Why this code shows warning? In C

Question

I can't able to understand,the reason behind the warning, displayed by the console....
https://code.sololearn.com/cUWwDo18D45h/?ref=app

Answer 1

a[5][7] is an array of 5 arrays
You can refer to the first one like that:
&a[0][0]
That is a pointer to the first element of the first array, or even like that:
a[0]
because the name of an array is a pointer to it's first element.
If you write &a[0] you have a pointer to the first element of your array of arrays, i.e. a pointer to a pointer to char.
But %s is expecting just a pointer to char

Answer 2

Same reason, a[0] is already an address, & a[0] asks for the address of an address, and so it gives you a warning

Answer 3

scanf("%s") needs a memory address.
a[0] is a variable that holds a memory address
a[0] gives a value, that value is the memory address of your array
&a[0] gives a memory address, the memory address of a[0].
So, scanf("%s", a[0]) puts %s in the array
scanf("%s", &a[0]) puts the %s in a variable that is supposed to hold a memory address, not %s
So it gives you error

Answer 4

Thank you very much Davide 😊

Answer 5

You are welcome 🤗

Answer 6

a is the name of the 2D array
a[0] is the name of the first array of the 2D array
Since an array name is a pointer to its first element, a = &a[0] and a[0] = &a[0][0]
scanf("%d", a); is the same of:
scanf("%d", &a[0]);
You can't do that, a is a pointer to a pointer to int, but %d expects a pointer to int
what you can do is
scanf("%d", a[0]); that is like:
scanf("%d", &a[0][0]);

Answer 7

Thank you very much Davide 😊😃

Answer 8

🤗

Answer 9

Thank you very much Angelo 😊
But can you tell me,what is the meaning of warning, displayed in console....

Answer 10

Thank you Angelo 😊

Answer 11

Davide and Angelo ,I have a another doubt😅Can you tell me ,
what does the "a" indicates in this below function......
I mean,it point's to,arrays which first element?
scanf("%d",a);//in two dimensional array
And also tell me difference between a[0] && a in two dimensional array.....