Thread: Character Array

Originally Posted by cstryx

You probably meant to do this:

Code:

char n[6];
printf("Enter you name");
scanf("%5s", n);

Note the importance of the '5' in "%5s": without it, your code would be vulnerable to buffer overflow. You should also check the return value of scanf: in this case it should return 1, otherwise you have an input error.

Furthermore, as Matticus implied in post #2 by observing that you are "confusing individual characters (%c) and strings (%s)", this would replace your for loop as you would be working with a string. Likewise, you do not need a second for loop for print the string once.

Also, you need to indent your code properly. This is especially important when your code involves loops and the like.

It's great that we have pretty much all bases covered now that the issues have been pointed out and laserlight was kind enough to provide further detail about them.

One thing I will add is about the "%5s". Some people get confused about what number should be put there. That 5 is for the actual visible char data that gets put into the string buffer before the null terminator. A format like %s means that the null terminator is appended automatically, so if you use %6s, it would add 6 characters to the buffer + 1 null terminator which would mean you would need a buffer of size 7 bytes, and you only have allocated enough for 6 chars. This is why the proper string is "%5s" and not "%6s".

I like to use macros to simplify this stuff so that I don't have to change the string manually.

Code:

#include <stdio.h>

#define STR_TOKEN(x) #x
#define STR(x) STR_TOKEN(x)
#define MAX_CHARS 5

int main(void)
{
  char n[MAX_CHARS + 1];
  printf("Enter you name");
  scanf("%" STR(MAX_CHARS) "s", n);
  printf("%s\n", n);
  return 0;
}