字符*和字节*相等性比较（数字和文本）

Question

I have the following a char* key which is a char array that can contain either integers or text. So the key can have value 3 or tom for example

I have a byte* data array which contains stored data. I need to test whether the key is equal to the data.

My logic is currently along the lines of:

int j = 0 ;
for (j = 0; j < len; j++) {
    sprintf(key_cmp, "%02x", (ulong)*data++);
}
if (!strcmp(key, key_cmp)) fprintf(stderr, "Equal \n");

I realise this code is incorrect as i am trying to print as hex rather than char here... but when I try to use %02x, garbage gets printed out.

How can I also ensure that 01 and 1 will be treated as equal? I realise that this may vary on byte ordering, hence I can't think of a general solution. I'd like to avoid using atoi so was wondering if there was another method (mostly because I have no real way of knowing whether the key is an integer or not)

Thanks

Answer 1

The guess is your problem is that %02x is the format for an int not a unsigned long - so you're on system where sizeof(int)!=sizeof(long) that will cause a problem.

See [Wiki][1] for a description of format specifiers.

Answer 2

Because char[] is 1 byte and byte[] is 1 byte it would be easier to compare them as byte arrays (unsigned char). Either will do the same thing without any special formatting.

memcmp(key,data,len)
strcmp(key,data,len)