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Challenges 1, 2, Fizz, 4, Buzz!

C (gcc), 113 Bytes i;main(){while(i++<100){char*h[]={"%d "," "},**p=h;i%3||(*p++="Fizz%2$s");i%5||(*p="Buzz ");printf(*h,i,h[1]);}} This compiles with several warnings, but no errors. Here'...

posted 4y ago by celtschk‭  ·  edited 4y ago by celtschk‭

Answer
#2: Post edited by user avatar celtschk‭ · 2020-11-20T14:18:54Z (almost 4 years ago)
Added ungolfed version of the code with explanation in the comments
  • # C (gcc), 113 Bytes
  • ```
  • i;main(){while(i++<100){char*h[]={"%d "," "},**p=h;i%3||(*p++="Fizz%2$s");i%5||(*p="Buzz ");printf(*h,i,h[1]);}}
  • ```
  • This compiles with several warnings, but no errors.
  • # C (gcc), 113 Bytes
  • ```
  • i;main(){while(i++<100){char*h[]={"%d "," "},**p=h;i%3||(*p++="Fizz%2$s");i%5||(*p="Buzz ");printf(*h,i,h[1]);}}
  • ```
  • This compiles with several warnings, but no errors.
  • Here's an ungolfed version with explanations:
  • ```
  • /* the following include is omitted; while using printf
  • without including it is not conforming to the current
  • C standard, with gcc it only generates a warning
  • and works flawlessly. Note that omission was valid K&R C */
  • #include <stdio.h>
  • /* Declare a global int variable. The golfed code omits the
  • type because of the old implicit-int rule, which only
  • generates a warning also in modern gcc. Also note that
  • global variables without initializer are zero-initialized;
  • this is still defined behaviour today. In this ungolfed
  • code I've added the implied initializer for clarity. */
  • int i=0;
  • /* The main function. Again, the golfed code makes use of the
  • old implicit int rule. */
  • int main()
  • {
  • /* loop up to 100; since we increment before entering the
  • loop body, the first number in the loop is 1. */
  • while(i++<100)
  • {
  • /* In this array of two strings, the first will be used as
  • format string to a printf later, while the second will
  • be the third argument for the same printf. As is, the
  • format string will tell printf to format the second
  • argument (which will be i) as number, followed by a space.
  • The third argument, a single-space string, is ignored
  • in that case. */
  • char *h[] = { "%d ", " " };
  • /* This second variable (which in the golfed version is
  • declared in the same declaration as h) points to the first
  • entry of h. It is used to change the entries of h, and
  • it is separate because pointer arithmetic is used to
  • modify different strings depending on the condition. */
  • char **p = h;
  • /* This is basically an if-not statement. If i%3 is not
  • nonzero (that is, if i is divisible by 3), this statement
  • replaces the format string with one that prints "Fizz"
  • followed by the content of the third argument of printf
  • as string, which is the second string in h. At this
  • point it contains just a space. Also, when replacing the
  • format string, the pointer p is incremented, so it then
  • points to the second element of h. */
  • i%3 || (*p++ = "Fizz%2$s");
  • /* This if-in-disguise handles divisibility by 5. It stores
  • a pointer to "Buzz " into whatever p points to. If i
  • is not divisible by 3, this still is the format string,
  • so if instructs the following printf to just print "Buzz "
  • and ignore any further arguments. Otherwise, p points
  • to the final string argument, so that after the "Fizz"
  • it prints "Buzz " instead of just a space. */
  • i%5 || (*p = "Buzz ");
  • /* This finally is the printf statement talked about
  • in the previous comments. Note that *h is equivalent
  • to h[0] */
  • printf(*h,i,h[1]);
  • }
  • }
  • ```
#1: Initial revision by user avatar celtschk‭ · 2020-11-19T19:40:40Z (almost 4 years ago)
# C (gcc), 113 Bytes
```
i;main(){while(i++<100){char*h[]={"%d "," "},**p=h;i%3||(*p++="Fizz%2$s");i%5||(*p="Buzz ");printf(*h,i,h[1]);}}
```
This compiles with several warnings, but no errors.