Does this code causes UB?

Question

I have checkd gcc and clang and both does not generate any warnings. I suppose that lifetime of temporary from foo() will be prolonged untill the end of full expression which is where semicolon in bar function call is located.

#include <iostream>
#include <string>

struct A
{
    std::string foo() const{ return "aaa"; }
};

void bar(const char* c) {
    std::cout << c;
}

int main()
{
    A a;
    bar(a.foo().c_str()); // Is this safe?
    bar(a.foo().substr().c_str()); // or this
}

yes, you're right, it's safe – Alexey Andronov Sep 27 '16 at 08:51 — Alexey Andronov, Sep 27 '16 at 08:51

Hatted Rooster · Accepted Answer · 2016-09-27T09:12:55.717

8

The temporary returned by foo() (and substr()) will continue to exist until the end of the bar call (after the chain of method calls), this is safe.

int main()
{
    A a;
    bar(a.foo().c_str()); 
    //temporary is destroyed here
    bar(a.foo().substr().c_str()); 
    // and here
}

Classical undefined behaviour case :

int main()
{
    A a;
    const char* charPtr = a.foo().c_str();
    printf("%s", charPtr);
}

Temporary std::string is created, a pointer to it's buffer is returned by c_str() and the temporary goes out of scope and is destroyed. charPtr now is a pointer pointing to an invalid location ( a dead std::string ).

edited Sep 27 '16 at 09:12

answered Sep 27 '16 at 09:01

Hatted Rooster

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UB example is not exactly right. charPtr remains valid until a non-const operation is called on a. – Trevir Sep 27 '16 at 09:17
@Trevir `printf` trying to use it to print it out should be just enough. – Hatted Rooster Sep 27 '16 at 09:17
How does printf interfer with object a? – Trevir Sep 27 '16 at 09:19
@Trevir It doesn't but it doesn't have to? `charPtr` is pointing to invalid memory, `printf` will try to deref that, that's undefined behaviour. – Hatted Rooster Sep 27 '16 at 09:20
as I pointed out, charPtr is valid until a non const operation is invoked on object a. – Trevir Sep 27 '16 at 09:21
2

@Trevir: There is UB unrelated to what happens to `a`. Lifetime of the `std::string` returned by `foo` ends before the call to `printf`. (and so `charPtr` is a dangling pointer). – Jarod42 Sep 27 '16 at 09:23
1

Yeah right. I oversaw foo() returning a temporary ... – Trevir Sep 27 '16 at 09:27

score 4 · Answer 2 · edited May 23 '17 at 12:02

4

bar(a.foo().c_str());
bar(a.foo().substr().c_str());

Short answer - Yes, both are safe.

What you're looking at is an rvalue.

An rvalue (so-called, historically, because rvalues could appear on the right-hand side of an assignment expression) is an xvalue, a temporary object or subobject thereof, or a value that is not associated with an object.

Read more: What are rvalues, lvalues, xvalues, glvalues, and prvalues?

edited May 23 '17 at 12:02

Community

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answered Sep 27 '16 at 09:02

Ivan Rubinson

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1

It being an rvalue isn't really the point – Lightness Races in Orbit Sep 27 '16 at 09:07
isnt? c_str() returns a pointer into an rvalue object. so it kinda matters. if you save this pointer, its dangerous! – jonas_toth Sep 27 '16 at 09:08

Does this code causes UB?

2 Answers2