this c programming has been giving me lot of grief...


why is

char x[10];

scanf("%s", x);
printf("%s", x);


this works?

BUT


int n ;
scanf("%i", n);
printf("%i", n);

this doesnt work, unless you change

scanf("%i", &n);



Is it because char is an array? But shouldnt pointers only be used when there is an array? actually when should you use a pointer?

Also if you want to use string in c, you have to do like char *string, so all strings are secretly arrays?

Also why is

char *s = "xyz";

printf("%s", s);

this is OK

but

printf("%s", s[1]);

This is NOT ok

and instead need to put the & like this to work

printf("%s", &s[1]);

??

In C, array is a kind of pointer, in short.

http://www.cs.ecu.edu/karl/3300/spr16/Note...ay/pointer.html

scanf("%i", &n);

This is dangerous and wrong, imagine you key in 55555
but n is only 4bytes.

n = value
&n = address

Think of it as "scan feed into value n" vs "scan feed into address n". Which do you think is correct?


You will typically use pointers in function parameters to avoid passing the size of the value into the function, using unnecessary memory.


Yes, its the same in almost any language. Strings are really just an array of chars, or maybe even a linked list.

%s is a string, which as you've learnt, is an array of characters, so its expecting a pointer/address. s[1] gives the value, not the address. If you changed %s to %c in printf, s[1] will work.

cannot scan feed into integer and must feed into address? why is it?

because the address is where the value is stored. How do you feed into a value?



so if you feed value 5 into 3, does that make sense? Or does it make more sense to say, scan feed (with value 5) into memory address 0x0001.

scanf() // read data and store into location. location => pointer.

* note 1 : string is array of characters.
* note 2 : scanf accept pointer, using variable name "x" or "&x" will send in pointer.

* note 3 : n is array of int (32 bits is from -2147483648 to 2147483647)
* note 4 : using "n" or "&n" will use pointer.
* note 5 : address of n[0] and n[1] different are 4, int32 use 4 bytes.

* note 6 : int num is a 32bit variable, default value assume is 0.
* note 7 : if scanf store value to "address 0" or "any other unknown address" will cause system crash.
* note 8 : for non-array variable, use &num to tell scanf store value to "this location"

basically u can think of any variable in C as a storage location in computer memory.

when u declare a variable, u are allocating a mempry storage for the variable, and the variable name refers to the value resides in that memory location.

when u use scanf, it needs a storage to store the input value, ie the address of a memory location, not the value residing in that location, hence the use of & as referring the address of the memory location/variable. with array, the variable name itself refers to the address of the start of the array (e.g string of chars) so u can use the array name itself as arg to scanf without the need to use &

in C, while types like int, double hold values, arrays are pointers (addresses in memory)

e.g.
int n = 5 (variable n holds integer 5)
printf(n) // prints value (5)
printf(&n) // prints 0x8f49be30 (some address in memory)

char x[10] = some letters
printf(x[n]) // prints letter (value) at specified index
printf(&x[n]) // prints pointer to specified index
printf(x) // prints pointer to the first index

As you know, arrays are indexed therefore you only get the value when you call its index. However, when calling x itself, it will return you the pointer of the address of the first index. So theres no such thing as &x (address of pointer)
x[n] returns value
x returns pointer
address : &x[0] is the same as x
address : &x[1] is the same as x+1
value : x[0] is the same as *x (dereference operator to get value pointed by pointer)
value : x[1] is the same as *(x+1)
I believe you understand how scanf works based on other comments here

This post has been edited by Horhorlidat: Nov 28 2021, 01:47 AM

.........

This post has been edited by FlierMate: Dec 31 2021, 05:35 PM