Question

此代码编译并执行。我知道在第一种情况下我们有未定义的行为。但在第二种情况下会发生什么呢？

#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输出在两种情况下都是“HELLO”，但我认为这是因为它没有清理原始内存。

在第二种情况下，临时对象到底是什么时候被破坏了？

Answer 1

实际上，这两种情况都没问题。

在第一种情况下，将临时绑定到const引用会将其生命周期延长到s的生命周期。因此，在main退出之前，它不会被销毁。

在第二种情况下，临时在包含它的完整表达式结束后被销毁。在这种情况下，它是函数调用。如果您将C字符串存储在遗忘bar的任何地方，然后尝试访问它，那么您的日期将是未定义的行为。

Answer 2

这两种情况都是明确定义的。要查看有问题的案例，请将c_str()的结果存储到std::string被破坏之后：

#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;
}

当我在Valgrind下运行时，它仅为变体情况（const char *s）提供错误：

==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)

什么时候在函数调用中发生临时对象的破坏？

2 个答案: