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Comment by nspattak | Hacker News Reader

I was so surprised by the first question that I actually tried it on my computer...

Am I the only one who tried compiling :

#include<stdio.h> #include<stdlib.h> int main(int argc, char *argv[]) { int i; int i=10; printf("i=%d\n", i); return 0; }

and got the redeclaration error I expected?

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The original article makes it clear that this is how things work with C globals, not locals. If you compile and run this program with gcc -Wall -pedantic x.c && ./a.out there will be no errors and it will emit i=10 as expected:

  int i;
  int i=10;
  #include<stdio.h>
  int main(int argc, char *argv[]) {
      printf("i=%d\n", i);
      return 0;
  }

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