How to convert uint32_t to string?

[ooosawaddee3]

Do you know how to convert uint32_t to string?

Thank you very much,

[cpjust]

char str[10];
uint32_t num = 50;

sprintf( str, "%u", num );

[mingerso]

Do you know how to convert uint32_t to string?

Thank you very much,

Another way without sprintf would be:

#include <stdint.h>
#include <stdio.h>

int main(void)
{
        uint32_t y;
        uint32_t x = 50;
        unsigned char buf[4];

        /* pack into buf string */
        buf[0] = x >> 24;
        buf[1] = x >> 16;
        buf[2] = x >> 8;
        buf[3] = x;

        /* convert the string back to a uint32_t */
        y = (buf[0] <<  24) | (buf[1] << 16) | (buf[2] << 8) | buf[3];

        printf("y: %d = %d\n", y, x);

        return 0;
}

[cas]

The easiest way is snprintf(), but don't use the %u specifier! There's no guarantee that an unsigned int is the same thing as a uint32_t. There are a couple of methods I'd suggest to do this, each of which is ugly in its own way:

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
int main(void)
{
  char str[11]; /* 11 bytes: 10 for the digits, 1 for the null character */
  uint32_t n = 12345;
  snprintf(str, sizeof str, "%lu", (unsigned long)n); /* Method 1 */
  snprintf(str, sizeof str, "%" PRIu32, n); /* Method 2 */
}

The first method converts the uint32_t to an unsigned long, which is guaranteed to be large enough to hold any possible uint32_t value. The second method uses a macro to get the proper conversion specifier.

The fact that %u might work on your system is no guarantee that it will work everywhere, so please don't rely on it. The methods I showed are admittedly uglier, but they're at least portable.

[matsp]

Another way without sprintf would be:

#include <stdint.h>
#include <stdio.h>

int main(void)
{
        uint32_t y;
        uint32_t x = 50;
        unsigned char buf[4];

        /* pack into buf string */
        buf[0] = x >> 24;
        buf[1] = x >> 16;
        buf[2] = x >> 8;
        buf[3] = x;

        /* convert the string back to a uint32_t */
        y = (buf[0] <<  24) | (buf[1] << 16) | (buf[2] << 8) | buf[3];

        printf("y: %d = %d\n", y, x);

        return 0;
}

That won't become a C style string tho' - it's just the binary data stored in 4 unsigned char's - a C style string has a 0-character for termination. Your method may well end up with a zero in the middle of the character array, and is by no means sure to have a zero at the end (and there is no space for a zero either).

--
Mats

[audinue]

char str[10];
uint32_t num = 50;

sprintf( str, "%u", num );

use...

sprintf( str, "%I64u", num);

[matsp]

use...

sprintf( str, "%I64u", num);

Ehm, that would almost certainly print rubbish if num is correctly defined as a 32-bit number - since it will use whatever is after the number in memory to use as the other 32-bit part of the 64-bit number. Aside from that, it's specific to Windows.

As an example:

#include <stdio.h>


int main()
{
  unsigned int x = 7;
  printf("x = %I64d\n", x);
  return 0;
}

prints:

x = 18035564108316679

Which doesn't look very much like 7 to me...

--
Mats

[mingerso]

That won't become a C style string tho' - it's just the binary data stored in 4 unsigned char's - a C style string has a 0-character for termination. Your method may well end up with a zero in the middle of the character array, and is by no means sure to have a zero at the end (and there is no space for a zero either).

--
Mats

Yes, you are right - I meant to mention that its another way to "put" a number into a "string" data type... but by no means the number represented as a string. Sorry for the confusion.