C++ std::move unexpected behaviour (moving string when intended for smart pointer)

Question

When I pass a string literal to a dynamically allocated unique_ptr and then store in vector (using std::move since there's apparently no other way), I get an odd return value when the string is fetched - it appears to have been cut/moved.

#include <iostream>
#include <vector>
#include <memory>

class Something {
private:
    const char* name;
public:
    Something(const char* name) {
        this->name = name;
    }

    void printDetails() {
        std::cout << this->name << std::endl;
    }
};

int main()
{
    std::vector<std::unique_ptr<Something>> stuff;

    for(int i = 0; i < 10; i++) {
        std::unique_ptr<Something> thing(new Something("A nice fun string" + i));
        stuff.push_back(std::move(thing));
    }

    for(uint i = 0; i < stuff.size(); i++) {
        stuff[i]->printDetails();
    }

    return 0;
}

Output:

A nice fun string
 nice fun string
nice fun string
ice fun string
ce fun string
e fun string
 fun string
fun string
un string
n string

You're storing a pointer to the address of the start of a string constant + i. — Jonathan Potter, Nov 27 '15 at 08:12
It has absolutely nothing to do with unique pointers. You should try to create an MCVE and see for yourself. — juanchopanza, Nov 27 '15 at 08:13
I see that you have learned some Java before. There are many things you need to unlearn. — molbdnilo, Nov 27 '15 at 08:17
@molbdnilo meh java is my main language. This was merely made as a test to move away from raw pointers then i ran into this — user3530525, Nov 27 '15 at 08:34

score 3 · Accepted Answer · edited May 23 '17 at 11:59

3

It's not moved. "ABC" is string literal, and actually is just a pointer char const *. Adding integer just moves it forward, so "ABC" + 1 acts like "BC".

C-style string literals are just char arrays, and arrays, when you access their contents, are just pointers to their first element. So adding integer th this pointer will work as suffix of your array.

What you want is probably concat your literal and integer, it's not so easy in C++. The fastest approach using C++11 looks terrible:

string("ABC") + to_string(i)

And, if you really want char * in the end, copy the resulting string of that (it's well described here)

(string("ABC") + to_string(i)).c_str()

edited May 23 '17 at 11:59

Community

1
1

answered Nov 27 '15 at 08:13

Lapshin Dmitry

1,084
9
28

4

I wouldn't store the result of `c_str()` from a temporary. – Bo Persson Nov 27 '15 at 08:23
2

"arrays are just pointers to their first element" - that's completely incorrect. – T.C. Nov 27 '15 at 08:25
@T.C. when you do aceess an array it's working like pointer to the first element. There are limitations: string literals are unchangable and you can't reassign that pointer. – Lapshin Dmitry Nov 27 '15 at 08:27
1

@LapshinDmitry arrays decay into pointers, but they are not pointers. – Simple Nov 27 '15 at 08:30
@Simple okay, I have not written the idea well, will edit. – Lapshin Dmitry Nov 27 '15 at 08:31
@Bo Persson: I wouldn't even store it from a non-temporary ;) – MikeMB Nov 27 '15 at 08:39

score 3 · Answer 2 · answered Nov 27 '15 at 08:20

3

This part causes the error:

("A nice fun string" + i)

i must be a char or a char* itself, you are increasing the index of "A nice fun string" by i.

answered Nov 27 '15 at 08:20

b.holz

237
3
17

C++ std::move unexpected behaviour (moving string when intended for smart pointer)

2 Answers2