炸弹实验，一个递归看不懂，求大神教【反汇编吧】

phase_4的密码应该是多少呢
我已经把语句的含义基本上写在后面了。
08048cbf <func4>:
8048cbf:57 push %edi
8048cc0:56 push %esi;assuming varible--a
8048cc1:53 push %ebx;assuming varible--b
8048cc2:83 ec 10 sub $0x10,%esp
8048cc5:8b 5c 24 20 mov 0x20(%esp),%ebx
8048cc9:8b 74 24 24 mov 0x24(%esp),%esi
8048ccd:85 db test %ebx,%ebx
8048ccf:7e 2c jle 8048cfd <func4+0x3e>; fd <=跳到最后
8048cd1:89 f0 mov %esi,%eax
8048cd3:83 fb 01 cmp $0x1,%ebx
8048cd6:74 2a je 8048d02 <func4+0x43>;结束
8048cd8:89 74 24 04 mov %esi,0x4(%esp)
8048cdc:8d 43 ff lea -0x1(%ebx),%eax;b-1
8048cdf:89 04 24 mov %eax,(%esp)
8048ce2:e8 d8 ff ff ff call 8048cbf <func4>; 递归
8048ce7:8d 3c 30 lea (%eax,%esi,1),%edi; edi=eax+esi=a+a
8048cea:89 74 24 04 mov %esi,0x4(%esp)
8048cee:83 eb 02 sub $0x2,%ebx;b-2
8048cf1:89 1c 24 mov %ebx,(%esp)
8048cf4:e8 c6 ff ff ff call 8048cbf <func4>;递归
8048cf9:01 f8 add %edi,%eax
8048cfb:eb 05 jmp 8048d02 <func4+0x43>
8048cfd:b8 00 00 00 00 mov $0x0,%eax
8048d02:83 c4 10 add $0x10,%esp
8048d05:5b pop %ebx
8048d06:5e pop %esi
8048d07:5f pop %edi
8048d08:c3 ret
08048d09 <phase_4>:
8048d09:83 ec 2c sub $0x2c,%esp
8048d0c:8d 44 24 18 lea 0x18(%esp),%eax
8048d10:89 44 24 0c mov %eax,0xc(%esp)
8048d14:8d 44 24 1c lea 0x1c(%esp),%eax
8048d18:89 44 24 08 mov %eax,0x8(%esp)
8048d1c:c7 44 24 04 71 a5 04 movl $0x804a571,0x4(%esp); %d %d
8048d23:08
8048d24:8b 44 24 30 mov 0x30(%esp),%eax;2
8048d28:89 04 24 mov %eax,(%esp);2
8048d2b:e8 30 fb ff ff call 8048860 <__isoc99_sscanf@plt>
8048d30:83 f8 02 cmp $0x2,%eax; MUST argu==2
8048d33:75 0c jne 8048d41 <phase_4+0x38>;bomb
8048d35:8b 44 24 18 mov 0x18(%esp),%eax;4
8048d39:83 e8 02 sub $0x2,%eax
8048d3c:83 f8 02 cmp $0x2,%eax
8048d3f:76 05 jbe 8048d46 <phase_4+0x3d>;<=
8048d41:e8 4f 05 00 00 call 8049295 <explode_bomb>
8048d46:8b 44 24 18 mov 0x18(%esp),%eax; 2
8048d4a:89 44 24 04 mov %eax,0x4(%esp); 0x4(esp)=4
8048d4e:c7 04 24 05 00 00 00 movl $0x5,(%esp)
8048d55:e8 65 ff ff ff call 8048cbf <func4>
8048d5a:3b 44 24 1c cmp 0x1c(%esp),%eax
8048d5e:74 05 je 8048d65 <phase_4+0x5c>
8048d60:e8 30 05 00 00 call 8049295 <explode_bomb>
8048d65:83 c4 2c add $0x2c,%esp
8048d68:c3 ret

吐血了，一直把两个数的输入顺序弄反了，是48 4但是我觉得真的应该是4 48才对呢，求解

楼主递归怎么写的
求

08048cc6 <func4>:
8048cc6:57 push %edi
8048cc7:56 push %esi
8048cc8:53 push %ebx
8048cc9:83 ec 10 sub $0x10,%esp
8048ccc:8b 5c 24 20 mov 0x20(%esp),%ebx
8048cd0:8b 74 24 24 mov 0x24(%esp),%esi
8048cd4:85 db test %ebx,%ebx
8048cd6:7e 2c jle 8048d04 <func4+0x3e>
8048cd8:89 f0 mov %esi,%eax
8048cda:83 fb 01 cmp $0x1,%ebx
8048cdd:74 2a je 8048d09 <func4+0x43>
8048cdf:89 74 24 04 mov %esi,0x4(%esp)
8048ce3:8d 43 ff lea -0x1(%ebx),%eax
8048ce6:89 04 24 mov %eax,(%esp)
8048ce9:e8 d8 ff ff ff call 8048cc6 <func4>
8048cee:8d 3c 30 lea (%eax,%esi,1),%edi
8048cf1:89 74 24 04 mov %esi,0x4(%esp)
8048cf5:83 eb 02 sub $0x2,%ebx
8048cf8:89 1c 24 mov %ebx,(%esp)
8048cfb:e8 c6 ff ff ff call 8048cc6 <func4>
8048d00:01 f8 add %edi,%eax
8048d02:eb 05 jmp 8048d09 <func4+0x43>
8048d04:b8 00 00 00 00 mov $0x0,%eax
8048d09:83 c4 10 add $0x10,%esp
8048d0c:5b pop %ebx
8048d0d:5e pop %esi
8048d0e:5f pop %edi
8048d0f:c3 ret
08048d10 <phase_4>:
8048d10:83 ec 2c sub $0x2c,%esp
8048d13:8d 44 24 18 lea 0x18(%esp),%eax
8048d17:89 44 24 0c mov %eax,0xc(%esp)
8048d1b:8d 44 24 1c lea 0x1c(%esp),%eax
8048d1f:89 44 24 08 mov %eax,0x8(%esp)
8048d23:c7 44 24 04 0f a3 04 movl $0x804a30f,0x4(%esp)
8048d2a:08
8048d2b:8b 44 24 30 mov 0x30(%esp),%eax
8048d2f:89 04 24 mov %eax,(%esp)
8048d32:e8 29 fb ff ff call 8048860 <__isoc99_sscanf@plt>
8048d37:83 f8 02 cmp $0x2,%eax
8048d3a:75 0c jne 8048d48 <phase_4+0x38>
8048d3c:8b 44 24 18 mov 0x18(%esp),%eax
8048d40:83 e8 02 sub $0x2,%eax
8048d43:83 f8 02 cmp $0x2,%eax
8048d46:76 05 jbe 8048d4d <phase_4+0x3d>
8048d48:e8 c8 03 00 00 call 8049115 <explode_bomb>
8048d4d:8b 44 24 18 mov 0x18(%esp),%eax
8048d51:89 44 24 04 mov %eax,0x4(%esp)
8048d55:c7 04 24 08 00 00 00 movl $0x8,(%esp)
8048d5c:e8 65 ff ff ff call 8048cc6 <func4>
8048d61:3b 44 24 1c cmp 0x1c(%esp),%eax
8048d65:74 05 je 8048d6c <phase_4+0x5c>
8048d67:e8 a9 03 00 00 call 8049115 <explode_bomb>
8048d6c:83 c4 2c add $0x2c,%esp
8048d6f:90 nop
8048d70:c3 ret

楼主，我的和你的几乎一样，我的怎么不正确的，求楼主帮一下忙

08048cb5 <func4>:
8048cb5:57 push %edi
8048cb6:56 push %esi
8048cb7:53 push %ebx
8048cb8:83 ec 10 sub $0x10,%esp
8048cbb:8b 5c 24 20 mov 0x20(%esp),%ebx
8048cbf:8b 74 24 24 mov 0x24(%esp),%esi
8048cc3:85 db test %ebx,%ebx
8048cc5:7e 2c jle 8048cf3 <func4+0x3e>
8048cc7:89 f0 mov %esi,%eax
8048cc9:83 fb 01 cmp $0x1,%ebx
8048ccc:74 2a je 8048cf8 <func4+0x43>
8048cce:89 74 24 04 mov %esi,0x4(%esp)
8048cd2:8d 43 ff lea -0x1(%ebx),%eax
8048cd5:89 04 24 mov %eax,(%esp)
8048cd8:e8 d8 ff ff ff call 8048cb5 <func4>
8048cdd:8d 3c 30 lea (%eax,%esi,1),%edi
8048ce0:89 74 24 04 mov %esi,0x4(%esp)
8048ce4:83 eb 02 sub $0x2,%ebx
8048ce7:89 1c 24 mov %ebx,(%esp)
8048cea:e8 c6 ff ff ff call 8048cb5 <func4>
8048cef:01 f8 add %edi,%eax
8048cf1:eb 05 jmp 8048cf8 <func4+0x43>
8048cf3:b8 00 00 00 00 mov $0x0,%eax
8048cf8:83 c4 10 add $0x10,%esp
8048cfb:5b pop %ebx
8048cfc:5e pop %esi
8048cfd:5f pop %edi
8048cfe:c3 ret
08048cff <phase_4>:
8048cff:83 ec 2c sub $0x2c,%esp
8048d02:8d 44 24 18 lea 0x18(%esp),%eax
8048d06:89 44 24 0c mov %eax,0xc(%esp)
8048d0a:8d 44 24 1c lea 0x1c(%esp),%eax
8048d0e:89 44 24 08 mov %eax,0x8(%esp)
8048d12:c7 44 24 04 db a2 04 movl $0x804a2db,0x4(%esp)
8048d19:08
8048d1a:8b 44 24 30 mov 0x30(%esp),%eax
8048d1e:89 04 24 mov %eax,(%esp)
8048d21:e8 3a fb ff ff call 8048860 <__isoc99_sscanf@plt>
8048d26:83 f8 02 cmp $0x2,%eax
8048d29:75 0c jne 8048d37 <phase_4+0x38>
8048d2b:8b 44 24 18 mov 0x18(%esp),%eax
8048d2f:83 e8 02 sub $0x2,%eax
8048d32:83 f8 02 cmp $0x2,%eax
8048d35:76 05 jbe 8048d3c <phase_4+0x3d>
8048d37:e8 e9 03 00 00 call 8049125 <explode_bomb>
8048d3c:8b 44 24 18 mov 0x18(%esp),%eax
8048d40:89 44 24 04 mov %eax,0x4(%esp)
8048d44:c7 04 24 07 00 00 00 movl $0x7,(%esp)
8048d4b:e8 65 ff ff ff call 8048cb5 <func4>
8048d50:3b 44 24 1c cmp 0x1c(%esp),%eax
8048d54:74 05 je 8048d5b <phase_4+0x5c>
8048d56:e8 ca 03 00 00 call 8049125 <explode_bomb>
8048d5b:83 c4 2c add $0x2c,%esp
8048d5e:c3 ret
求支援

你的func4参数是7,4 我的5,4

这应该是liunx上的AT&T反汇编，并不是windows上的反汇编

怎么算的参数值？？？递归函数伪代码能贴出来吗？我和楼主一样

一个递归
int func(int num1, int num2){
if(num1<=0) return num2;
return func(num1-1)+num2+func(num1-2)
}

谢谢楼主……不过好像格式有点问题？我改了一下，但还是不对

int func(int num1, int num2){
if(num1<=1) return num2;
return func(num1-1,num2)+func(num1-2,num2);
}

ai 还是不行
num的范围应该没错啊，(1,4]

而且给的定值是6也代进去了递归边界我看有<=0和=1就把他们合在一起

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炸弹实验，一个递归看不懂，求大神教

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