector's functions to add things to it. If you pass a pointer to the first element of the vector to a function, it won't be calling the vector's functions, it'll just write to it like it's a raw array.

Apparrently its working without the '\0'.
At least i dont have garbage at the end and nothing missing either.

If things like the following work:

Code:

string s = &v[0];

then this can only be because of pure (un)luck. If you are going to treat it as a C-style string, it has to be null-terminated. If you copy from the string to the vector, then string.end() or string.size() won't include the null-terminator (which doesn't have to be there).

Thank you, anon.
You sure know how to recognize different types of trees from quite a long way away.

I will '\0' terminate it then.

And if i concatenate a vector element to a string like this, do i need to '\0' terminate it too?

Code:

int main()
{
    string  str1 =  "Hello ";
    string  str2 =  "World";
    vector <char> vec( str2.begin(), str2.end());

    str1 += &vec[0];

    cout  << str1 << endl;

    return 0;
}

Originally Posted by Ducky

And if i concatenate a vector element to a string like this, do i need to '\0' terminate it too?

Code:

int main()
{
    string  str1 =  "Hello ";
    string  str2 =  "World";
    vector <char> vec( str2.begin(), str2.end());

    str1 += &vec[0];

    cout  << str1 << endl;

    return 0;
}

Yes, because with the code you have, there is no guarantee that &vec[0] is null terminated.

Thank you Bithub!

Can it be done simpler than this?

Code:

 
str1 += &vec[0];

str1 += '\0';

Originally Posted by Ducky

Thank you Bithub!

Is it ok like this too?

Code:

 str1 += '\0';

You need to null terminate the vector before you add it to the string, not after. A push_back('\0') on the vector before adding it to the string should do the trick.

Thanks Daved!

You are revealing details about vectors that doesnt really explained anywhere. Interesting.

std::string does not require null termination. The question of how to properly "null terminate" the vector is moot.

Convert vector of char to string:

Code:

std::string foo( vec.begin(), vec.end() );

Convert string to vector of char:

Code:

std::vector< char > foo( str.begin(), str.end() );

And yes, when you call c_str() on the resulting string, it will be null terminated automatically. std::string is nothing like a C-string.

If you just wanted a C string for reading, you'd use string::c_str(). The reason for using a vector is to pass it to a C function for writing into a C string, and for that you'd need to add a '\0' char to the end (assuming it's going to append instead of just overwriting the string).