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如下内容:

My question is how I would go about converting something like:

    int i = 0x11111111;

到一个字符指针?我尝试使用 itoa() 函数，但它给了我一个浮点异常.

to a character pointer? I tried using the itoa() function but it gave me a floating-point exception.

推荐答案

itoa 是非标准的.远离.

一种可能性是使用 sprintf 和正确的十六进制格式说明符，即 x 并执行:

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);

然而，这个计算 value 数组大小的问题.您必须进行一些猜测，FAQ 12.21 是一个很好的起点.

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.

在任何基本 b 中表示一个数字所需的字符数可以通过以下公式近似:

The number of characters required to represent a number in any base b can be approximated by the following formula:

⌈log_b(n + 1)⌉

⌈log_b(n + 1)⌉

如果需要，再添加几个来保存 0x，然后您的 BIG_ENOUGH 就准备好了.

Add a couple more to hold the 0x, if need be, and then your BIG_ENOUGH is ready.

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