这个函数调用如何工作??? [英] How this function call works???

问题描述

有些人可以帮我解决这里发生的事情...


谢谢..


char sc [] =

" \ x31 \ xc0" / * xor％eax，％eax * /

" \ x50" / * push％eax * /

" \ xx68 \ x2f \ x2f \ x73 \ x68" / * push $ 0x68732f2f * /

" \ xx68 \ x2f \ x62 \ x69 \ x6e" / * push $ 0x6e69622f * /

" \ x89 \ xe3" / * mov％esp，％ebx * /

" \ x50" / * push％eax * /

" \ x53" / * push％ebx * /

" \ x89 \ xe1" / * mov％esp，％ecx * /

" \ x31 \ xd2" / * xor％edx，％edx * /

" \ xb0 \ x0b" / * mov $ 0xb，％al * /

" \ xcd \ x80" ;; / * int $ 0x80 * /


main（）

{

void（* fp）（void）; //这一行发生了什么


fp =（void *）sc;

fp（）;

}


-

没有Linux的一天是没有学习的一天！

- Java B？y


---

发表于 news：//freenews.netfront.net

投诉 ne ** @ netfront.net

could some body help me what''s happening here...

thanks..

char sc[] =
"\x31\xc0" /* xor %eax, %eax */
"\x50" /* push %eax */
"\x68\x2f\x2f\x73\x68" /* push $0x68732f2f */
"\x68\x2f\x62\x69\x6e" /* push $0x6e69622f */
"\x89\xe3" /* mov %esp,%ebx */
"\x50" /* push %eax */
"\x53" /* push %ebx */
"\x89\xe1" /* mov %esp,%ecx */
"\x31\xd2" /* xor %edx,%edx */
"\xb0\x0b" /* mov $0xb,%al */
"\xcd\x80"; /* int $0x80 */

main()
{
void (*fp) (void); // what is happening at this line

fp = (void *)sc;
fp();
}

--
A day spent without Linux is a day spent without learning!
-- Java B?y


---
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Complaints to ne**@netfront.net

推荐答案

0x68732f2f * /

" \ x68 \ x2f \ x62 \ x69 \ x6e" / * push

0x68732f2f */
"\x68\x2f\x62\x69\x6e" /* push

0x6e69622f * /

" \ x89 \ xe3" / * mov％esp，％ebx * /

" \ x50" / * push％eax * /

" \ x53" / * push％ebx * /

" \ x89 \ xe1" / * mov％esp，％ecx * /

" \ x31 \ xd2" / * xor％edx，％edx * /

" \ xb0 \ x0b" / * mov

0x6e69622f */
"\x89\xe3" /* mov %esp,%ebx */
"\x50" /* push %eax */
"\x53" /* push %ebx */
"\x89\xe1" /* mov %esp,%ecx */
"\x31\xd2" /* xor %edx,%edx */
"\xb0\x0b" /* mov

0xb，％al * /

" \ xcd \ x80" ;; / * int

0xb,%al */
"\xcd\x80"; /* int

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