char variable[#] - confusion

[Chris2222]

Okay I'm just learning so bear with me.

Suppose i have some code like this:
[code]
.
.
char name[4];
printf("Please enter your name:");
scanf("%s",&name);
printf("Your name is %s",name);
.
.
[\code]
Now I've set aside 4 characters for the name variable. However when I type in a name like Christopher I get the full name back.

Well haven't I specifically set the size for 4 characters? So why do I get 11 characters back in the output? Shouldn't my name be truncated after the 4th character?

[Chris2222]

Okay I'm just learning so bear with me.

Suppose i have some code like this:

.
.
char name[4];
printf("Please enter your name:");
scanf("%s",&name);
printf("Your name is %s",name);
.
.

Now I've set aside 4 characters for the name variable. However when I type in a name like Christopher I get the full name back.

Well haven't I specifically set the size for 4 characters? So why do I get 11 characters back in the output? Shouldn't my name be truncated after the 4th character?

[robatino]

You just got lucky. You actually only allocated enough space for a 3-character name (3 characters plus the trailing \0) so writing anything bigger is undefined behavior. The extra space you wrote into just happened not to be used by anything else at the time. It's safer to use fgets() instead of scanf() to enforce not reading a longer string than you have allocated.

http://en.wikipedia.org/wiki/Fgets

[Prelude]

>Shouldn't my name be truncated after the 4th character?
C isn't that user-friendly, you get exactly what you ask for, even if it wasn't what you wanted. In your case, run this to see where those extra characters are stored:

#include <stdio.h>

int main ( void )
{
  char left[256] = {0};
  char name[4];
  char right[256] = {0};

  printf ( "Enter your name: " );
  fflush ( stdout );
  scanf ( "%s", name );

  printf ( "left:  |%s|\n", left );
  printf ( "name:  |%s|\n", name );
  printf ( "right: |%s|\n", right );

  return 0;
}

Now try to think of how many ways that could break your program.

[QuantumPete]

Well haven't I specifically set the size for 4 characters? So why do I get 11 characters back in the output? Shouldn't my name be truncated after the 4th character?

Writing past the end of an array results in that lovely concept of "Undefined Behaviour". since it is undefined, anything could happen, including that everything works just fine. Just don't count on it.

QuantumPete

[HowardL]

Maybe an illustration would help.

Think of you memory as looking like this where each 0 is one byte:
...0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0...

Each of those bytes has an address.
When you start the program the OS will set up an address to store your array:
char name[4];

          |4 bytes| other potential data
          |       |
...0 0 0 0|0 0 0 0|0 0 0 0 0 0 0 0 0 0 0 0 ...
           \
            the starting address of array 'name'

You can show this address like this:
printf("%p", (void*)name);

.....
When you type in Christopher (and hit enter)

scanf("%s", &name);

...will store the characters entered and end it with a '\0'
(called a 'null' which is actually just zero)

          |4 bytes| other potential data
          |       |
...0 0 0 0|C h r i|s t o p e r 0 0 0 0 0 0 ...
                               ^
                               zero added by scanf()

As was said you will sometimes get away with writing more elements than are allocated.
That does not mean that you should not worry about it because it works THIS time.
If another strings data is stored right next to 'name' writing too many characters will ruin that data as the Preludes example will show. . It is up to us as programmers to BE SURE we're staying 'in bounds'. . Just think, it could a simple string or it could be your savings balance!
Howard();

[Chris2222]

Okay thanks.

So by specifying name[4] it'll be stable up to 3 characters otherwise, as you have said, they sit in undefined territory and I'll get unpredicatable results with anything beyond the 3 character mark.

Before, if you wrote more than specified, I thought the computer would automatically allocate more characters to it. Sort of like when a float is multiplied by an int and the answer is forced to become a floating variable - the computer automatically floats the answer but of course in this case that is not the case.
Thanks for clearing that up.

[Prelude]

>I thought the computer would automatically allocate more characters to it.
"Automatically" isn't something you should expect as a C programmer. The world is your oyster, but you have to dive for it, yank it up, shuck it, cook it, and swallow it by yourself. And it's still kinda slimy.

[robatino]

> I thought the computer would automatically allocate more characters to it.

In C++ there's a string class which can grow automatically as needed, which in particular lets you input arbitrary-length strings without having to specify an upper size limit in advance. In C, the best you can do easily is to use something like fgets() to make sure that your input doesn't overwrite your fixed-size buffer - but if you exceed it it will be truncated (which at least is better than a buffer overrun).