Reputation: 143
I need to convert u16(unsigned int -2 byte) value into string (not ascii). How to convert unsigned int(u16) into string value(char *)?
Upvotes: 5
Views: 51236
Reputation: 980
/* 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:
char *str = (char *) (intptr_t) my_uint16;
Or, if you are after a string that is at the same address:
char *str = (char *) &my_uint16;
Update: For completeness, another way of presenting an uint16_t
is as a series of four hexadecimal digits, requiring 4 chars. Skipping the WE_REALLY_WANT_A_POINTER
ordeal, here's the code:
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];
Upvotes: 10
Reputation: 95405
A uint16_t
value only requires two unsigned char
objects to describe it. Whether the higher byte comes first or last depends on the endianness of your platform:
// 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
Upvotes: 3
Reputation: 9578
You can use sprintf
:
sprintf(str, "%u", a); //a is your number ,str will contain your number as string
Upvotes: 1
Reputation: 4314
It's not entirely clear what you want to do, but it sounds to me that what you want is a simple cast.
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.
Upvotes: 1