I'm trying to figure out what all of the elements of this objdump -d mean.
我想弄清楚这个objdump -d的所有元素是什么意思。
for example i have:
例如我有:
08048b50 <phase_1>:
8048b50: 83 ec 1c sub $0x1c,%esp
8048b53: c7 44 24 04 68 a2 04 movl $0x804a268,0x4(%esp)
8048b5a: 08
8048b5b: 8b 44 24 20 mov 0x20(%esp),%eax
8048b5f: 89 04 24 mov %eax,(%esp)
8048b62: e8 63 04 00 00 call 8048fca <strings_not_equal>
8048b67: 85 c0 test %eax,%eax
8048b69: 74 05 je 8048b70 <phase_1+0x20>
8048b6b: e8 f5 06 00 00 call 8049265 <explode_bomb>
8048b70: 83 c4 1c add $0x1c,%esp
8048b73: c3 ret
specifically im not sure what the 1st and center columns are telling me
具体地说，我不知道第一列和中间列告诉了我什么
1 个解决方案
#1
3
The first column tells you the memory addresses where the code will be located at runtime.
第一列告诉您代码将位于运行时的内存地址。
The second column has the hex version of the executable instruction.
第二列包含可执行指令的十六进制版本。
The third (and forth) columns, have a disassembled version of the second column. i.e. opcode and operands.
第三列(和第四列)具有第二列的分解版本。即操作码和操作数。
#1
3
The first column tells you the memory addresses where the code will be located at runtime.
第一列告诉您代码将位于运行时的内存地址。
The second column has the hex version of the executable instruction.
第二列包含可执行指令的十六进制版本。
The third (and forth) columns, have a disassembled version of the second column. i.e. opcode and operands.
第三列(和第四列)具有第二列的分解版本。即操作码和操作数。