通过读取输出了解C中的malloc（），realloc（）和free（）

Question

我需要写一篇关于如何实现内存管理的作业，以便理解此代码输出中的几个非零数字代表什么。

我确实知道malloc（）函数保留了指定字节数的内存块。并且，它返回类型为void的指针，该指针可以转换为任何形式的指针。我还知道，如果动态分配的内存不足或超出要求，则可以使用realloc（）函数更改以前分配的内存的大小。

这是我必须分析的代码：

#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;

}

这是没有任何其他输入的输出文件：

... 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 

以下是使用输入“ 64”进行编译时的输出：

... 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q指向的内存中有一个p指向的东西。

在分析输出时，我应该重点关注什么，最重要的是，这些数字相对于malloc，realloc和free的定义代表什么？

在此先感谢您的帮助。

Answer 1

malloc可能分配了超过所需的 n 个字节。它必须做一些簿记，以便例如free知道该块有多大才能完全释放它。

malloc是特定于实现方式的，它可以在free,realloc之前读取的某个已知结构之前添加每个块，或者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”或类似的方法。