What is the need of hh and h format specifiers?

问题

In the code below mac_str is char pointer and mac is a uint8_t array:

 sscanf(mac_str,"%x:%x:%x:%x:%x:%x",&mac[0],&mac[1],&mac[2],&mac[3],&mac[4],&mac[5]);

When I try the above code it gives me a warning:

warning: format ‘%x’ expects argument of type ‘unsigned int *’, but argument 8 has type ‘uint8_t *’ [-Wformat]

but I saw in some code they specified

sscanf(str,"%hhx:%hhx:%hhx:%hhx:%hhx:%hhx",&mac[0],&mac[1],&mac[2],&mac[3],&mac[4],&mac[5]);

which doesn't give any warning but both are working the same.

What's the need of using hhx instead of just x?

回答1:

&mac[0] is a pointer to an unsigned char.¹%hhx means the corresponding arguments points to an unsigned char. Use square pegs for square holes: the conversion specifiers in the format string must match the argument types.

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¹ Actually, &mac[0] is a pointer to a uint8_t, and %hhx is still wrong for uint8_t. It “works” in many implementations because uint8_t is the same as unsigned char in many implementations. But the proper format is "%" SCNx8, as in:

#include <inttypes.h>
…
scanf(mac_str, "%" SCNx8 "… rest of format string", &mac[0], … rest of arguments);

回答2:

hh is a length modifier that specifies the destination type of the argument. The default for conversion format specifier x is unsigned int*. With hh, it becomes unsigned char* or signed char*.

Refer to the table herein for more details.

回答3:

hhx converts input to unsigned char, while x converts to unsigned int. And since uint8_t is typedef to unsigned char, hhx fixes warning.

来源：https://stackoverflow.com/questions/21782244/what-is-the-need-of-hh-and-h-format-specifiers