sscanf question (getting seg fault)

[Pointextder]

char* user = NULL;

sscanf(optarg, "%s", user); //optarg from getopt()

Anyone see anything wrong with those 2 lines? I'm getting a segfault I think is coming from that line. Thanks.

[Philip Nicoletti]

You did not allocate any space for user.

char user[20]; // make sure big enough

[Pointextder]

I guess my understanding of pointers isn't very good.

For a pointer to char like I have above, when can I point it to a string?

For example is the following correct:

char *user = "Tom";

//later in code I would redefine user
user = "Ted";

Thanks.

[Paul McKenzie]

Originally posted by Pointextder
For a pointer to char like I have above, when can I point it to a string?

In C++, a string has a special meaning -- a std::string.

What you want is to have an array of char that is big enough to hold all the characters (including the NULL) that you expect to recieve.

char user[500];
sscanf(optarg, "%s", user);

For example is the following correct:

char *user = "Tom";

It is "correct", but it is incorrect for the purposes that you wanted to use "user" for. The reason why is that

char *user = "Tom";

is called a string-literal. You cannot change the value of a string-literal (it is undefined behavior to do so), and that is what you would have attempted to do with your call to sscanf().

Again, you need a modifiable buffer of characters, and give the address of the first character of this modifiable buffer. My example showed an array, but there are other ways to create such a buffer (malloc(), new[], etc.).

Regards,

Paul McKenzie

[Pointextder]

I'm getting an incompatible types in assignment when I attempt to do the following:

char username[40] = "default_user";
...
username = optarg; //optarg from call to getopt


If I declare username as char username*; the assignment will work but I can't change the value of it.

[Kheun]

You need to use strcpy().

char username[40] = "default_user";
...


// Must enable the length of optarg is equal or less than username (40 characters including NULL)
strcpy(username, optarg);

[Paul McKenzie]

Originally posted by Pointextder
I'm getting an incompatible types in assignment when I attempt to do the following:

char username[40] = "default_user";
...
username = optarg; //optarg from call to getopt


If I declare username as char username*; the assignment will work but I can't change the value of it.

To change an array (which is what username is), you can't do it by using operator =. Arrays are not copyable or assignable (by definition). You need to change each element of the array in a loop, or a function that effectively changes the values using some sort of loop. To do this, there are functions such as strcpy(), memset(), etc. that work with buffers and arrays.

When you changed username to a pointer, all you did was assign one pointer to another pointer.

I believe you need to get a good C or C++ tutorial on how to handle NULL-terminated strings and pointers, or if you're using C++, use std::string and not use pointers. If you don't understand the basics of pointers, you will undoubtedly come across more problems.

Regards,

Paul McKenzie

[Gorgor]

> In C++, a string has a special meaning -- a std::string.

Are you sure? I've been programming in C++ for 10 years now, and when a programmer says "string", they still always mean an array of characters.

If they mean std:string, then they say std:string.

[gstercken]

Originally posted by Gorgor
Are you sure? I've been programming in C++ for 10 years now, and when a programmer says "string", they still always mean an array of characters.

Those are C programmers - C++ programmers mean std::string. Using a char array was an anachronism from the moment when std::string had been included in the C++ standard library.