C Programming Tutorial 18, Arrays pt.3

[ketz]

One of the best videos on c programming on Youtube...It's an absolute must for student having difficulty with c programming.

[windday]

Gosh, you help me out in doing my assignment very well, thanks dude ! I like your tutorial..

[TaliWalt]

Love your tutorials! Question regarding this code:

char name[20];
printf("please enter your name: ");
scanf("%s", &name);


My complier gives me the following warning:

warning: format ‘%s’ expects type ‘char *’, but argument 2 has type ‘char (*)[20]

So I searched around, and found one suggestion online, which worked for me, lose the '&' in the scanf line.

[gs]

You should get a lot more views! You're amazing

[arenagods]

Please make a tutorial how to compare strings in (if, else ) function

[TaliWalt]

(this is a continuation of my previous post)

Here is the explanation given for getting rid of the '&':

"%s" means that scanf is expecting a pointer to the first element of a char array. me is an object array and could evaluated as pointer. So that's why you can use me directly without adding &. Adding & to me will be evaluated to ‘char (*)[20]’ and your scanf is waiting char *

So, how is it possible that one line of code works for you, but not for me? Is it a compiler thing?

[massimogiardina]

You forgot to mention that there's a \0 at the end of every string

[muhammadsada]

That's pretty much it for this tutorial..so, see you guys!

[OneTimWhatley]

so that's pretty much it for this tutorial, so see you guys!

[iTzAdamX]

I fucked up in the video, I shouldn't have put the & there. The thing is though, arrays in C are kinda weird. &myArray is actually the same thing as just plain old myArray. The 'error' that you're getting is actually just a warning, so you can ignore it.

[m.mohammad]

Why cant i do this


char newWord[10];
printf("Enter a new sentence: ");
scanf("%s", &newWord);
printf("%s", newWord);