dev-cpp-users

[Dev-C++] strncpy output probs

[<Den...@ao...>]

Hi there, 
im having problems on a small program that I'm using for a course to display 
the properties of strncpy.
The trouble is having compiled the code I keep getting a strange output,
my code is

#include <stdio.h>
#include <string.h>

#define SIZE_OF 15

int main()
{
char FirstName[SIZE_OF], SecondName[SIZE_OF], 
FirstMix[SIZE_OF],SecondMix[SIZE_OF];
  // used SIZE_OF instead of declaring each array size individualy

printf("Please enter the name JACK\n");
scanf("%s", &FirstName);

printf("Now please enter the name ROBINSON\n");
scanf("%s", &SecondName);

strncpy(FirstMix,FirstName,3);
printf("FirstMix=%s\n", FirstMix);

strncat(FirstMix,SecondName,3);
printf("FirstMix=%s\n", FirstMix);


return 0 ;
}


and my output is 


Please enter the name JACK
JACK
Now please enter the name ROBINSON
ROBINSON
FirstMix=JAC?-?@
FirstMix=JAC?-?@ROB

has anybody an idea as to where the extra characters are coming from????
thanks in advance. 

Re: [Dev-C++] strncpy output probs

[Daniel K. O. <dan...@ya...>]

Den...@ao... wrote:

> strncpy(FirstMix,FirstName,3);

(...)

> has anybody an idea as to where the extra characters are coming from????


They are garbage of uninitialized memory.
If you read the documentation for strncpy() you'll see that it copies 3 
characters when you call it.
None of your 3 characters are the '\0' terminator, because the '\0' it's 
beyond the 3 chars you are copying (it's the 5th char in "JACK", and 9th 
in "ROBINSON"). So you get a non-terminated string. When using strncpy() 
you have always to terminate it by hand.


Daniel K. O.