I am facing a strange behavior with a char* variable
user_message* parseMessage(char *incoming_msg, uint64_t size)
{
user_message* msg = calloc(1, sizeof(user_message));
printf("value: %s\n", incoming_msg);
return msg;
}
void start_server()
{
char* msg = "1|david|pwd|";
printf("msg: %s\n", msg);
parseMessage(&msg, 12);
}
The output :
msg: 1|david|pwd|
value: �[
I struggle to figure out what is wrong in my code.
parseMessage(&msg, 12); should be parseMessage(msg, 12);
The &msg means you're passing a pointer to msg, which will be a char ** when the function takes only a char *
The first parameter of the function parseMessage and the expression used as an argument for this parameter have different types.
The parameter is declared as having the type char *
user_message* parseMessage(char *incoming_msg, uint64_t size)
while the expression used as an argument has the type char **.
char* msg = "1|david|pwd|";
//...
parseMessage(&msg, 12);
So in fact instead of outputting the string pointed to by the pointer msg you are trying to output as a string what is stored in memory occupied by the variable msg and what is followed it
You need to call the function at least like
parseMessage( msg, 12 );
Also it would be better at least to declare the first function parameter with the qualifier const if the passed string is not changed within the function
user_message* parseMessage( const char *incoming_msg, uint64_t size);
&msg yields a char **, or a pointer-to-pointer-to-char.
Both your parseMessage function, and the printf specifier %s expect a char *.
You don't need the address-of operator here
parseMessage(msg, 12);