(lldb)以十六进制打印无符号长整型

Question

我正在尝试调试我的Objective-C程序,我需要以unsigned long long十六进制打印我的变量.我正在使用lldb调试器.

要short以十六进制打印,您可以使用:

(lldb) type format add --format hex short
(lldb) print bit
(short) $11 = 0x0000

但是,我无法让它发挥作用unsigned long long.

// failed attempts:
(lldb) type format add --format hex (unsigned long long)
(lldb) type format add --format hex unsigned long long
(lldb) type format add --format hex unsigned decimal
(lldb) type format add --format hex long long
(lldb) type format add --format hex long
(lldb) type format add --format hex int

我在模拟器上运行iOS应用程序,如果这有任何区别.

Answer 1

您可以使用格式字母.链接到GDB文档(也适用于LLDB):http://www.delorie.com/gnu/docs/gdb/gdb_55.html

(lldb) p a
(unsigned long long) $0 = 10
(lldb) p/x a
(unsigned long long) $1 = 0x000000000000000a

Answer 2

type format add期望类型名称为单个单词 - 如果参数是多个单词,则需要引用该参数.例如

   2    {
   3      unsigned long long a = 10;
-> 4      a += 5;
   5      return a;
   6    }
(lldb) type form add -f h "unsigned long long"
(lldb) p a
(unsigned long long) $0 = 0x000000000000000a
(lldb)