如何在十六进制的uint32_t上执行sprintf并将其存储到字符串中？

Question

I am running a timer to calculate the efficiency of my program and I need to output it to a hyperterminal. So, I need to do a sprintf on a uint32_t hexidecimal to store it into a string. But I keep getting an error on the %08X . So what should I use instead? I have tried using %ll , %lu but the warning is still there.

volatile char str_cycles=0;
volatile uint32_t total_cycles = 0x00ffffff - current_time;
sprintf(str_cycles, "%08X",total_cycles);

Can anyone help me with this?

Answer 1

You should be getting an error because str_cycles is shown as volatile char str_cycles; (a single character).

You should be using <inttypes.h> and:

char str_cycles[16];

snprintf(str_cycles, sizeof(str_cycles), "%08" PRIX32, total_cycles);

Generally, snprintf() should be preferred over sprintf() , but if you have enough space allocated it is perfectly OK to use sprintf() .

I'm not convinced about the volatile qualifier either; it is your job to determine why you have that and whether it is correct and whether it matters. Generally, you do not want a volatile string; it makes using it unreliable. (This answer first omitted the volatile on str_cycles , then added it, and has now omitted it again.)

Answer 2

Use fixed width print specifiers.

#include <inttypes.h>
char buf[9];
sprintf(buf, "%08" PRIX32,total_cycles);

Answer 3

Cast to an int or unsigned first. This is the type expected by %X :

sprintf(str_cycles, "%08X", (unsigned) total_cycles);