C Programming - Functions - Discussion
#include<stdio.h>
int i;
int fun1(int);
int fun2(int);
int main()
{
extern int j;
int i=3;
fun1(i);
printf("%d,", i);
fun2(i);
printf("%d", i);
return 0;
}
int fun1(int j)
{
printf("%d,", ++j);
return 0;
}
int fun2(int i)
{
printf("%d,", ++i);
return 0;
}
int j=1;
Step 1: int i; The variable i is declared as an global and integer type.
Step 2: int fun1(int); This prototype tells the compiler that the fun1() accepts the one integer parameter and returns the integer value.
Step 3: int fun2(int); This prototype tells the compiler that the fun2() accepts the one integer parameter and returns the integer value.
Step 4: extern int j; Inside the main function, the extern variable j is declared and defined in another source file.
Step 5: int i=3; The local variable i is defines as an integer type and initialized to 3.
Step 6: fun1(i); The fun1(i) increements the given value of variable i prints it. Here fun1(i) becomes fun1(3) hence it prints '4' then the control is given back to the main function.
Step 7: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.
Step 8: fun2(i); The fun2(i) increements the given value of variable i prints it. Here fun2(i) becomes fun2(3) hence it prints '4' then the control is given back to the main function.
Step 9: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.
Hence the output is "4 3 4 3".
As you said int j is global but value of i is given so how can we assume it to be the value of j? Neither we have any pointer associated. Please tell me.
According to you, the value of j is discarded but here int fun1(int j).
For int j how can we take the value from another variable i? Can you explain?
In the fun1 and fun2 the value of 3 is pre-incremented and then displayed but the value of "i" in main remains 3 because the value of i = 3 and its scope is local.