您可以使用十六进制一个char **申报数据？ [英] can you declare data in a char** using hex?

问题描述

可以说我有这个

char *something[] = { 
    "/bi", 
    "-c", 
    "5", 
    NULL, 
    NULL 
};

但我要声明其十六进制，我将如何做到这一点;编译器不断对我示数出来：

but I want to declare it in hex, how would I do this; compiler keeps erroring out on me:

char *something[] = { 
    {0x2f,0x62,0x69},
    {0x2d,0x63},
    {0x35},
    {0x00}, 
    {0x00}
};

添加别的东西来此，为0x00始终为空？并始终为0x00转化为对系统NULL其中null是-1例如？

to add something else to this, is 0x00 ALWAYS null? does 0x00 always translate to NULL on systems where NULL is -1 for example?

推荐答案

您可以在字符串中使用十六进制转义序列。例如：

You can use hexadecimal escape sequences within a string literal. For example:

char *something[] = { 
    "\x2f\x62\x69",
    "\x2d\x63"
}; 

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