c++ string constructor under the hood

Question

I have a simple program: ``` #include #include #include using namespace std; string read0() { int length = 4; char *cstr = new char[length]; string str(cstr); delete[] cstr; return str; } string read1() { int length = 4; char cstr[length]; memset(cstr, '-', 4); string str(cstr); return str; } string read2() { const char* cstr = "abcd"; string str(cstr); return str; } ``` In all 3 functions above, for constructing a string, they `call basic_string( const CharT* s, const Allocator& alloc = Allocator()`. When I use valgrind/massif to check the heap usage, function read0 only uses 4 bytes (from `new`), but read1 and read2 both use 29 bytes. Here's some detail output of massif: For read0: 16.67% (4B) (heap allocation functions) malloc/new/new[], --alloc-fns, etc. ->16.67% (4B) 0x400A0B: read0() (main.cpp:10) ->16.67% (4B) 0x400BC8: main (main.cpp:40) For read1 and read2: 72.50% (29B) (heap allocation functions) malloc/new/new[], --alloc-fns, etc. ->72.50% (29…

Accepted Answer

Someone please correct me if I'm wrong, but I think I know why this is.

In `read0`, you do:

```
char *cstr = new char[length];
string str(cstr);
```

You don't initialize `cstr` at all, so it could be undefined. The string constructor takes a null-terminated c-string and copies it up until the null terminator. I think it finds it at the very first element, so it's only copying the pointer that would point to such a string, which presumably takes up 4 bytes.

I think your `read1` is similar. It ends up finding a null terminator at the some point past the string itself, since it's not null-terminated, which ends up being 29 bytes.

I don't know why `read2` is doing the same thing though, to be honest. It could be I'm wrong about the reason for `read1` above and that 29 bytes (minus the 4 characters/bytes in the c-string) is the minimal operating cost of a string in your architecture and compiler's implementation of the STL.

In any case, to narrow down the various possibilities, I suggest you null-terminate the strings in `read0` and `read1` and try your experiment again by allocating one extra element and setting the last element to `'\0'` or by using the alternative string constructor that takes an additional second parameter which denotes how many characters to copy:

```
string(non_null_terminated_string, this_many_characters_to_copy)
```

c++ string constructor under the hood

Solution

Problem