C语言 如何将无符号整数（u16）转换为字符串值（char *）？
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How to convert unsigned int(u16) into string value(char *)?
提问by Sujatha
I need to convert u16(unsigned int -2 byte) value into string (not ascii). How to convert unsigned int(u16) into string value(char *)?
我需要将 u16(unsigned int -2 byte) 值转换为字符串（不是 ascii）。如何将无符号整数（u16）转换为字符串值（char *）？
回答by Nerius
/* The max value of a uint16_t is 65k, which is 5 chars */
#ifdef WE_REALLY_WANT_A_POINTER
char *buf = malloc (6);
#else
char buf[6];
#endif
sprintf (buf, "%u", my_uint16);
#ifdef WE_REALLY_WANT_A_POINTER
free (buf);
#endif
Update: If we do not want to convert the number to text, but to an actual string (for reasons that elude my perception of common sense), it can be done simply by:
更新：如果我们不想将数字转换为文本，而是转换为实际的字符串（出于我对常识的看法），可以简单地通过以下方式完成：
char *str = (char *) (intptr_t) my_uint16;
Or, if you are after a string that is at the same address:
或者，如果您要查找位于同一地址的字符串：
char *str = (char *) &my_uint16;
Update: For completeness, another way of presenting an uint16_tis as a series of four hexadecimal digits, requiring 4 chars. Skipping the WE_REALLY_WANT_A_POINTERordeal, here's the code:
更新：为了完整起见，另一种表示 an 的方式uint16_t是一系列四个十六进制数字，需要 4 个字符。跳过WE_REALLY_WANT_A_POINTER考验，这是代码：
const char hex[] = "0123456789abcdef";
char buf[4];
buf[0] = hex[my_uint16 & f];
buf[1] = hex[(my_uint16 >> 4) & f];
buf[2] = hex[(my_uint16 >> 8) & f];
buf[3] = hex[my_uint16 >> 12];
回答by dreamlax
A uint16_tvalue only requires two unsigned charobjects to describe it. Whether the higher byte comes first or last depends on the endiannessof your platform:
一个uint16_t值只需要两个unsigned char对象来描述它。高字节是第一个还是最后一个取决于您平台的字节序：
// if your platform is big-endian
uint16_t value = 0x0A0B;
unsigned char buf[2];
buf[0] = (value >> 8); // 0x0A comes first
buf[1] = value;
// if your platform is little-endian
uint16_t value = 0x0A0B;
unsigned char buf[2];
buf[0] = value;
buf[1] = (value >> 8); // 0x0A comes last
回答by One Man Crew
You can use sprintf:
您可以使用sprintf：
sprintf(str, "%u", a); //a is your number ,str will contain your number as string
回答by Henrik
It's not entirely clear what you want to do, but it sounds to me that what you want is a simple cast.
不完全清楚你想做什么，但在我看来你想要的是一个简单的演员。
uint16_t val = 0xABCD;
char* string = (char*) &val;
Beware that the string in general is not a 0-byte terminated C-string, so don't do anything dangerous with it.
请注意，字符串通常不是以 0 字节结尾的 C 字符串，所以不要对它做任何危险的事情。
