gdb 如何简化内存地址/偏移量？

Question

This is my sample c code

user@linux:~$ gdb -q hello
Reading symbols from hello...done.
(gdb) 
(gdb) list 
1   #include<stdio.h>
2   
3   int main()
4   {
5       printf("Hello World!\n");
6       return 0;
7   }
8   

And this is it's assembly code.

(gdb) disassemble main 
Dump of assembler code for function main:
   0x000000000000063a <+0>: push   %rbp
   0x000000000000063b <+1>: mov    %rsp,%rbp
   0x000000000000063e <+4>: lea    0x9f(%rip),%rdi        # 0x6e4
   0x0000000000000645 <+11>:    callq  0x510 <puts@plt>
   0x000000000000064a <+16>:    mov    $0x0,%eax
   0x000000000000064f <+21>:    pop    %rbp
   0x0000000000000650 <+22>:    retq   
End of assembler dump.
(gdb) 

The memory address contains 18 character and most of them is number 0 .

Instead of displaying all numbers, would it be possible to simplify it?

Let say 0x63a instead of 0x000000000000063a

Answer 1

The memory address contains 18 character and most of them is number 0

There is no "the memory" here. You are apparently talking about displaying addresses.

Let say 0x63a instead of 0x000000000000063a

You are on a 64-bit system, and every address is exactly 64-bits. Displaying addresses as something other than a 64-bit number would be very confusing.

PS You have a position-independent executable. It doesn't actually run at address 0x000000000000063a . If you use start and disas main , you will get a very different result.