C ++中的sizeof()运算符是做什么的 [英] what is sizeof() operator doing in C++

问题描述

C中的 sizeof()运算符在编译时给出其操作数的大小.它不评估其操作数.例如

The sizeof() operator in C gives the size of its operand at compile time. It does not evaluate its operand. For example,

int ar1[10];
sizeof(ar1) // output 40=10*4
sizeof(ar1[-1]) // output 4
int ar2[ sizeof(ar1) ]; // generate an array of 40 ints.

谈到C ++模板类时，我发现了一些奇怪的结果.

When it came to C++ template class, I find some strange result.

template<typename T>
struct S{
    T a;
};

sizeof( S<int> )       // output 4
sizeof( S<bool> )      // output 1
sizeof( vector<int> )  // output 24
sizeof( vector<char> ) // output 24
sizeof( vector<bool> ) // output 40

我猜vector或其他STL容器上的 sizeof 取决于特定的环境.

I guess the sizeof on vector or other STL container depends on specific environment.

问题1.如何在C/C ++中实现 sizeof ?它不能是运行时函数.是宏吗?(我在在线教学视频中学到了什么).如果它是一个宏，它的 #define 是什么样的?什么时候执行 sizeof()?

Question 1. How is sizeof implemented in C/C++? It cannot be a run-time function. Is it a macro? (what I learned in a online tutorial vedio). If it is a Macro, what the #define of it looks like? When the sizeof() is executed?

问题2.如果我在 struct S 的定义中添加成员方法 void f(){} . sizeof(S< int>)仍为4.结构体的大小不应该增加吗?

Question 2. If I add a member method void f(){} to the definition of struct S. The sizeof(S<int>) is still 4. Shouldn't the size of the struct increase?

问题3..STL容器是模板类.以 vector 为例，它具有12个成员属性/类型和许多成员方法?容易解释 sizeof(S< int>)的输出.但是我发现很难解释 sizeof(vector< int>)的输出.模板类应在编译时实例化，并且编译器应完全了解该类的大小，即 vector< int> .所以 sizeof()操作员应该知道.

Question 3. STL containers are template classes. Take vector for example, it has 12 member attributes/types and many member methods? It is easy to explain the output of sizeof( S<int> ). But I find it hard to explain the output of sizeof( vector<int> ). Template class should be instantiated at compile-time and the compiler should have total knowledge of the size of the class, i.e. vector<int>. So should sizeof() operator know.

推荐答案

按照问题1 : sizeof 由编译器实现和评估.它不是宏，并且始终提供编译时结果.从概念上讲，您可以想象编译器用数字替换每个 sizeof .

As per Question 1: sizeof is implemented and evaluated by the compiler. It is not a macro, and it always provides a compile-time result. Conceptually, you can imagine that the compiler replaces every sizeof with a number.

根据问题2 : sizeof 计算一个 S 实例占用的存储量.方法不按实例存储，而仅按字段存储(因为每个实例存在一次).但是，一种方法确实会占用某个位置的静态存储空间，以保存该功能的机器代码.

As per Question 2: sizeof counts the amount of storage that one instance of S occupies. A method does not take per-instance storage, only fields do (since they exist once per instance). A method does, however, occupy static storage somewhere to hold the machine code for the function.

根据问题3 :对于 sizeof(vector< int>)，编译器将计算 vector< int> 的大小为此实例化.如果由于 vector 可以具有可变大小而感到困惑:的确如此，但是多余的存储空间是从堆中分配的，因此不会反映在 sizeof 的应用结果中到 vector .

As per Question 3: For sizeof(vector<int>) the compiler computes the size of the vector<int>, which it instantiates in order to do so. If you are confused because a vector can be of variable size: that is true, but the extra storage is allocate from the heap and thus not reflected in the result of sizeof applied to a vector.

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