问题描述
如果我们有这个代码:
int foo=100;
int& reference = foo;
int* pointer = &reference;
引用的数据和指针的数据没有实际的二进制差异.(它们都包含 foo 在内存中的位置)
There's no actual binary difference in the reference's data and the pointer's data. (they both contain the location in memory of foo)
第 2 部分
那么指针和引用之间的所有其他差异在哪里(在这里讨论)进来?编译器是否强制执行它们，或者它们实际上是汇编级别上的不同类型的变量?换句话说，以下是否产生相同的汇编语言?
So where do all the other differences between pointers and references (discussed here) come in? Does the compiler enforce them or are they actually different types of variables on the assemebly level? In other words, do the following produce the same assembly language?
foo=100;
int& reference=foo;
reference=5;
foo=100;
int* pointer=&foo;
*pointer=5;
推荐答案
理论上，它们可以以不同的方式实现.
Theoretically, they could be implemented in different ways.
实际上，我见过的每个编译器都编译指向相同机器代码的指针和引用.区别完全在于语言层面.
In practice, every compiler I've seen compiles pointers and references to the same machine code. The distinction is entirely at the language level.
但是，就像 cdiggins 所说的那样，您不应该依赖这种概括，直到您验证它对您的编译器和平台来说是正确的.
But, like cdiggins says, you shouldn't depend on that generalization until you've verified it's true for your compiler and platform.
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