free() ：无效指针以及如何重用当前释放的指针

Question

我想将我的指针指向另一个对象并释放前一个对象的内存。这就是我所做的，但我有一个无效的指针错误。正确的做法是什么？

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char const *argv[])
{
    int number = 5;
    int number2 = 10;

    // Give memory manually 
    int *pointer2number = (int *) malloc(sizeof(int));
    pointer2number = &number;

    printf("Number is %i\n", number);       
    printf("Address of number is %p\n", pointer2number);

    // Free memory 
    free(pointer2number);
    pointer2number = (int *) malloc(sizeof(int));
    pointer2number = &number2;
    printf("New number is %i\n", *pointer2number);
    return 0;
}

Answer 1

int *pointer2number = (int *) malloc(sizeof(int));
pointer2number = &number;

你将number的地址赋值给指针pointer2number，并用它替换之前存储在pointer2number中的已分配动态内存的地址。

第二次尝试 number2 也是如此。

任何尝试 free() 之前未由内存管理函数分配的内存，都会调用未定义的行为：

"The free function causes the space pointed to by ptr to be deallocated, that is, made available for further allocation. If ptr is a null pointer, no action occurs. Otherwise, if the argument does not match a pointer earlier returned by a memory management function, or if the space has been deallocated by a call to free or realloc, the behavior is undefined."

Source: ISO/IEC 9899:2018 (C18), Section 7.22.3.3/2

作为副作用，之前分配的动态内存也不会 free()。

想多了你程序的逻辑！

free() ：无效指针以及如何重用当前释放的指针

free() : invalid pointer and how to reuse a currently freed pointer

c

free