Assigning format specifier using pointer inside printf() function

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

#include<开发者_开发百科;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);}

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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).

---

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

---

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?