cannot convert from 'const char ' to 'char ' for std::string::c_str

Question

This gives the error: cannot convert from 'const char *' to 'char *'.

class Mock
{
public:
    ...
    static void func(char **result)
    {
          *result = (resultsI++)->c_str();
    }
static std::vector<std::string> results;
static std::vector<std::string>::iterator resultsI;
};

std::vector<std::string> Mock::results;
std::vector<std::string>::iterator Mock::resultsI;

How can I validly get rid of this error without changing the interface to the function func? The implementer of this interface:

void (func*)(char **result)

forgot to use const char** in the signature. I can't change it.

Remember this is a mock and I'm only used in my unit tests.

You could (but shouldn't) cast away the constness (const_cast<char*>((resultsI++)->c_str());) — Ryan Guthrie
– Ryan Guthrie, Commented Nov 16, 2012 at 16:24
const is there for your own protection. By casting it away, you lose that protection. — wardd
– wardd, Commented Nov 16, 2012 at 17:13

Mark Ransom · Accepted Answer · 2012-11-16 16:25:05Z

6

Try:

*result = &(*resultsI++)[0];

Although this isn't guaranteed to work prior to C++11 it is known to be OK on most or all current compilers.

The danger is that if the function tries to change the length of the string, you could get some nasty errors. Changing individual characters should be OK.

answered Nov 16, 2012 at 16:25

Mark Ransom

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5 Comments

Kos Over a year ago

So only C++11 has specified that std::string must have its memory contiguous?

John Dibling Over a year ago

@Kos: That's correct. Although most C++03 implementations I'm familiar with actually do use contiguous storage, this wasn't guaranteed in the C++03 Standard. When I need C++03 guaranteed contigious storage, I typically use vector<char> rather than string

Mark Ransom Over a year ago

@Kos, the guarantee was added to C++11 after a survey of all the committee members showed that their std::string implementations were already compliant.

Mark Ransom Over a year ago

@JohnDibling, I've used vector<char> too, it depends on the task. Sometimes it's great and sometimes not having the string methods makes it tedious.

Loki Astari Over a year ago

The danger here is that any modification to the original string can cause it to move and their is no guarantee that it is '\0' terminated which may be a problem if you pass it to a C-String function. If I was doing something this dangerous I like to make it obvious that it is dangerous. By using a const_cast<char*>((resultsI++)->c_str()) you make it stick out because we know we should not be using const cast. That way you know your code is bad and it sticks out.

Douglas Leeder · Accepted Answer · 2012-11-16 16:24:15Z

4

In test code, and if you are certain that the user of the interface isn't going to mutate the char*, maybe you could use a const_cast?

That assumes the caller doesn't take ownership of the char *; if that is the case, then you should make a copy.

answered Nov 16, 2012 at 16:24

Douglas Leeder

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Comments

Praetorian · Accepted Answer · 2012-11-16 16:24:54Z

4

If you're absolutely certain that the interface function will not modify the string, you can use

*result = const_cast<char *>((resultsI++)->c_str());

to remove constness.

Otherwise, another option is to switch from using std::vector<std::string> to std::vector<std::vector<char>> but then you'll have to make sure you properly null terminate the strings.

answered Nov 16, 2012 at 16:24

Praetorian

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