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exvor

12-09-2005 03:17 PM

C string and printf question

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

Don't believe everything you read...

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

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 :p

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.

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