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/*
* rl.c - Reverse a list
*/
#include <stdlib.h>
#include <stdio.h>
struct node {
int data;
struct node *next;
};
struct node *create_list(int array[], int length) {
int i = 0;
struct node *head, *p;
head = (struct node *) malloc( sizeof(struct node) * length );
for ( i=0; i<length-1; i++ ) {
p = &(head[i]);
p->data = array[i];
p->next = &(head[i+1]);
}
p = &(head[i]);
p->data = array[i];
p->next = NULL;
return head;
}
struct node *reserve_list(struct node *list) {
struct node *p, *prev, *next;
for (p = list, prev = NULL; p; p = next) {
next = p->next;
p->next = prev;
prev = p;
}
return prev;
}
void free_list(struct node *list) {
free(list);
}
int main()
{
struct node *p;
int source[10] = {1,2,3,4,5,6,7,8,9,10};
p = create_list(source, 10);
p = reserve_list(p);
free_list(p);
}
gcc -o rl rl.c
$ ./rl
*** glibc detected *** ./rl: double free or corruption (out): 0x0000000000bf90a0 ***
======= Backtrace: =========
/lib/x86_64-linux-gnu/libc.so.6(+0x7ec66)[0x7f04f776bc66]
./rl[0x400668]
./rl[0x4006e9]
/lib/x86_64-linux-gnu/libc.so.6(__libc_start_main+0xed)[0x7f04f770e78d]
./rl[0x400489]
======= Memory map: ========
00400000-00401000 r-xp 00000000 08:51 794284 /media/DATA/projects/c-sample/rl
00600000-00601000 r--p 00000000 08:51 794284 /media/DATA/projects/c-sample/rl
00601000-00602000 rw-p 00001000 08:51 794284 /media/DATA/projects/c-sample/rl
00bf9000-00c1a000 rw-p 00000000 00:00 0 [heap]
7f04f74d7000-7f04f74ec000 r-xp 00000000 08:01 529232 /lib/x86_64-linux-gnu/libgcc_s.so.1
7f04f74ec000-7f04f76eb000 ---p 00015000 08:01 529232 /lib/x86_64-linux-gnu/libgcc_s.so.1
7f04f76eb000-7f04f76ec000 r--p 00014000 08:01 529232 /lib/x86_64-linux-gnu/libgcc_s.so.1
7f04f76ec000-7f04f76ed000 rw-p 00015000 08:01 529232 /lib/x86_64-linux-gnu/libgcc_s.so.1
7f04f76ed000-7f04f78a2000 r-xp 00000000 08:01 529443 /lib/x86_64-linux-gnu/libc-2.15.so
7f04f78a2000-7f04f7aa2000 ---p 001b5000 08:01 529443 /lib/x86_64-linux-gnu/libc-2.15.so
7f04f7aa2000-7f04f7aa6000 r--p 001b5000 08:01 529443 /lib/x86_64-linux-gnu/libc-2.15.so
7f04f7aa6000-7f04f7aa8000 rw-p 001b9000 08:01 529443 /lib/x86_64-linux-gnu/libc-2.15.so
7f04f7aa8000-7f04f7aad000 rw-p 00000000 00:00 0
7f04f7aad000-7f04f7acf000 r-xp 00000000 08:01 529435 /lib/x86_64-linux-gnu/ld-2.15.so
7f04f7caf000-7f04f7cb2000 rw-p 00000000 00:00 0
7f04f7ccc000-7f04f7ccf000 rw-p 00000000 00:00 0
7f04f7ccf000-7f04f7cd0000 r--p 00022000 08:01 529435 /lib/x86_64-linux-gnu/ld-2.15.so
7f04f7cd0000-7f04f7cd2000 rw-p 00023000 08:01 529435 /lib/x86_64-linux-gnu/ld-2.15.so
7fff6c313000-7fff6c334000 rw-p 00000000 00:00 0 [stack]
7fff6c3ff000-7fff6c400000 r-xp 00000000 00:00 0 [vdso]
ffffffffff600000-ffffffffff601000 r-xp 00000000 00:00 0 [vsyscall]
Aborted
p = reserve_list(p);
/*
* rl.c - Reverse a list
*/
#include <stdlib.h>
#include <stdio.h>
struct node {
int data;
struct node *next;
};
struct node *create_list(int array[], int length) {
int i = 0;
struct node *head, *p;
head = (struct node *) malloc( sizeof(struct node) * length );
for ( i=0; i<length-1; i++ ) {
p = &(head[i]);
p->data = array[i];
p->next = &(head[i+1]);
}
p = &(head[i]);
p->data = array[i];
p->next = NULL;
return head;
}
struct node *reserve_list(struct node *list) {
struct node *p, *prev, *next;
for (p = list, prev = NULL; p; p = next) {
next = p->next;
p->next = prev;
prev = p;
}
return prev;
}
void free_list(struct node *list) {
free(list);
}
int main()
{
struct node *p, *p1;
int source[10] = {1,2,3,4,5,6,7,8,9,10};
p = create_list(source, 10);
p1 = reserve_list(p);
free_list(p);
}
#include <stdio.h>
#include <stdlib.h>
int main(int argc, _TCHAR* argv[])
{
char *p =(char *)malloc(10);
p++;
free(p);
return 0;
}
这个的报错结果和你的一致,我申请了10个字节空间,释放之前把P的位置向后移动了1个字节,在free的时候,系统会free10个字节的空间,但实际已经free了第十一个字节,free一个没有malloc的空间,这报错,应该能理解了吧//将c:\\tmp文件夹下的所有文件的内容全部放到用malloc分配的内存中
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <io.h>
struct FB {
char fn[256];
size_t fl;
char *b;
struct FB *next;
struct FB *prev;
} *fh,*fb,*ft;
char ln[256];
char fpn[256];
FILE *af;
FILE *f;
int L,n;
int main() {
system("dir /b /a-d c:\\tmp\\*.* >c:\\allfn.txt");
af=fopen("c:\\allfn.txt","r");
if (NULL==af) {
printf("Can not open file c:\\allfn.txt!\n");
return 1;
}
fh=NULL;
fb=NULL;
n=0;
while (1) {
if (NULL==fgets(ln,256,af)) break;
L=strlen(ln);
if ('\n'==ln[L-1]) ln[L-1]=0;
printf("read %s\n",ln);
strcpy(fpn,"c:\\tmp\\");
strcat(fpn,ln);
ft=(struct FB *)malloc(sizeof(struct FB));
if (NULL==ft) {
printf("Can not malloc ft!\n");
fclose(af);
return 2;//之前的malloc在main退出后由操作系统自动free
}
printf("ft[%d]==%p\n",n,ft);
strcpy(ft->fn,fpn);
f=fopen(fpn,"rb");
if (NULL==f) {
printf("Can not open file %s!\n",fpn);
fclose(af);
return 3;//之前的malloc在main退出后由操作系统自动free
}
ft->fl=_filelength(fileno(f));
ft->b=malloc(ft->fl);
if (NULL==ft->b) {
printf("Can not malloc ft->b!\n");
fclose(f);
fclose(af);
return 4;//之前的malloc在main退出后由操作系统自动free
}
printf("ft[%d]->b==%p\n",n,ft->b);
if (ft->fl!=fread(ft->b,1,ft->fl,f)) {
printf("fread error!\n");
fclose(f);
fclose(af);
return 5;//之前的malloc在main退出后由操作系统自动free
}
fclose(f);
ft->next=NULL;
if (NULL==fh) {
ft->prev=NULL;
fh=ft;
} else {
fb->next=ft;
ft->prev=fb;
}
fb=ft;
n++;
}
fclose(af);
printf("-----list-----\n");
for (ft=fh;NULL!=ft;ft=ft->next) {
printf("%8d %s\n",ft->fl,ft->fn);
if (NULL!=ft) fb=ft;
}
printf("-----free-----\n");
n--;
if (NULL!=fh) {
for (ft=fb->prev;NULL!=ft;ft=ft->prev) {
if (NULL!=ft->next->b) {
printf("ft[%d]->b==%p\n",n,ft->next->b);
free(ft->next->b);
}
if (NULL!=ft->next) {
printf("ft[%d]==%p\n",n,ft->next);
free(ft->next);
}
n--;
}
if (NULL!=fh->b) {
printf("ft[0]->b==%p\n",fh->b);
free(fh->b);
}
printf("ft[0]==%p\n",fh);
free(fh);
}
return 0;
}
//C:\tmp\tmp\Debug>dir /a-d c:\tmp
// 驱动器 C 中的卷是 C_HD5_1
// 卷的序列号是 1817-D526
//
// c:\tmp 的目录
//
//找不到文件
//
//C:\tmp\tmp\Debug>tmp
//找不到文件
//-----list-----
//-----free-----
//
//C:\tmp\tmp\Debug>dir /a-d c:\tmp
// 驱动器 C 中的卷是 C_HD5_1
// 卷的序列号是 1817-D526
//
// c:\tmp 的目录
//
//2011-06-30 18:04 44,840 my_c.rar
//2011-06-30 17:18 1,036 err.frm
//2011-06-30 14:32 14,243 出租.txt
//2011-06-28 12:08 23,681 MSDN98书签.txt
// 4 个文件 83,800 字节
// 0 个目录 17,041,870,848 可用字节
//
//C:\tmp\tmp\Debug>tmp
//read my_c.rar
//ft[0]==00421800
//ft[0]->b==00520068
//read err.frm
//ft[1]==00421670
//ft[1]->b==0052AFC0
//read 出租.txt
//ft[2]==00421530
//ft[2]->b==00378F28
//read MSDN98书签.txt
//ft[3]==004213F0
//ft[3]->b==0052B3F8
//-----list-----
// 44840 c:\tmp\my_c.rar
// 1036 c:\tmp\err.frm
// 14243 c:\tmp\出租.txt
// 23681 c:\tmp\MSDN98书签.txt
//-----free-----
//ft[3]->b==0052B3F8
//ft[3]==004213F0
//ft[2]->b==00378F28
//ft[2]==00421530
//ft[1]->b==0052AFC0
//ft[1]==00421670
//ft[0]->b==00520068
//ft[0]==00421800
//
//C:\tmp\tmp\Debug>
int main()
{
struct node *p;
int source[10] = {1,2,3,4,5,6,7,8,9,10};
p = create_list(source, 10); //在这里p指针式指向的申请的堆的首地址
p = reserve_list(p); //这里执行完成以后,p指针指向的申请的堆的最后一个节点
free_list(p); //释放堆空间的时候,只释放了最后一个节点,别告诉我释放空间的时候能够智能化到倒着释放空间的
}
申请的空间中,每个节点对应的数据和地址没有改变,改变的只是链表顺序中的每个指针
在最后一步释放空间的时候,实际上就是释放的最后一个节点的空间,前面9个节点的空间都没有释放,你这个报错应该是内存泄露的报错吧?
菜鸟理解,高手勿喷!
for (p = list, prev = NULL; p; p = next) {
next = p->next;
p->next = prev;//bug在这个地方,执行第一次循环就错了
prev = p;
}