Assigning format specifier using pointer inside printf() function

Question

I could not understand how the below program code outputs that value.Please help me to understand.

#include<stdio.h>
char*s="char*s=%c%s%c;main(){printf(s,34,s,34);}";
int main()
{
        printf(s,34,s,34);
        return 0;
}

output:

char*s="char*s=%c%s%c;main(){printf(s,34,s,34);}";main(){printf(s,34,s,34);}

Sorry this is not a macro function. I was going through bunch of these questions and have given the heading wrongly. — Angus
– Angus, Commented Aug 16, 2011 at 16:59
Thanks for making me to understand about how the output got arrrived. — Angus
– Angus, Commented Aug 16, 2011 at 17:22

Mark Wilkins · Accepted Answer · 2011-08-16 16:59:44Z

2

It isn't actually a macro in use here. It is just a simple call to printf. The first parameter to printf is the format string. In this case it is the value defined in the global variable s. The format characters %c%s%c are supplied by parameters 34,s,34". So the string is just printed in its entirety because of the %s format character. And the two 34 values are printed as double quote characters (via %c).

answered Aug 16, 2011 at 16:59

Mark Wilkins

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Comments

Oliver Charlesworth · Accepted Answer · 2011-08-16 17:04:21Z

1

Your printf statement is effectively equivalent to:

printf("char*s=%c%s%c;main(){printf(s,34,s,34);}", 34, s, 34);
               ^ ^ ^

I've marked the conversion specifiers with ^. These get replaced with, respectively:

" - the ASCII character corresponding to 34
the contents of *s
" - the ASCII character corresponding to 34

edited Aug 16, 2011 at 17:04

answered Aug 16, 2011 at 16:58

Oliver Charlesworth

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Comments

JohnBrighton · Accepted Answer · 2011-08-16 16:58:00Z

0

You are inserting as an argument your format string, so the output is correct. The %s is replaced with the actual format. PS: Where are macros?

answered Aug 16, 2011 at 16:58

JohnBrighton

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1 Comment

JohnBrighton Over a year ago

Correct, I forgot to write it.

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