Copy Char to char*

I have a question for C programming. I have a loop where I generate at each iteration a new char. Each generated char I want to append to a string (char *) This is my code:

char *string[20];
string[0]=0;

for (p = 0; p < length(message) - 1; p = p + 2)
{
...
char cc="";
cc = (char) strtol(pp, pp, 16);
char *tt[2];
tt[0]="";
*tt=(char *)cc;
strcat(string,cc);
}

I know that strcat uses char*. So I tried to copy the content of the char i wanna开发者_运维问答 append to a new pointer. I have probably done sth wrong...i'm not very experienced. Thank you

Reason for strtol: I have a char* msg, that holds a hexa number. I want to process that hexa number but when I try to do for instance shifting (<<16) is shifts 16 bits (multiplies by 2^16 ) the decimal ascii value of the hexa character, instead of shifting the character itself. So, I converted the hexa number to a character, so that when I try to do shifting it will shift the correct value. I know it' wierd. Cann you help me come up with sth better?

unsigned char *octets0"3F214365876616AB15387D5D59";
crc ^= ((*octets++) << 16);

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Several issues.

First problem is that you've created an array of char pointers, not a char array. You want:

char string[20];

In C, an array implicitly "degrades" into a pointer to its first element, so if you pass just string it's a char* ( string == &string[0] )

From there ... no idea what pp is, but strtol() returns a long int which you're trying to assign to a char - that's not going to work the way you think. You're short about 7 bytes.

(Edited to define "work" as in "Not what you think")

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Try not to cast between char* and char – it does not do anything useful and your compiler should not let you do that, anyway. If you really want to use strcat, that should be

char cc; // no ="" here!
...
tt[0] = cc;
strcat(string, tt)

But it would be much easier to just do what cnicutar posted above (I was going to type something like that here, too).