Assigning a const char[] do you need to put in a null term

Question

We are setting a char [] to some hex values i.e.

char [] test1 = {0x30,0x31,0x32,0x33,0x34,0x35};

Then we copy it into a string using

string teststring(test1, sizeof(test1));

Is the array suppose to be null terminated? or is the way we do the assignment, C++ is smart enough to know that its null terminated and have it appended anyhow?

Answer 1

Using this string constructor you don't need to add an extra ending 0. However, I'd add the ending 0 just in case, or so that this variable could be used in other contexts where pointers to chars are used. In this case, the sizeof should be modified:

string teststring(test1, sizeof(test1) - 1);

Answer 2

Since you are using the sizeof operator and supplying the length of the array, you should not need to add the NULL.

You can find the API for the constructors here. It mentions this explicitly.

As mentioned in other solutions however, if you decided to create an array of wchar_ts then you would need to modify your supplied length argument to the constructor as follows:

sizeof(test1) / sizeof(wchar_t)

This is because the sizeof operator returns the size in bytes of the target, not the number of elements. For your current question using char does not have this requirement since the size of a char is defined by the C++ to be 1, thus division is not needed.

Answer 3

It is not null-terminated. Nulls have no special meaning for std::string. You can do something like this freely:

#include <string>
#include <iomanip>
#include <iostream>
int main()
{
    char test1 [] = {0x30, 0x31, 0, 0x33, 0, 0x35};
    std::string teststring(test1, sizeof(test1));
    for(size_t i = 0; i<teststring.size(); ++i)
        std::cout << std::hex << std::showbase << (int)teststring[i] << ' ';
    std::cout << '\n';
}

Although, of course, if I used std::string teststring(test1); as the constructor in this example, the resulting string would have been 2 characters long.