Discussion
What will be the output of the following code?
#include <stdio.h>
int main()
{
extern int a;
static char j = 'E';
printf("%c %d", ++j, ++a);
return 0;
}
Answer: B
Correct answer: F 1
++j takes the static char from 'E' to 'F', which %c prints as the letter. extern int a; declares a variable defined at file scope elsewhere; such an object has static storage duration and is therefore zero-initialised, so ++a gives 1. Output: F 1.
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