C* && C[400]


 
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# 1  
Old 08-06-2011
C* && C[400]

I am normally a Java programmer, super noob in C. I got a noob question here

Code:
 
char input[400];
printf("Please enter the country name!");
scanf("%s", &input); //& character is essential for scanf() %f = float, %

There error it say
warning: format ‘%s’ expects type ‘char *’, but argument 2 has type ‘char (*)[400]’
I dont understand %s is for string rite?? char make up string then why it require char *?? char* is a pointer?
# 2  
Old 08-06-2011
When using in functions calls the name of an array is equivalent to the pointer to the first element of this array. You should use just "input" in scanf (and btw it's better to use "%399s" or fgets).
And yes, strings in C are pointers to char.
# 3  
Old 08-06-2011
Sry but i dont get what you mean. In scanf i did try what you say using "input" instead of "&input". What is the & for?? I put the & because i read a book saying that
& character is essential for scanf()
# 4  
Old 08-06-2011
& is for getting the address of a variable. But a pointer is already an address. So for example, you need it for an int variable, but if you have already a pointer to int, then you just name this pointer in function arguments without &.
Code:
int x;
int *ip = &x;
scanf("%d", ip)

With char buffers:
Code:
char input [400];
char *inputp = input; 
scanf("%s399", inputp);

# 5  
Old 08-06-2011
Ok i get what your trying to say but char input [400] is not a pointer why cant i
Code:
scanf("%s", &input);

and also what is
Code:
%s399

i understand that
Code:
 
%f

is float then
Code:
%s

should be string??
# 6  
Old 08-06-2011
In C, variables can be accessed in 2 ways: directly, or as pointers, where a pointer is, basically, the address where the memory reserved for that variable starts. Now here's the funny thing: there aren't really any strings in C, but arrays of characters, and most functions dealing with them require the last character to be the null character ('\0'). And even that's not really correct, because really there aren't arrays either. Declaring an array tells the compiler to reserve a certain amount of memory, and the variable really is the pointer to the first address in that block.

So you using &input doesn't mean "pass the pointer", but rather "pass the pointer to the pointer", which is 1 indirection too much.

And I think yazu meant %399s, which means "A string of characters, 399 at most" (because the 400th should be null).
This User Gave Thanks to pludi For This Post:
# 7  
Old 08-06-2011
Quote:
char input [400] is not a pointer
In function arguments It is a pointer.
Quote:
what is %s399
It is the maximum number of chars that you can get in the buffer (1 is needed for the last '\0'). So if you input more you can't overflow the buffer.
Quote:
%s should be string
There is no "string" type in C.

PS K&R is a wonderful book. One of the best programming books I've ever read.

---

Oh... Not again. Of course, I meant "%399s" instead of "%s399"... Smilie

Last edited by yazu; 08-07-2011 at 02:27 AM..
This User Gave Thanks to yazu For This Post:
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