Why printf runs into error with n format specifier?

Question

I have some C code with a %n printf:

#include <stdio.h>
int main(){
    int i, j;
    i = printf( "something%n", &j );
    return i + j; 
}

There is no error in compilation and execution. printf prints the string "something".

Why my printf returns -1 and why it doesn't save the n parameter in j?

Here a photo while debugging..

Answer 1

If you are using the Microsoft C compiler and running your program on Windows, then by default, using %n with printf will indeed fail. This is because Microsoft views %n as a security risk and disables it by default in printf and related print formatting functions.

To enable %n add the following line early on in your program:

_set_printf_count_output(1);

This is documented in the important note on %n in the following help article:

Format specification syntax: printf and wprintf functions https://docs.microsoft.com/en-us/cpp/c-runtime-library/format-specification-syntax-printf-and-wprintf-functions

Answer 2

I compiled your code and verified, you can check it here: https://onlinegdb.com/BJIF5EUOI

Result:

Seems it works fine according to documentation ( http://www.cplusplus.com/reference/cstdio/printf/ )

printf returned total number of characters written, result stored at i variable

%n stored number of characters written so far under j

main() returned i + j - so 2x total number of characters written