如何与string :: copy C ++相反？

Question

How to copy std::string to specific index in char array in C++ ?

example :

    std::string str = "aaaabbb";
    unsigned char arr[3];
    str.copyToFunction(arr,4,3);

result :

    arr value should be = "bbb".

Thanks.

Edit:

Does it works with std::array ? or there is another way to do it?

Answer 1

std::copy(str.begin() + 4, str.begin() + 4 + 3, arr);

std::copy documentation

Answer 2

Try this:

std::string str = "aaaabbb";
std::vector<char> myvector (7);
std::copy ( str, str+7, myvector.begin()+4, myvector.end() );

Answer 3

Without bounds checking (which you probably need to do first):

std::copy(str.begin() + 4, str.end(), arr);

Note there is no null termination, which you probably want to add:

auto end = std::copy(str.begin() + 4, str.end(), arr);
*end++ = '\0';

Of course arr must now have length 4 at least.

Answer 4

If you want that the array indeed keep string literal "bbb" then you have to declare the array like

    unsigned char arr[4];

Otherwise if you do not need to append the array with the terminating zero then the array can be declared as you did.

You can use for example a standard C function declared in header <cstring> .

#include <cstring>

//...


memcpy( arr, str.c_str() + 4, 3 );

Or you can use standard algorithm std::copy declared in header <algorithm>

#include <algorithm>

//...

std::copy( str.begin() + 4, str.begin() + 7, arr );

Or you can rewrite the same using standard function std::next declared in header <iterator>

#include <algorithm>
#include <iterator>    
//...

std::copy( std::next( str.begin(), 4 ), std::next( str.begin(), 7 ), arr );

If you need that the array would contain a string then as I said early you have to append zero

    unsigned char arr[4];

    // using some method of copying

    arr[3] = '\0;'

Answer 5

Why not just use substr ? Very good at chopping up strings