一段代码的疑问（1）——unsigned与signed

时间：2014-08-14 14:06:59 阅读：458 评论：0 收藏：0 [点我收藏+]

现象：

先来看一段代码：

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这段代码的输出结果是：

-84

4294967264

分析：

xiaoqiang@dev:~/cpp$ g++ -g c212.cc -o temp
xiaoqiang@dev:~/cpp$ ls
c143.cc  c144.cc  c212.cc  temp

可以看到多出一个temp文件

xiaoqiang@dev:~/cpp$ gdb temp
GNU gdb (Ubuntu/Linaro 7.4-2012.04-0ubuntu2.1) 7.4-2012.04
Copyright (C) 2012 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  Type "show copying"
and "show warranty" for details.
This GDB was configured as "x86_64-linux-gnu".
For bug reporting instructions, please see:
<http://bugs.launchpad.net/gdb-linaro/>...
Reading symbols from /home/xiaoqiang/cpp/temp...done.
(gdb) list
1       #include <iostream>
2(gdb) b 6
Breakpoint 1 at 0x4007ea: file c212.cc, line 6.

3       int main(){4               unsigned u = 10;5               int i = -42;6               std::cout << i + i << std::endl;7               std::cout << u + i << std::endl;8               return 0;9       }(gdb)

进入到gdb调试模式我们来看个究竟

(gdb) b 6
Breakpoint 1 at 0x4007ea: file c212.cc, line 6.

开始运行，就会发现程序停在了第6行

(gdb) r
Starting program: /home/xiaoqiang/cpp/temp 
warning: no loadable sections found in added symbol-file system-supplied DSO at 0x7ffff7ffa000

Breakpoint 1, main () at c212.cc:6
6               std::cout << i + i << std::endl;
(gdb)

我们来看看i的二进制值是多少

(gdb) print /t i
$1 = 11111111111111111111111111010110

在使用unsigned int 和 int类型变量运算的时候，首先会将int类型转换成（看成）unsigned int类型，上面两个值在内存中相加为：

11111111 11111111 11111111 01100000

十六进制为 FFFFFFE0

十进制为4294967264

练习：

看一下下面代码猜一下最终输出结果：

xiaoqiang@dev:~/cpp$ gdb temp
GNU gdb (Ubuntu/Linaro 7.4-2012.04-0ubuntu2.1) 7.4-2012.04
Copyright (C) 2012 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  Type "show copying"
and "show warranty" for details.
This GDB was configured as "x86_64-linux-gnu".
For bug reporting instructions, please see:
<http://bugs.launchpad.net/gdb-linaro/>...
Reading symbols from /home/xiaoqiang/cpp/temp...done.
(gdb) list
1       #include <iostream>
2
3       int main(){
4               unsigned u1 = 42, u2 = 10;
5               std::cout << u1 - u2 << std::endl;
6               std::cout << u2 - u1 << std::endl;
7               return 0;
8       }
(gdb)

看一下u1和u2的二进制表示

Breakpoint 1, main () at c213.cc:5
5               std::cout << u1 - u2 << std::endl;
(gdb) next
32
6               std::cout << u2 - u1 << std::endl;
(gdb) next
4294967264
7               return 0;
(gdb) print /t u2
$2 = 1010
(gdb) print /t u1
$3 = 101010
(gdb)

差值为11111111 111111111 11111111 11100000

如果以unsigned int 类型显示则是 4294967264，如果以int类型显示（有符号）则为-32.

一段代码的疑问（1）——unsigned与signed

现象：

分析：

练习：

相关知识：