Error: Incompatible types in assignment

[Chrisch]

I am using C language, I have a small problem, it gives me the following error message: "incompatible types in assignment"

char * address (int octet1, int octet2,int octet3)
{
char *ad [64]
if octet 1> =1 && (octet 1<=26))
{
ad = strcat (octet1. "0.0.0.");
else if (octet 1<=128) && (octet 1<=192)&& (octet 1<=255)
{
ad = strcat (octet1, octet2 "0.0.0.");
}
return *ad;

The function is supposed to send me the address network (concatenated with the bytes passed as parameters ...) or it does not work very well ...Thank you in advance for your help!

[Zecho]

Malloc allocates memory and returns a pointer to the beginning of that block of memory. The malloc (memory allocation) is used to dynamically allocate memory at run time. Possible uses for this function are: Read records of an unknown length.
Read an unknown number of database records.The malloc() function shall allocate unused space for an object whose size in bytes is specified by size and whose value is unspecified.

[Walby]

You can pass to the function strcat more than two parameters. But I remind you of the prototype of the function strcat:
char * strcat (char * string1, const char * string2);
Even if the problem is not there, it is also possible that the compiler rattle as you pass it in the int parameters, or it expects pointers to char.

[Aberto]

strcat is not used like that ...
It only take 2 and 2 char pointers as parameters. The strcat() function will concatenate two strings by placing the second input on the end of the first.
You need to convert the int to char * (using sprintf for example), then concatenate example Byte1 with Byte2, retrieve the string in the ad, and then concatenate with ad octet3, etc. ...

You make a confusion by declaring ad. To declare a string is
char d [64] = "";
or
char * d = malloc (64 * sizeof (* ad));