hacktricks/binary-exploitation/heap/double-free.md

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双重释放

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基本信息

如果释放一块内存超过一次可能会破坏分配器的数据并为攻击打开大门。事情是这样发生的当您释放一块内存时它会回到空闲块的列表中例如“fastbin”。如果连续两次释放相同的内存块则分配器会检测到这一点并报错。但是如果在两次释放之间释放另一块内存块,则会绕过双重释放检查,导致数据损坏。

现在,当您请求新的内存(使用malloc)时,分配器可能会给您一个已经释放两次的块。这可能导致两个不同的指针指向相同的内存位置。如果攻击者控制其中一个指针,他们可以更改该内存的内容,这可能会导致安全问题,甚至允许他们执行代码。

示例:

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

int main() {
// Allocate memory for three chunks
char *a = (char *)malloc(10);
char *b = (char *)malloc(10);
char *c = (char *)malloc(10);
char *d = (char *)malloc(10);
char *e = (char *)malloc(10);
char *f = (char *)malloc(10);
char *g = (char *)malloc(10);
char *h = (char *)malloc(10);
char *i = (char *)malloc(10);

// Print initial memory addresses
printf("Initial allocations:\n");
printf("a: %p\n", (void *)a);
printf("b: %p\n", (void *)b);
printf("c: %p\n", (void *)c);
printf("d: %p\n", (void *)d);
printf("e: %p\n", (void *)e);
printf("f: %p\n", (void *)f);
printf("g: %p\n", (void *)g);
printf("h: %p\n", (void *)h);
printf("i: %p\n", (void *)i);

// Fill tcache
free(a);
free(b);
free(c);
free(d);
free(e);
free(f);
free(g);

// Introduce double-free vulnerability in fast bin
free(h);
free(i);
free(h);


// Reallocate memory and print the addresses
char *a1 = (char *)malloc(10);
char *b1 = (char *)malloc(10);
char *c1 = (char *)malloc(10);
char *d1 = (char *)malloc(10);
char *e1 = (char *)malloc(10);
char *f1 = (char *)malloc(10);
char *g1 = (char *)malloc(10);
char *h1 = (char *)malloc(10);
char *i1 = (char *)malloc(10);
char *i2 = (char *)malloc(10);

// Print initial memory addresses
printf("After reallocations:\n");
printf("a1: %p\n", (void *)a1);
printf("b1: %p\n", (void *)b1);
printf("c1: %p\n", (void *)c1);
printf("d1: %p\n", (void *)d1);
printf("e1: %p\n", (void *)e1);
printf("f1: %p\n", (void *)f1);
printf("g1: %p\n", (void *)g1);
printf("h1: %p\n", (void *)h1);
printf("i1: %p\n", (void *)i1);
printf("i2: %p\n", (void *)i1);

return 0;
}

在这个例子中填充了tcache与多个已释放的块后代码释放了块h,然后是块i,然后再次释放h,导致双重释放错误。这打开了重新分配时接收重叠内存地址的可能性,意味着两个或更多指针可以指向相同的内存位置。通过一个指针操纵数据然后可以影响另一个,从而创建了严重的安全风险和潜在的利用可能。

执行后,注意**i1i2获得了相同的地址**

初始分配:
a: 0xaaab0f0c22a0
b: 0xaaab0f0c22c0
c: 0xaaab0f0c22e0
d: 0xaaab0f0c2300
e: 0xaaab0f0c2320
f: 0xaaab0f0c2340
g: 0xaaab0f0c2360
h: 0xaaab0f0c2380
i: 0xaaab0f0c23a0
重新分配后:
a1: 0xaaab0f0c2360
b1: 0xaaab0f0c2340
c1: 0xaaab0f0c2320
d1: 0xaaab0f0c2300
e1: 0xaaab0f0c22e0
f1: 0xaaab0f0c22c0
g1: 0xaaab0f0c22a0
h1: 0xaaab0f0c2380
i1: 0xaaab0f0c23a0
i2: 0xaaab0f0c23a0

参考

从零开始学习AWS黑客技术成为专家 htARTE (HackTricks AWS Red Team Expert)!

支持HackTricks的其他方式