Member Avatar for himanjim 
#include<stdio.h>
char *str="char*str=%c%c%c;main(){printf(str,34,str,34);}";
void main()
{
printf(str,34,str,34);
}

gives the O/P

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

Can't say how??
No formatting string is used in printf() like %s or so.

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  • Latest Post Latest Post by Ancient Dragon

Dani AI

Generated

This is a classic self-reproducing program (a quine). correctly identified it as such, and was right to point out that the first argument to printf is the format string — it does not have to be a literal. What can make the output look odd is a mismatch between format conversions and the actual argument types: using %c tells printf to fetch an int and print a single character. Passing a char * where an int is expected is undefined behavior, so different compilers or runs can produce different results.

A portable, easy-to-read way to write this kind of quine is to use %s for the embedded string and %c only for the quote characters. Use int main(void) and return 0 to be standard-conforming:

#include <stdio.h>

int main(void) {
    const char *s = "const char *s = %c%s%c; int main(void) { printf(s, 34, s, 34); return 0; }";
    printf(s, 34, s, 34);
    return 0;
}

Troubleshooting tips:

  • If you see strange symbols, remember it’s almost always undefined behavior from a type mismatch. %p prints pointer values if you need to inspect addresses.
  • Compile with warnings (gcc -Wall -Wextra -Wformat) to catch many format issues when the format string is a literal.
  • Never rely on matching a pointer to %c — either use %s for strings or %p to examine pointers.
    These points explain why the program’s output can vary and how to make a correct, portable version.

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Member Avatar for Ancient Dragon

The O/P you posted is wrong (the actual o/p may be slightly different from one computer to another because the second %c in the format string.)

char*str="∟";main(){printf(str,34,str,34);}

The formatting string is the first argument to printf() -- it does not have to be a string literal and can contain other characters besides %s, %c %f etc. printf() just displays the characters that are not followed immediately by '%' character.

Member Avatar for Salem

Congratulations on finding a Quine.

Just substiute for readability

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

The first one contains 3 % conversions for the 3 params which follow
The second one is just a string which happens to contain % characters
The 'magic' is that it's the same string.

Member Avatar for Ancient Dragon

Congratulations on finding a Quine.

Just substiute for readability

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

The first one contains 3 % conversions for the 3 params which follow
The second one is just a string which happens to contain % characters
The 'magic' is that it's the same string.

Sorry, but I don't see any magic in it at all. The 3d argument is just used by printf() as an address of a string displayed as a character not as a string. He could have substituted "Hello World" there and the o/p would have been similar.

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