85

Compiling with gcc-7.1 with the flag -std=c++17, the following program raises an error:

#include <string_view>
void foo(const char* cstr) {}
void bar(std::string_view str){
    foo(str);
}

The error message is

In function 'void bar(std::string_view)':
error: cannot convert 'std::string_view {aka std::basic_string_view<char>}' to 'const char*' for argument '1' to 'void foo(const char*)'
 foo(str);

I'm surprised there is no conversion to const char* because other libraries (abseil, bde), provide similar string_view classes which implicitly convert to const char*.

0

3 Answers 3

114

A std::string_view doesn't provide a conversion to a const char* because it doesn't store a null-terminated string. It stores a pointer to the first element, and the length of the string, basically. That means that you cannot pass it to a function expecting a null-terminated string, like foo (how else are you going to get the size?) that expects a const char*, and so it was decided that it wasn't worth it.

If you know for sure that you have a null-terminated string in your view, you can use std::string_view::data.

If you're not you should reconsider whether using a std::string_view in the first place is a good idea, since if you want a guaranteed null-terminated string std::string is what you want. For a one-liner you can use std::string(object).c_str() (note: the return value points to a temporary std::string instance that will get destroyed after the end of the expression!).

Sign up to request clarification or add additional context in comments.

4 Comments

The explanation for why std::string_view doesn't provide a conversion is great, but only offers a provisional answer to the question.
Your recommendation about std::string(object).data() is a bad one, because in most cases the data will point to unallocated storage.
@rems4e An alarm went off for me as well when I saw std::string(object).data() recommended without the important caveat that the returned pointer is to data inside the temporary std::string object, which lasts only til the end of the largest enclosing expression.
Please remove std::string(object).data() from the answer or provide an easily visible warning. The std::string(object) here gets deallocated, so the resulting C string is undefined. I've seen this snippet being pasted all over the internet and in various code bases. This answer has caused an immense amount of damage.
61

Simply do a std::string(string_view_object).c_str() to get a guaranteed null-terminated temporary copy (and clean it up at the end of the line).

This is required because string view doesn't guarantee null termination. You can have a view into the middle of a longer buffer, for example.

If this use case is expensive and you have proven it to be a bottleneck, you can write an augmented string_view that tracks if it is null terminated (basically, if it was constructed from a raw char const*).

Then you can write a helper type that takes this augmented string_view and either copies it to a std::string or stores the augmented string_view directly, and has an implicit cast-to-char const* that returns the properly null-terminated buffer.

Then use that augmented helper type everywhere in your code base instead of string_view, possibly augmenting string view interaction with std string as well to catch the cases where you have a view that goes to the end of the std string buffer.

But really, that is probably overkill.

A better approach is probably rewriting the APIs that take const char* to take string_view.

3 Comments

by " (and clean it up at the end of the line).", I can continue use the member of that Rvalue til the end of scope that std::string(string_view_object).c_str() sitting, right?
@sandthorn No, until the end of the full expression, not scope.
@sandthorn If you want the string to stick around, you'd maybe prefer std::string TempVar(string_view_object); then where you need it in scope, TempVar.c_str().
12

You can call foo(std::string(str).c_str()).

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.