C编程:将Hex Int转换为Char * [英] C Programming: Convert Hex Int to Char*
问题描述
我的问题是我将如何转换类似:
int i = 0x11111111;
到字符指针?我尝试使用itoa()函数,但它给了我一个浮点异常。
是非标准的。远离。
一种可能性是使用 sprintf
和hexa的正确格式说明符 x
and do:
char str [BIG_ENOUGH + 1];
sprintf(str,%x,value);但是,这个计算的问题是值的大小$
数组。您必须使用一些猜测,常见问题12.21 是一个很好的起点。
表示任何基本 b
中的数字所需的字符数可以通过以下公式近似:
⌈ log b (n + 1)⌉
如果需要,再添加一对以保存 0x
,然后您的 BIG_ENOUGH
已准备就绪。
My question is how I would go about converting something like:
int i = 0x11111111;
to a character pointer? I tried using the itoa() function but it gave me a floating-point exception.
解决方案 itoa
is non-standard. Stay away.
One possibility is to use sprintf
and the proper format specifier for hexa i.e. x
and do:
char str[ BIG_ENOUGH + 1 ];
sprintf(str,"%x",value);
However, the problem with this computing the size of the value
array. You have to do with some guesses and FAQ 12.21 is a good starting point.
The number of characters required to represent a number in any base b
can be approximated by the following formula:
⌈logb(n + 1)⌉
Add a couple more to hold the 0x
, if need be, and then your BIG_ENOUGH
is ready.
这篇关于C编程:将Hex Int转换为Char *的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!