Question
#include<stdio.h>
int main ()
{
extern int a;
printf ("%d\n", a);
}
int a = 20;
What is the output of the program?
#include<stdio.h>
int main ()
{
extern int a;
printf ("%d\n", a);
}
int a = 20;
Answer: Option A
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extern int a; indicates that the variable a is defined elsewhere, usually in a
separate source code module.
printf("%d`setminus`n", a); it prints the value of local variable int a = 20. Because, whenever
there is a conflict between local variable and global variable, local variable gets the
highest priority. So it prints 20.
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