如何在记忆后释放分配的指针

Question

解

显然，根据free（a）和memset（a，0，malloced_size）之间的区别是什么 （实际标题是“ c中的Free function”，但我觉得不够具体），我需要将数据设置为0我释放它之前/之后实际上释放它，就好像它从未分配过任何数据一样

更新：固定

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

uintptr_t round_up(uintptr_t value, uintptr_t size)
{
    fprintf(stderr, "called round_up\nreturning %p\n", (void *) (value ? size * ((value + (size - 1)) / size) : size));
    return value ? size * ((value + (size - 1)) / size) : size;
}

int read_fast_verify(const char *src, int len_of_source, char **dest, char ** a, int requested_len) {
    *a = malloc(requested_len+4096);
    if (len_of_source < requested_len) memcpy(*a, src, len_of_source);
    else memcpy(*a, src, requested_len);
    *dest = memmove((void *)round_up((uintptr_t)*a, 4096), src, requested_len);
    return requested_len;
}

void __lseek_string__(char **src, int len, int offset) {
    memmove(*src, *src+offset, len);
}

char * string1 = "hello";
char * string2;
char * s;
int main(void) {
    read_fast_verify(string1, strlen(string1), &string2, &s, (strlen(string1) + 5));
    __lseek_string__(&string2, strlen(string1), 5);
    free(s);
}

问题

记忆后如何释放已分配的指针，因为如果我使用NULL dest作为返回值，则记忆似乎会出现段错误（例如char * dest = memmove（...）;）

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

uintptr_t round_up(uintptr_t value, uintptr_t size)
{
    fprintf(stderr, "called round_up\nreturning %p\n", (void *) (value ? size * ((value + (size - 1)) / size) : size));
    return value ? size * ((value + (size - 1)) / size) : size;
}

int read_fast_verify(const char *src, int len_of_source, char **dest, char ** a, int requested_len) {
    *a = malloc(requested_len+4096);
    if (len_of_source < requested_len) memcpy(*a, src, len_of_source);
    else memcpy(*a, src, requested_len);
    *dest = memmove((void *)round_up((uintptr_t)*a, 4096), *dest, requested_len);
    return requested_len;
}

void __lseek_string__(char **src, int len, int offset) {
    memmove(*src, *src+offset, len);
}

char * string1 = "hello";
char * string2;
char * s;
int main(void) {
    read_fast_verify(string1, strlen(string1), &string2, &s, (strlen(string1) + 5));
    __lseek_string__(&string2, strlen(string1), 5);
free(s);
}

输出

    Starting program: /home/arch/universal-dynamic-loader/loader/test_case 
    called round_up
    returning 0x55555555a000

    Program received signal SIGSEGV, Segmentation fault.
    0x00007ffff7f2d3b0 in __memmove_ssse3_back () from /usr/lib/libc.so.6

但是，如果我照常做

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

uintptr_t round_up(uintptr_t value, uintptr_t size)
{
    fprintf(stderr, "called round_up\nreturning %p\n", (void *) (value ? size * ((value + (size - 1)) / size) : size));
    return value ? size * ((value + (size - 1)) / size) : size;
}

int read_fast_verify(const char *src, int len_of_source, char **dest, char ** a, int requested_len) {
    *dest = malloc(requested_len+4096);
    if (len_of_source < requested_len) memcpy(*dest, src, len_of_source);
    else memcpy(*dest, src, requested_len);
    *dest = memmove((void *)round_up((uintptr_t)*dest, 4096), *dest, requested_len);
    return requested_len;
}

void __lseek_string__(char **src, int len, int offset) {
    memmove(*src, *src+offset, len);
}

char * string1 = "hello";
char * string2;
char * s;
int main(void) {
    read_fast_verify(string1, strlen(string1), &string2, &s, (strlen(string1) + 5));
    __lseek_string__(&string2, strlen(string1), 5);
    free(string2);
}

然后我得到这个

called round_up
returning 0x564c86f54000
free(): invalid pointer
Aborted (core dumped)

和来自valgrind

==19175== HEAP SUMMARY:
==19175==     in use at exit: 4,106 bytes in 1 blocks
==19175==   total heap usage: 1 allocs, 1 frees, 4,106 bytes allocated
==19175== 
==19175== Searching for pointers to 1 not-freed blocks
==19175== Checked 68,008 bytes
==19175== 
==19175== 4,106 bytes in 1 blocks are possibly lost in loss record 1 of 1
==19175==    at 0x4837757: malloc (vg_replace_malloc.c:299)
==19175==    by 0x10923C: read_fast_verify (test_case.c:13)
==19175==    by 0x109357: main (test_case.c:28)
==19175== 
==19175== LEAK SUMMARY:
==19175==    definitely lost: 0 bytes in 0 blocks
==19175==    indirectly lost: 0 bytes in 0 blocks
==19175==      possibly lost: 4,106 bytes in 1 blocks
==19175==    still reachable: 0 bytes in 0 blocks
==19175==         suppressed: 0 bytes in 0 blocks
==19175== 
==19175== ERROR SUMMARY: 2 errors from 2 contexts (suppressed: 0 from 0)
==19175== 
==19175== 1 errors in context 1 of 2:
==19175== Invalid free() / delete / delete[] / realloc()
==19175==    at 0x4838904: free (vg_replace_malloc.c:530)
==19175==    by 0x109388: main (test_case.c:30)
==19175==  Address 0x4a30000 is 4,032 bytes inside a block of size 4,106 alloc'd
==19175==    at 0x4837757: malloc (vg_replace_malloc.c:299)
==19175==    by 0x10923C: read_fast_verify (test_case.c:13)
==19175==    by 0x109357: main (test_case.c:28)
==19175== 
==19175== ERROR SUMMARY: 2 errors from 2 contexts (suppressed: 0 from 0)

因为我正在尝试使其内存安全，因为当读取.so（例如libc 2.28）时，我的动态链接器当前正在使用2.7 gb（和4 GB以上的共享内存），因此根本不应该使用它。

Answer 1

您应该存储先前的值，并在移动成功完成后释放。 现在，您只需丢弃先前的地址即可。

同样不确定在此示例中是否需要移动到此处，因为您将始终指向第一个指针内的相对地址并返回该地址，除了原始指针外，没有空闲指针。

另请参见memmove是否实际上“移动”了一块内存并在源头留下了零？