2-3 questions .

tushar_pandey · 07-25-2012, 10:08 PM

what is the difference among

printf("gcchen");
&
printf("hello""NevemTeve");
&
printf("pan64","NevemTeve");

and

"its my thinking that , printf works on base address" if i am correct than please say ::: yes/no !
now i tell you why i am thinking this , because printf(2+"hello");
so here printf() is fetching the base address of "hello" string and adding "+2" means 8 bytes in it !
and the output is :: llo

pixellany · 07-25-2012, 10:24 PM

What language are you working in?

tushar_pandey · 07-25-2012, 10:31 PM

i am working in c ,

pixellany · 07-25-2012, 10:36 PM

Is this legal syntax in C?:

Code:

printf(2+"hello")

pixellany · 07-25-2012, 10:42 PM

I think the C printf format is:

Code:

printf("<format string>", "<string to print>")
OR
printf("<format string>", <variable>)

None of your examples match that format

tushar_pandey · 07-25-2012, 10:42 PM

@pixellany , please watch this

#include<stdio.h>

int main()
{

printf(2+"hello");

printf("\n");
return 0 ;
}

daa@daa-Aspire-5740:~/c$ ./a.out
llo

pixellany · 07-25-2012, 10:48 PM

Interesting.....do you have a reference that discusses that syntax and why it would be used?

It does appear that it takes the base address of the string and adds 2 to it......Have you tried 1+, 3+, etc.?

tushar_pandey · 07-25-2012, 11:02 PM

Quote:

Originally Posted by pixellany

Interesting.....do you have a reference that discusses that syntax and why it would be used?

i am solving a question , which is present in my book "test your c skills , yahswant kanitkar" . but there is no solution for it .

and if i find the meaning of " printf("hello") , the working behind this statement , than it is easy !

NevemTeve · 07-25-2012, 11:20 PM

> printf("gcchen");

Perfectly legal

> printf("hello""NevemTeve");

Perfectly legal, the two string literal concatenated by the compiler.

> printf("pan64","NevemTeve");

Perfectly legal, but compilation warning may arise: there is no %-sequences in format string, so the second argument is ignored.

> printf(2+"hello");

Perfectly legal, can be written as: printf("hello"+2);
or printf("llo");

or:
const char hellostr[]="hello";
printf (hellostr+2);

or
const char hellostr[]="hello";
printf (&hellostr[2]);

tushar_pandey · 07-25-2012, 11:37 PM

Sorry , but how can you say that printf("gcchen"); is perfectly legal , and if it is perfectly legal than why this statement does not need the format specifier !

because in

> printf("pan64","NevemTeve");

Perfectly legal, but compilation warning may arise: there is no %-sequences in format string, so the second argument is ignored.

this statement it needs format specifier .

tushar_pandey · 07-26-2012, 12:12 AM

Sir , "Sorry , but how can you say that printf("gcchen"); is perfectly legal , and if it is perfectly legal than why this statement does not need the format specifier !" // this line is clear , after reading this syntax "String that contains the text to be written to stdout . It can optionally contain embedded format tags that are substituted by the values specified in subsequent argument(s) and formatted as requested." from "http://v2.cplusplus.com/reference/cl...cstdio/printf/"

but why this is illegal .... printf("hello","nevemteve");

SIG_SEGV · 07-26-2012, 01:51 AM

Hello tusshar,
check out this pgm.... find out printf's working..........

#include <stdio.h>
int main ()
{
printf ("\ngcchen");
printf ("\ngcc%xhen"); // will get junk value o/p for %x here
printf("\npan64","NevemTeve");
printf("\npan64%s","NevemTeve");
}

pan64 · 07-26-2012, 03:09 AM

Quote:

Originally Posted by tushar_pandey

but why this is illegal .... printf("hello","nevemteve");

No, it is not illegal, it works, just the second argument ("nevemteve") will not be used at all, will be ignored.
You can also try: printf("hello", 1,2,54,7, "fe","ignored", "some more", "anything") and any number of arguments are legal, but will be ignored.