Question
#include<stdio.h>
int main()
{
int fun(int);
int i = fun(10);
printf("%d\n", --i);
return 0;
}
int fun(int i)
{
return (i++);
}
What will be the output of the program?
#include<stdio.h>
int main()
{
int fun(int);
int i = fun(10);
printf("%d\n", --i);
return 0;
}
int fun(int i)
{
return (i++);
}
Answer: Option A
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Step 1: int fun(int); Here we declare the prototype of the function fun().
Step 2: int i = fun(10); The variable i is declared as an integer type and
the result of the fun(10) will be stored in the variable i.
Step 3: int fun(int i){ return (i++); } Inside the fun() we are returning a value
return(i++). It returns 10. because i++ is the post-increement operator.
Step 4: Then the control back to the main function and the value 10 is assigned
to variable i.
Step 5: printf("%dn", --i); Here --i denoted pre-increement. Hence it prints the value 9.
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