printf() vs puts() behavior

When I do the following

m-net% more mod.c

Code:

---

#include <stdio.h>

int main (void)
{
  char str[] = "Aamit";
  *str='R';
  puts(str);
  return 0;
}

---

The output is fine.

Code:

---

m-net% gcc -g -Wall mod.c -o mod
m-net% ./mod
Ramit

---

However, when I replace puts() with printf()

m-net%m-net% more mod.c

Code:

---

#include <stdio.h>

int main (void)
{
  char str[] = "Aamit";
  *str='R';
  printf("%s \n " ,str);
  return 0;
}

---

I get some junk afterwards.

Code:

---

m-net% gcc -g -Wall mod.c -o mod
m-net% ./mod
Ramit
 %                                                                             
m-net%

---

Whis is this?

*str == str[0]
Pointers point to the first element of an array. So when you dereferenced str you were assigning 'R' as the first element of the array.

printf prints all the characters in an array until it reaches the \0 character. You need to null out the rest of the characters in the array if you do not want them to print.

Then how comes this doesn't happen with puts()?

So the real problem is the percent prompt at the end yes?

Well I'm not sure it's entirely printf's fault, but it does have a different job then puts. puts doesn't flush stdout after printing, so it could be that you are only seeing this after the program has closed and the buffers have been flushed in the first example.

Code:

---

>  printf("%s \n " ,str);

---

Try removing the space after the newline in the above printf(), and see if you still have the problem.

What is your prompt set to?

Quote:

---

Originally Posted by gwarf420

printf prints all the characters in an array until it reaches the \0 character. You need to null out the rest of the characters in the array if you do not want them to print.

---

The string is already null terminated... The *str line doesn't change that.

> Then how comes this doesn't happen with puts()?
Because puts() puts a new line on the end for you...

Swoopy has already answered your question IMO.