array subscript operator in for loop counter

Question

I am learning C. In the following code why does replacing "*ptr_str" with "ptr_str[i]" in the for loop truncate?

/* 13L01.c: Initializing strings */
#include <stdio.h>
main()
{
 char str1[] = {'A', ' ',
 's', 't', 'r', 'i', 'n', 'g', ' ',
 'c', 'o', 'n', 's', 't', 'a', 'n', 't', '\0'};
  char str2[] = "Another string constant";
  char *ptr_str;
  int i;

  /* print out str2 */
  for (i=0; str1[i]; i++)
      printf("%c", str1[i]);
      printf("\n");
  /* print out str2 */
  for (i=0; str2[i]; i++)
      printf("%c", str2[i]);
      printf("\n");
   /* assign a string to a pointer */
   ptr_str = "Assign a strings to a pointer.";
   for (i=0; *ptr_str; i++)
       printf("%c", *ptr_str++);
   return 0;

}

Bill Forster · Accepted Answer · 2010-11-13 00:26:41Z

2

   // A, ok
   while (*ptr_str)
       printf("%c", *ptr_str++);

   // B, also ok
   for (i=0; ptr_str[i]; i++)
       printf("%c", ptr_str[i]);

   // C, works but ugly
   for (i=0; *ptr_str; i++)
       printf("%c", *ptr_str++);

C is your form, it is flawed because i is doing nothing here, so A is an improved version. If you want to use i, do it like B. If you use both i and ptr_str, and increment both of them in the loop, nothing good will happen. Increment one or the other.

answered Nov 13, 2010 at 0:26

Bill Forster

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5 Comments

Chris Lutz Over a year ago

I answered, then reread your answer and figured out that you had already answered it, just with fewer words. So +1

Bill Forster Over a year ago

@gmiket Truncation is absolutely not a necessary consequence of prt_str[i] as the test in the for loop. See my B) snippet, which works fine.

Bill Forster Over a year ago

@Chris, thanks your answer is good too, so +1 for you from me too!

gmiket Over a year ago

// C, works but ugly for (i=0; *ptr_str; i++) printf("%c", *ptr_str++;); seems to me a pointless construction. Why not just for (; *ptr_str;) printf("%c", *ptr_str++);

Bill Forster Over a year ago

@gmiket I agree it's a pointless contruction, I only mentioned it because that is the code you posted !!!! Your suggestion of for (; *ptr_str;) printf("%c", *ptr_str++); is just a slightly uglier version of my while (*ptr_str) printf("%c", *ptr_str++); Both are fine and will work.

kichik · Accepted Answer · 2013-01-20 23:49:12Z

2

Because you're advancing ptr_str and then treating it as an array and testing if it's pointing to NULL on the i'th member. You're basically testing if ptr_str[i+i] is NULL instead of ptr_str[i].

edited Jan 20, 2013 at 23:49

answered Nov 13, 2010 at 0:09

kichik

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4 Comments

Chris Lutz Over a year ago

ASCII NUL ('\0') is not NULL ((void *)0).

Jens Gustedt Over a year ago

@Chris: NULL is not necessarily (void*)0 it may be 0 or 0L or equivalent even in plain C.

Chris Lutz Over a year ago

@Jens - I know, but @kichik wants the NUL character, which is '\0'. Even if using NULL might compile and work, it's still important to differentiate the two (especially to beginners).

Jens Gustedt Over a year ago

@Chris: Sure. Just wanted to state the "real" value of NULL, namely that there isn't :) at least nothing reliable.

Chris Lutz · Accepted Answer · 2010-11-13 01:08:57Z

Your question is a bit unclear, but I think I've got it:

In the following code why does replacing "*ptr_str" with "ptr_str[i]" in the for loop truncate?

I think you mean changing this:

for (i=0; *ptr_str; i++)
    printf("%c", *ptr_str++);
return 0;

to this:

for (i=0; ptr_str[i]; i++)
    printf("%c", *ptr_str++);
return 0;

The second one truncates because you're advancing i and ptr_str, so the modified starting position of ptr_str plus the modified starting position of i ends up cutting you off too soon (or worse, having an odd number of characters and overflowing into data that isn't yours). The second example that truncates is equivalent to:

for (i=0; ptr_str[i * 2]; i++)
    printf("%c", ptr_str[i]);
return 0;

Now do you see why it truncates?

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