108

参考js0n.c

代码语法如下:

    static void *gostruct[] =
    {
        [0 ... 255] = &&l_bad,
        ['\t'] = &&l_loop, [' '] = &&l_loop, ['\r'] = &&l_loop, ['\n'] = &&l_loop,
        ['"'] = &&l_qup,
        [':'] = &&l_loop, [','] = &&l_loop,
        ['['] = &&l_up, [']'] = &&l_down, // tracking [] and {} individually would allow fuller validation but is really messy
        ['{'] = &&l_up, ['}'] = &&l_down,
        ['-'] = &&l_bare, [48 ... 57] = &&l_bare, // 0-9
        [65 ... 90] = &&l_bare, // A-Z
        [97 ... 122] = &&l_bare // a-z
    };

........
.......

l_bad:
    *vlen = cur - json; // where error'd
    return 0;

........
........

谁能解释这里正在做什么?语法[0 ... 255]&&l_bad在这里做什么?

4

1 回答 1

109

...是 GCC 提供的扩展

https://gcc.gnu.org/onlinedocs/gcc/Designated-Inits.html#Designated-Inits

要将一系列元素初始化为相同的值,请编写[first ... last] = value. 这是一个 GNU 扩展。例如,

 int widths[] = { [0 ... 9] = 1, [10 ... 99] = 2, [100] = 3 };

&&是另一个扩展

https://gcc.gnu.org/onlinedocs/gcc/Labels-as-Values.html#Labels-as-Values

您可以使用一元运算符获取当前函数(或包含函数)中定义的标签的地址&&。该值具有类型void *。该值是一个常量,可以在该类型的常量有效的任何地方使用。例如:

 void *ptr;
 /* ... */
 ptr = &&foo;
于 2015-05-22T04:55:22.943 回答