将uint8_t转换为ascii字符串C

Question

Given the following function

UART_write(UART_Handle handle, const void *buffer, size_t size);

I want to send via uart a int8_t value ( log it )

What i tried:

int8_t value;
UART_write(uart, value, strlen(value));

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const char *echoPrompt = (char *)value;
UART_write(uart, echoPrompt, sizeof(echoPrompt));

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const char echoPrompt2[] = {value};
UART_write(uart, echoPrompt2, sizeof(echoPrompt2));

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const char* buff = value;
UART_write(uart, value, strlen(value));

The best i got is logging the hex value

Exemple of how the uart_write function works: In orded to log "12" what I need to do is

   const uint8_t value[] = {0x31, 0x32};
   UART_write(uart, value, sizeof(value));

So my question is, how to log my int8_t variable ( I need to log negative numbers as well)

Answer 1

You will need to convert your integer to string.

snprintf is a standard way to do this, if your libc provides it.

Answer 2

Convert uint8_t to an ascii string C

Determine the maximum string size needed for any value of that type. Is there a better way to size a buffer for printing integers?

#define UINT_BUFFER10_SIZE(type) (1 + (CHAR_BIT*sizeof(type)*LOG10_2_N)/LOG10_2_D + 1)

Form the buffer

char buf[UINT_BUFFER10_SIZE(value)];

"Print" the uint8_t to the buffer.

int len = sprintf(buf, "%d", value);
// or pedantically
int len = snprintf(buf, sizeof buf, "%" PRId8, value);  // see <inttypes.h>
assert(len >= 0 && (unsigned)len < sizeof buf);

Send it

UART_write(uart, buf, len);

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how to log my int8_t variable

#define INT_BUFFER10_SIZE(type) (2 + ((CHAR_BIT*sizeof(type)-1)*LOG10_2_N)/LOG10_2_D + 1)
char buf[INT_BUFFER10_SIZE(ivalue)];
int len = sprintf(buf, "%d", ivalue);
UART_write(uart, buf, len);

---

IMO, code should add a helper function to send a string

void UART_write_str(UART_Handle handle, const char *str) {
  UART_write(uart, str, strlen(str));
}