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.

CodePudding user response：

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);

CodePudding user response：

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 *

CodePudding user response：

&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);