在没有new的情况下为指针赋值

Question

根据我对 C++ 的基本了解，我假设以下代码会出现运行次错误。因为编译器没有为y指针分配任何space，我应该在给y指针赋值之前添加y = new int;。

我错了还是编译器隐式为 y 指针分配了 space？（我用 Dev-C++ 4.9.9.2 编译了我的代码。）

#include <cstdlib>
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{   
    int* x;
    int* y;               
    x = new int;          
    *x = 42;              
    cout << *x << "\n";   
    *y = 13;              
    cout << *y << "\n";   
}

Answer 1

第 4.1 节指出：

An lvalue (3.10) of a non-function, non-array type T can be converted to an rvalue. If T is an incomplete type, a program that necessitates this conversion is ill-formed. If the object to which the lvalue refers is not an object of type T and is not an object of a type derived from T, or if the object is uninitialized, a program that necessitates this conversion has undefined behavior. If T is a non-class type, the type of the rvalue is the cv-unqualified version of T. Otherwise, the type of the rvalue is T.

未定义意味着任何事情都有可能发生——无法保证。

来自维基Making pointers safer

A pointer which does not have any address assigned to it is called a wild pointer. Any attempt to use such uninitialized pointers can cause unexpected behavior, either because the initial value is not a valid address, or because using it may damage other parts of the program. The result is often a segmentation fault, storage violation or wild branch (if used as a function pointer or branch address).

Answer 2

Am I wrong or compiler has allocate space for y pointer implicitly?

没有，这样的赋值是未定义的行为。这意味着它可以长时间工作并且不会引起任何问题，但突然会使您的应用程序崩溃。变量 y 实际上被分配了一些随机值，*y=13; 将 13 分配给了一些随机内存地址，它可以是某个结构或堆栈的有效内存地址。如果您最初将 y 初始化为 nullptr（或 0、NULL）int* y=nullptr;，那么您应该会遇到应用程序崩溃。

在没有new的情况下为指针赋值

Assigning value to pointer without new

c++

pointers

memory-management

heap-memory