snprintf does not stores last character in buffer?

Question

I have written a simple small program to read a file with following format using snprintf,

skip first 15 chars , next 9 chars are sequence number, next 2 char is message and so on.   

I am interested in sequence number and message i.e. from char no 16 to 26;

below is the program. It does not reads last char for each field. It reads 8 chars instead of 9 for sequence number and and 1 byte instead of 2 for message.

#include<stdio.h>

typedef struct
{
  char seqno[9];
  char msg[2];
}Header_T;

int main()
{
  char buf[64]={'\0'};
  FILE *fp;
  int i = 0;
  Header_T hdr1;
  int skipbytes = 15;

  fp=fopen("asdf", "r");
  if (fp == NULL)
  {
    printf("FILE OPEN ERROR\n");
  }
  printf("--sequence--msg--\n");
  while( fgets( buf, sizeof(buf), fp ) != NULL )
  {
    i=skipbytes;
    snprintf(hdr1.seqno, 9, "%s", (buf+i));
    i+=sizeof(hdr1.seqno);
    snprintf(hdr1.msg, 2, "%s", (buf+i));
    i=0;

    printf("--%s--%s--\n", hdr1.seqno, hdr1.msg);
    memset(buf, '\0', 64 );
  }
fclose(fp);
return 0 ;
}

---

part of file contents is as follows

201301082323458000000001H QB234
201301082323558000000002J QB234
201301082323658000000003N QB234
201301082323758000000004JRQB234
201301082333458000000010JSQB234

so expected output is

--sequence--msg--
--000000001--H --
--000000002--J --
--000000003--N --
--000000004--JR--
--000000010--JS--

But Instead I am getting output as

--seqno--msgtype--
--00000000--H--
--00000000--J--
--00000000--N--
--00000000--J--
--00000001--J--

Can anyone explain this behavior and how to fix it?

Instead of snprintf I tried the same program using for loop character-by-character assignment and program works fine; but for that I need to add some fillers for byte alignment in structure.

I also tried using pragma pack() but it also makes no difference.

I am using gcc 4.4.3 on ubuntu 64 bit machine

Answer 1

From cppreference

int snprintf ( char * s, size_t n, const char * format, ... );

n: Maximum number of bytes to be used in the buffer.

The generated string has a length of at most n-1, leaving space for the additional terminating null character.

So if you expect 9 characters, you should pass n as 10, instead of 9:

snprintf(hdr1.seqno, 10, "%s", (buf+i));

Header_T needs to get changed accordingly:

typedef struct
{
  char seqno[10];
  char msg[3];
}Header_T;

Answer 2

You need to have an extra character to store the trailing '\0' character, so your struct should look like

typedef struct
{ char seqno[10];
  char msg[3];
}Header_T;

EDIT: since this is a simple data transfer, why not use memcpy, like:

memset(hdr1, 0, sizeof(hdr1));
memcpy(hdr1.seqno, &buf[skipbytes], 9);
memcpy(hdr1.msg, &buf[skipbytes + 9], 2);

Answer 3

#include<stdio.h>

typedef struct {
    char seqno[9+1];//+1 for EOS('\0')
    char msg[2+1];
} Header_T;

int main(void){
    char buf[64]={'\0'};
    FILE *fp;
    Header_T hdr1;

    fp=fopen("asdf", "r");
    if (fp == NULL) {
        printf("FILE OPEN ERROR\n");
        return 1;
    }
    printf("--sequence--msg--\n");
    while( fgets( buf, sizeof(buf), fp ) != NULL ){
        sscanf(buf, "%*15c%9c%2c", hdr1.seqno, hdr1.msg);
        hdr1.seqno[sizeof(hdr1.seqno)-1] = hdr1.msg[sizeof(hdr1.msg)-1] = '\0';
        printf("--%s--%s--\n", hdr1.seqno, hdr1.msg);
        //memset(buf, '\0', 64 );
    }
    fclose(fp);
    return 0 ;
}

Answer 4

This anwser is after the two anwser above.They all give a some right suggestions . I just spliced them together .

First , use :

typedef struct
{ char seqno[10];
  char msg[3];
}Header_T;       //so we have the memory for '\0'

Then , use :

   while( fgets( buf, sizeof(buf), fp ) != NULL )
   {
        i=skipbytes;
        snprintf(hdr1.seqno, 10, "%s", (buf+i));//here we read 9 number and a '\0'
        i+=sizeof(hdr1.seqno) -1;               //the  '\0' is not part of buffer.
        snprintf(hdr1.msg, 3, "%s", (buf+i));   //also for '\0'
        i=0;

        printf("--%s--%s--\n", hdr1.seqno, hdr1.msg);
        memset(buf, '\0', 64 );
   }