如何转换矢量<string>到 char* arr[]?</string>

Question

Recently I want to convert vector to char* array[].

So I had found the solution. But It was not safety way..

Here is my code

char** arr = new char* [4];

vector<string> vv;

// setting strings
for (int i = 0 ;i < 4; i++)
{
    vv.emplace_back("hello world");
}
// convert
for (int i = 0; i < 4; i++)
{
    arr[i] = new char[vv[i].size()];
    memcpy(arr[i], vv[i].c_str(), vv[i].size());
}

// prints
for (int i = 0; i < 4; i++)
{
    cout << arr[i] << endl;
}

// output
// hello world羲羲?솎솨
// hello world羲羲拂솽솨
// hello world羲羲
// hello world羲羲?펺솨

// delete memorys
for (unsigned index = 0; index < 4; ++index) {
    delete[] arr[index];
}

delete []arr;

Why does it happen string crash?? Is there no safe way anymore?

Answer 1

When you use memcpy , arr[i] is not guaranteed to be a null-terminated string. To treat arr[i] as null terminated string, as in

cout << arr[i] << endl;

causes undefined behavior.

---

You need couple of minor changes.

1. Allocate one more byte of memory.
2. Use strcpy instead of memcpy .

arr[i] = new char[vv[i].size() + 1];
strcpy(arr[i], vv[i].c_str());

Answer 2

There's an easier way of doing this if you can maintain the lifetime of your initial vector .

for (int i = 0; i < 4; i++)
{
    arr[i] = vv[i].c_str();
}

This way you won't allocate no additional memory, however, you'd have to keep in mind that once your initial vector gets destroyed, that array will be corrupted as well. But if you need such conversion for some simple synchronous api call within the same thread, that would do the trick.

Answer 3

In such cases I usually use this ugly construct.

arr[i] = new char[vv[i].size() + 1];
arr[i][vv[i].copy(arr[i], vv[i].size())] = 0;