hh 和 h 格式说明符需要什么? [英] What is the need of hh and h format specifiers?

问题描述

在下面的代码中，mac_str 是字符指针，mac 是 uint8_t 数组:

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]);

当我尝试上面的代码时，它给了我一个警告:

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]

但我在他们指定的一些代码中看到了

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]);

没有发出任何警告，但两者的工作方式相同.

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

使用 hhx 而不是 x 有什么需要?

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

推荐答案

&mac[0] 是一个指向 unsigned char 的指针.¹ %hhx 表示对应的参数指向一个unsigned char.方孔使用方钉:格式字符串中的转换说明符必须与参数类型匹配.

&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.

¹ 其实&mac[0]是一个指向uint8_t的指针，而%hhxuint8_t 仍然是错误的.它在许多实现中有效"，因为 uint8_t 在许多实现中与 unsigned char 相同.但正确的格式是 "%" SCNx8，如:

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);

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