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So there is a program that I am working on, that requires me to access data from a char array containing hex values. I have to use a function called func(), in this example, in order to do access the data structure. Func() contains 3 pointer variables, each of different types, and I can use any of them to access the data in the array. Whichever datatype I choose will affect what values will be stored to the pointer. Soo heres the code:

unsigned char data[]
{
    0xBA, 0xDA, 0x69, 0x50,
    0x33, 0xFF, 0x33, 0x40,
    0x20, 0x10, 0x03, 0x30,
    0x66, 0x03, 0x33, 0x40,
}

func()
{
   unsigned char *ch;
   unsigned int i*;
   unsigned short* s;
   unsigned int v;  

   s = (unsigned short*)&data[0];
   v = s[6];
   printf("val:0x%x \n",v);
}

Output:

Val:0x366

The problem with this output is that it should be 0x0366 with the zero in front of the 3, but it gets cut off at the printf statement, and I'm not allowed to modify that. How else could I fix this?

4

1 回答 1

4

Use a format that specifies leading zeros: %04x.

Without changing the format passed to printf or replacing it entirely I'm afraid there's no way to affect the output.

于 2013-10-10T05:36:54.160 回答