C string and printf question

exvor · 12-09-2005, 03:17 PM

Ok in my book here it says doing the following would be bad

Code:

 
 
char test[] =" test string"; 
  
 printf("%c\n",test[5]);

I would suppose you would have to do the following

Code:

 

char test[] =" test string"; 

   char gdb; 
   
   gdb = test[5]; 
   printf("%c\n",gdb);

now is the latter correct because when I tested this it
had the same result both ways and GCC gave no erros
with -Wall

of course this isent compleate code but its kinda a
simple question. If the first code is incorrect can you give me a reason why it would be bad to do that as well.
Sorry just trying to understand.

graemef · 12-09-2005, 04:01 PM

Not quite certain what the author was getting at. Maybe it was because of the potential confusion between a character data type '%c' and the variable being a string. This could lead to confusion when revisiting the code.

What was the chapter of the book covering and was there anything else to explain why it was 'bad'?

graeme.

paulsm4 · 12-09-2005, 04:20 PM

If that's really what the author said, I'd consider getting a refund ;-)

dmail · 12-09-2005, 07:51 PM

Quote:

Originally Posted by graemef

Not quite certain what the author was getting at. Maybe it was because of the potential confusion between a character data type '%c' and the variable being a string. This could lead to confusion when revisiting the code.
graeme.

This is what i think.
"%c" is a character where as you want a string of characters "%s".
O no you dont lol. your just using one char, so i dont really know.;(

exvor · 12-10-2005, 08:19 AM

Its a book by H.M. Deitel/P.J. Deitel C how to program second edition.

lol no refund needed book was only $4

its an older text book that i picked up somewhere in lack of being able to find anything else

anway im in the section for Formated input/output and its one of the commen programming errors that they spek of

Quote:

Commen programming error 9.3
Using %c to print the first character of a string. The conversion specification %c expects a char argument. A string is a pointer to char. i.e.. a char *.

in looking at that maybe there talking about wierdness such as doing this.

Code:

 
   char test[] ="this is a test"; 

   printf("%c\n",test);

in that case i would see why it would be wrong but what i was doing earlier as being perfectly ok.

graemef · 12-10-2005, 08:44 AM

Yes the way you coded it was correct. Omitting the element as in

Code:

printf("%c\n",test);

would have been wrong and I think that is the common error they are talking about. You don't end up printing the first elemnt of the array but part of the address.

graeme.

arunvk · 12-10-2005, 09:52 AM

i guess the second is recommended because its easier to debug a program.

and also for the sake of readability i guess. there r many other reasons but i think this is the most logical explaination

Lsatenstein · 12-10-2005, 10:53 AM

The message about being wrong is a warning. That warning is based on human error.
Suppose you had a define
MYTEST = "test string"

and in the code you had
char test[] = MYTEST;

Your printf would work ok, as YOU know that MYTEST is long enough.
But suppose, at 1am in the morning, in a rush to get a fix out, you had changed
char test[] = "test";

Where would your expression point to?

printf("%c\n",test[5]);

To avoid errors, you can do what you want to do, but do lint your program before moving it to production. Lint is a wonderful detecter of pitfalls.