通过读取输出了解 C 中的 malloc()、realloc() 和 free()答案

【问题标题】：Understanding malloc(), realloc() and free() in C from reading an output通过读取输出了解 C 中的 malloc()、realloc() 和 free()
【发布时间】：2019-09-19 22:54:53
【问题描述】：

我需要编写一个关于如何实现内存管理的分配，以便了解此代码输出中的少数非零数字代表什么。

我知道 malloc() 函数会保留一块指定字节数的内存。并且，它返回一个 void 类型的指针，该指针可以转换为任何形式的指针。我也知道，如果动态分配的内存不足或超出要求，你可以使用 realloc() 函数改变之前分配的内存大小。

这是我要分析的代码：

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

/*** Just playing with the malloc(), realloc(), free()
 *** in order to guess how memory management
 *** is implemented on this machine. If you get SEGMENTATION
 *** FAULT while addressing unallocated memory, just run
 *** the program with different "min" and/or "max" values,
 *** explicitly given on the command line through argv[]
 *** NOTICE: the default values are appropriate for the 32bit systems
 *** available in the labs ***/


void showmem (unsigned char *ptr, int min, int max, char name) {
    int i;
    for (i = min; i < 0; i++)           
      printf ("%hhu ",ptr[i]);          
    printf ("*%c=%hhu ",name,*ptr);     
    for (i = 1; i <= max; i++)          
      printf ("%hhu ",ptr[i]);
    printf ("\n\n");
}


int main(int argc, char**argv) {
    unsigned char *p, *q, *o;
    int sz=1, min=-8, max=60;

    if ( argc > 1 )             
        sscanf(argv[1],"%d",&sz);       
    if ( sz <= 0 )
        sz = 1;
    else if ( sz > 300 )
        sz = 300;
    if ( argc > 2 )             
        sscanf(argv[2],"%d",&min);
    if ( min > -1 )
        min = -1;
    else if ( min < -50 )
        min = -50;
    if ( argc > 3 )             
        sscanf(argv[3],"%d",&max);
    if ( max < sz )
        max = sz;
    else if ( max > (sz+100) )
        max = sz+100;

    printf("... allocating %d bytes to p[] (show memory from p[%d] to p[%d])\n\n",sz,min,max);
    p = (unsigned char*)malloc(sz);     
    if ( p == NULL ) {
        perror ("Error allocating p\n");
        return -1;
    }
    showmem (p,min,max,'p');

    printf("... allocating %d bytes to q[]\n\n",sz);
    q = (unsigned char*)malloc(sz);     
    if ( q == NULL ) {
        perror ("Error allocating q\n");
        return -1;
    }
    showmem (p,min,max,'p');
    showmem (q,min,max,'q');

    sz += 10;                   
    printf("... reallocating p[] to %d bytes (show old p[], new p[], and q[])\n\n",sz);
    o = p;
    p = (unsigned char*)realloc((void*)p,sz);   
    showmem (o,min,max,'o');
    showmem (p,min,max,'p');
    showmem (q,min,max,'q');

    sz += 15;                   
    printf("... reallocating p[] to %d bytes\n\n",sz);
    p = (unsigned char*)realloc((void*)p,sz);   //void e' l'indirizzo di memoria. sz e' la nuova dimensione
    showmem (o,min,max,'o');
    showmem (q,min,max,'q');
    showmem (p,min,max,'p');

    sz -= 25;                   
    printf("... reallocating p[] to %d bytes\n\n",sz);
    p = (unsigned char*)realloc((void*)p,sz);
    showmem (o,min,max,'o');
    showmem (q,min,max,'q');
    showmem (p,min,max,'p');

    printf("... freeing p\n\n");
    free((void*)p);     `       
    showmem (o,min,max,'o');
    showmem (q,min,max,'q');
    showmem (p,min,max,'p');

    printf("... freeing q\n\n");
    free((void*)q);
    showmem (o,min,max,'o');
    showmem (q,min,max,'q');
    showmem (p,min,max,'p');

    printf("... freeing old p\n\n");
    free((void*)o);
    showmem (o,min,max,'o');
    showmem (q,min,max,'q');
    showmem (p,min,max,'p');

    return 0;

}

这是在没有任何其他输入的情况下编译文件的输出：


... allocating 1 bytes to p[] (show memory from p[-8] to p[60])

33 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 129 253 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... allocating 1 bytes to q[]

33 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 97 253 1 0 0 

33 0 0 0 0 0 0 0 *q=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 97 253 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... reallocating p[] to 11 bytes (show old p[], new p[], and q[])

33 0 0 0 0 0 0 0 *o=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 97 253 1 0 0 

33 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 97 253 1 0 0 

33 0 0 0 0 0 0 0 *q=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 97 253 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... reallocating p[] to 26 bytes

33 0 0 0 0 0 0 0 *o=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 0 0 0 0 

33 0 0 0 0 0 0 0 *q=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

49 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 253 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... reallocating p[] to 1 bytes

33 0 0 0 0 0 0 0 *o=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 0 0 0 0 

33 0 0 0 0 0 0 0 *q=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

49 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 253 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... freeing p

33 0 0 0 0 0 0 0 *o=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 0 0 0 0 

33 0 0 0 0 0 0 0 *q=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

49 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 253 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... freeing q

33 0 0 0 0 0 0 0 *o=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 0 0 0 0 0 0 0 112 210 6 212 50 86 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 0 0 0 0 

33 0 0 0 0 0 0 0 *q=112 210 6 212 50 86 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

49 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 253 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... freeing old p

33 0 0 0 0 0 0 0 *o=144 210 6 212 50 86 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 33 0 0 0 0 0 0 0 112 210 6 212 50 86 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 0 0 0 0 

33 0 0 0 0 0 0 0 *q=112 210 6 212 50 86 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

49 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 49 253 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

这是使用输入“64”编译时的输出：


... allocating 64 bytes to p[] (show memory from p[-8] to p[64])

81 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... allocating 64 bytes to q[]

81 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

81 0 0 0 0 0 0 0 *q=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... reallocating p[] to 74 bytes (show old p[], new p[], and q[])

81 0 0 0 0 0 0 0 *o=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

97 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

81 0 0 0 0 0 0 0 *q=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... reallocating p[] to 89 bytes

81 0 0 0 0 0 0 0 *o=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

81 0 0 0 0 0 0 0 *q=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

113 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... reallocating p[] to 64 bytes

81 0 0 0 0 0 0 0 *o=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

81 0 0 0 0 0 0 0 *q=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

81 0 0 0 0 0 0 0 *p=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... freeing p

81 0 0 0 0 0 0 0 *o=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

81 0 0 0 0 0 0 0 *q=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

81 0 0 0 0 0 0 0 *p=112 210 66 144 174 85 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... freeing q

81 0 0 0 0 0 0 0 *o=0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

81 0 0 0 0 0 0 0 *q=16 211 66 144 174 85 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

81 0 0 0 0 0 0 0 *p=112 210 66 144 174 85 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

... freeing old p

81 0 0 0 0 0 0 0 *o=192 210 66 144 174 85 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

81 0 0 0 0 0 0 0 *q=16 211 66 144 174 85 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

81 0 0 0 0 0 0 0 *p=112 210 66 144 174 85 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

我确实注意到在 q 指向的内存中如何有 p 指向的内容的一部分。

在分析输出时我应该关注什么，最重要的是，这些数字在 malloc、realloc 和 free 的定义中代表什么？

提前感谢您的帮助。

【问题讨论】：

首先要知道malloc分配的内存没有以任何方式初始化，它的内容是indeterminate（可以看成是随机的或者垃圾）。你需要知道的第二件事是你永远不应该重新分配给你传递给realloc的指针，如果realloc失败，它将返回一个空指针，但不会释放第一个参数指向的内存（导致丢失原始指针时内存泄漏）。
此外，C 中的所有指针都可以隐式转换为void *，void * 可以隐式转换为任何其他指针。这就是为什么你shouldn't cast the result of malloc (or it siblings)。
第一次调用p = realloc(p, ...);后，o无效，或指向与p相同的内存。所以你不应该打电话给free(o)。你有三个free()，但只有两个malloc()。

标签： c memory memory-management malloc heap-memory

【解决方案1】：

malloc 分配的字节数可能超过所需的 n 个字节。它必须做一些簿记，以便例如free 知道块有多大才能完全释放它。

malloc 如何做到这一点是特定于实现的，它可以通过free,realloc 可以在此处读取的某些已知结构为每个块添加前缀，或者它可以保留某种带有块信息的可搜索结构。这需要进行大量优化，因此可能会变得相当复杂，请尝试搜索“内存分配算法”或类似的内容。

【讨论】：