临时对象的销毁究竟何时在函数调用中发生？

Question

This code compiles and executes. I know that we have undefined behaviour in the first case. But what happens exactly in the second case?

#include <string>
#include <iostream>
#include <cstdio>

std::string foo() {
    return "HELLO";
}

void bar(const char *p) {
    std::printf("%s\n", p);
}


int main() {

    // FIRST CASE:
    // I know this is bad, because after the assignment
    // the variable returned by foo() is destroyed and we
    // have a bad reference.
    const std::string &s = foo(); 
    bar(s.c_str());


    // SECOND CASE:
    // But what about that ? I don't know exactly if the 
    // object is alive after the call to c_str() 
    bar(foo().c_str());

    return 0;
}

GCC output is in both cases "HELLO" but I think that's because it's not cleaning the raw memory.

In the second case when exactly is the temporary object destroyed?

Answer 1

Actually, both of these cases are okay.

In the first case, binding the temporary to a const reference extends its lifetime to that of s . So it won't be destroyed until main exits.

In the second case, the temporary is destroyed after the end of the full-expression containing it. In this case, it is the function call. If you stored that C string anywhere which outlived bar and then tried to access it, then you're due a date with undefined behaviour.

Answer 2

Both of those cases are well-defined. To see a problematic case, store the result of c_str() until after the std::string is destructed:

#include <string>
#include <cstdio>

std::string foo() {
    return "HELLO";
}

void bar(const char *p) {
    std::printf("%s\n", p);
}

int main() {
    {
        // FIRST CASE:
        // This is okay, because the reference is const, so the object is alive
        // until s goes out of scope.
        const std::string &s = foo();
        bar(s.c_str());
    }

    {
        // VARIANT FIRST CASE:
        // This is bad; the pointer is dangling
        const char *s = foo().c_str();
        bar(s);
    }

    {
        // SECOND CASE:
        // Is the object still alive after the call to c_str()?  Yes, it's alive
        // until after bar() has returned.
        bar(foo().c_str());
    }

    return 0;
}

When I run this under Valgrind, it gives errors only for the variant case (with const char *s ):

==9981== Invalid read of size 1
==9981==    at 0x4C2E0E2: strlen (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==9981==    by 0x543EC7B: puts (ioputs.c:36)
==9981==    by 0x400937: bar(char const*) (37946437.cpp:9)
==9981==    by 0x4009AA: main (37946437.cpp:25)
==9981==  Address 0x5aabcf8 is 24 bytes inside a block of size 30 free'd
==9981==    at 0x4C2C2BC: operator delete(void*) (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==9981==    by 0x4F058FD: std::basic_string<char, std::char_traits<char>, std::allocator<char> >::~basic_string() (in /usr/lib/x86_64-linux-gnu/libstdc++.so.6.0.22)
==9981==    by 0x40099E: main (37946437.cpp:24)