nullptr引用未定义的行为在C + +？ [英] Are nullptr references undefined behaviour in C++?

问题描述

以下代码愚弄了 nullptr 指针和引用：

The following code fools around with nullptr pointer and reference:

#include <cstdio>

void printRefAddr(int &ref) {
    printf("printAddr %p\n", &ref);
}

int main() {    
    int *ip = nullptr;
    int &ir = *ip;

    // 1. get address of nullptr reference
    printf("ip=%p &ir=%p\n", ip, &ir);

    // 2. dereference a nullptr pointer and pass it as reference
    printRefAddr(*ip);

    // 3. pass nullptr reference
    printRefAddr(ir);

    return 0;
}

问题：在C ++标准中， 1..3有效代码或未定义的行为？

Question: In C++ standards, are commented statements 1..3 valid code or undefined behavior?

对于不同版本的C ++，这是相同或不同的（旧的版本当然会使用 0 literal而不是 nullptr 关键字）？

Is this same or different with different versions of C++ (older ones would of course use 0 literal instead of nullptr keyword)?

奖金问题：是否有已知的编译器/优化选项，这实际上会导致上面的代码做某事意外/崩溃？例如，是否有任何编译器的标志，它将生成 nullptr 的隐含断言，其中引用被初始化，包括从 * ptr传递引用参数？

Bonus question: are there known compilers / optimization options, which would actually cause above code to do something unexpected / crash? For example, is there a flag for any compiler, which would generate implicit assertion for nullptr everywhere where reference is initialized, including passing reference argument from *ptr?

奇怪，没有什么意外的输出示例：

An example output for the curious, nothing unexpected:

ip=(nil) &ir=(nil)
printAddr (nil)
printAddr (nil)

推荐答案

2.解除引用一个nullptr指针并将其作为引用传递

// 2. dereference a nullptr pointer and pass it as reference

解除空指针<行为，因此无论您是以引用还是通过值传递它，事实是您已经取消引用它，因此调用UB，意味着从那时起所有投注都将关闭。

Dereferencing a null pointer is Undefined Behaviour, and so whether you pass it as a reference or by value, the fact is that you've dereferenced it and therefore invoked UB, meaning from that point on all bets are off.

您已经在这里调用UB：

You've already invoked UB here:

int &ir = *ip; //ip is null, you cannot deref it without invoking UB.

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