Copying a char* into a pointer to an char[] in a header file

[quasi epoch]

Hello, I am currently stuck trying to figure out why strncpy(user->name, username, sizeof(username)); gives me a segfault. user comes from:

    char name[64];
    ...```
and username:
```char *username = argv[1];```
the first line of code above is the one giving me a segfault, and I do not know why, everything I have read tells me that it should work.

[waxen urchin]

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[lusty shadow]

sizeof(username) is the size of a pointer to character

aside from that, i don't think strncpy is what you want here

[quasi epoch]

what would you use?

[lusty shadow]

strcpy + a length check

[quasi epoch]

a length check?

[lusty shadow]

something like

if(strlen(argv[1])>=sizeof(user->name)){
    /*name is too long to fit into your buffer, exit now with an error message of your choosing*/
}
strcpy(user->name,argv[1]);

[quasi epoch]

so like user->name[10] = '\0';
?

I see

I will try that, thank you

still givin me segfualt sadly :/

[lusty shadow]

did you actually initialize user to something meaningful?

[quasi epoch]

wdym?

[lusty shadow]

user is a pointer, so it ought to point at something for you to use it

[quasi epoch]

wait give me a sec

[lusty shadow]

and do you also provide an argument for argv[1]

[quasi epoch]

okay so I do not initialize it to something meaningful as you put it, i just did struct user_t *user

and yes i did

but why would what I initialize user to matter?

[lusty shadow]

user is just a pointer to some user_t

[quasi epoch]

yes

[lusty shadow]

it doesn't provide any store for the user_t itself

[quasi epoch]

wait

😐

[lusty shadow]

you will probably be fine without a pointer and just making user of type user_t

[quasi epoch]

🫥

welp

thank you

!solved