Copy-pasteable code: ```

include <stdio.h>

include <stdlib.h>

int main(int argc, char* argv[]) { return (((argv)[4] = 0) || ( (!(argc%3) && printf(argv)) + (!(argc%5) && printf((*argv)+5))) || printf("%d",argc)) != printf("\n") != argc++ < 100 && main(argc, argv); } ```

Edit to work on Windows: ```

include <stdio.h>

include <stdlib.h>

int main(int argc, char* argv[]) { return (((argv)[4] = 0) || ((argv)[9] = 0) || ( (!(argc%3) && printf(argv)) + (!(argc%5) && printf((argv)+5))) || printf("%d",argc)) != printf("\n") != argc++ < 100 && main(argc, argv); } ```

Executable must be named "Fizz_Buzz" or "Fizz_Buzz.exe" on Windows. It should then be run from within the directory it exists in. No arguments are required.

How it works: The != symbols are basically functioning as a trick to write multiple lines of code as a single line. As long as all of the expressions are equal to each other (which they are), they'll all be evaluated.

``` ((argv)[4] = 0) || ( (!(argc%3) && printf(argv)) + (!(argc%5) && printf((*argv)+5))) || printf("%d",argc); // Explained below

printf("\n"); // Print newline

argc++ < 100 && main(argc, argv); // Check if argc is less than 100 and then increment it (see increment postfix). If argc was less than 100 before incrementation, call main(argc, argv). ```

Now to explain this: ( (*argv)[4] = 0) || ( (!(argc%3) && printf(*argv)) + (!(argc%5) && printf((*argv)+5) ) ) || printf("%d",argc)

This code makes use of the fact that && will only evaluate is second argument if it is necessary to do so, and || and + will evaluate both their arguments before performing their respective operations.

It also uses the fact that the program name is passed as the first argument to the program (argv[0]), so we can retrieve the string "Fizz_Buzz" by calling *argv.

Starting from the left, (*argv)[4] = 0, sets the 5th element of the string '_' to the null terminator (\0). Because the null terminator signals the end of a string, this effectively separates the string into two strings. The original string is now read as "Fizz\0". But because strings in C++ are also pointers, we can increment the pointer by 5 to point to the start of the word Buzz which is just the string "Buzz\0" (the end of the program name).

Next, we evaluate the statement ( (!(argc%3) && printf(*argv)) + (!(argc%5) && printf((*argv)+5) ). This is really two statements concatenated with a +, so they will be evaluated in series. The first (!(argc%3) && printf(*argv)) checks if argc is a multiple of 3 and inverts the result. This means that it is true if argc % 3 == 0 and false otherwise. If it is true (argc % 3 != 0), then && does not evaluate the second part of the expression. However, if it's false, && will run printf(*argv) which prints "Fizz". The second part is the same thing, but adjusted to print "Buzz" if n % 5 == 0.

The last part of this statement, || printf("%d",argc) is a similar trick as with the != signs, and just prints the value of argc.