[Solved] Please explain the difference in the printfs below

by


%x format specifier experts the argument to be of type unsigned int.

In your case,

 printf("%x\n",(const uint8_t)0x0D);
 printf("%x\n",0x0D);

arguments will be promoted (default promotion rule) to match the type, but in case of

 printf("%x\n",(const uint8_t *)0x0D);  //supplying a pointer
 printf("%x\n",(uint8_t *)0x0D);        //supplying a pointer

You’ll invoke undefined behavior, as per C11, chapter §7.21.6.1

[…] If any argument is
not the correct type for the corresponding conversion specification, the behavior is
undefined. […]

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solved Please explain the difference in the printfs below