char* 字符串到 C 中的十六进制数组

Question

I have char * para.mac , string c8934641d0b7 stored in it.

I want to convert it to hex array int m[6] , make

m[0]=0xc8;
m[1]=0x93;
...
m[5]=0xb7;

I tried to do it with:

int i;
int m[6];
sscanf(para.mac, "%x%x%x%x%x%x", &m[0], &m[1], &m[2], &m[3], &m[4], &m[5]);

for(i=0;i<6;i++)
    printf("%i  \r\n",m[i]);

but it seems doesn't working.

and I don't know why the result shows to me like:

1178718391  
0  
0  
134448233  
8  
134272063  

I think it should be

200(0xc8)
147(0x93)
70(0x46)
65(0x41)
208(0xd0)
183(0xb7)

so what's the problem?

Answer 1

You're on the right track, if I'm not mistaken.

Consider trying: "%2x%2x%2x%2x%2x%2x" , which means that each hex digit will only consume at most 2 characters.

Answer 2

You are using the only %x in the sscanf function.

  sscanf(buf, "%x%x%x%x%x%x", &m[0], &m[1], &m[2], &m[3], &m[4], &m[5]);

Which will get only character of hexadecimal. But it should take two characters to convert it. You can change the sscanf function like this:

 sscanf(buf, "%2x%2x%2x%2x%2x%2x", &m[0], &m[1], &m[2], &m[3], &m[4], &m[5]);

Now as per your string : c8934641d0b7 The first %2x will get c8 and the second %2x will get 93 , the third will get 46 fourth will get 41 fifth will get d0 and finally fifth will get b7 .

So it will work fine now.