C Programming - Expressions - Discussion
Discussion Forum : Expressions - Find Output of Program (Q.No. 9)
9.
What will be the output of the program?
#include<stdio.h>
int main()
{
int i=3;
i = i++;
printf("%d\n", i);
return 0;
}
Discussion:
70 comments Page 2 of 7.
Meghana said:
1 decade ago
@Anjena the answer will be:.
7 9 5 7 (evaluate from right to left).
7 9 5 7 (evaluate from right to left).
Srini said:
1 decade ago
Step 1: int i=3
Step 2: i=i++ post increment so still i=3 in this statement
Step 3: after completion of the above statement i will be increment by 1 so i becomes 4
step 4: in printf statement i value is printed as 4
Step 2: i=i++ post increment so still i=3 in this statement
Step 3: after completion of the above statement i will be increment by 1 so i becomes 4
step 4: in printf statement i value is printed as 4
Arnab Sanyal (IIT, BOMBAY) said:
1 decade ago
i is initialised to 3. We know that every variable can hold one value at a time. So only after assigning 3, with post increment operator i is assigned to 4.
Edwin Jose said:
1 decade ago
In post increment i value get incremented in the next step only.
Govindaraju said:
1 decade ago
i=i++; => i =3;
After executing this statement the i value is i=i.
+1 => 4.
As it is post decrement operator the value will be changed after the execution of the statement.
After executing this statement the i value is i=i.
+1 => 4.
As it is post decrement operator the value will be changed after the execution of the statement.
Dixit said:
1 decade ago
First i = 3 will be assigned to i (i = i++).
Then i will be incremented to 4.
Then i will be incremented to 4.
OmPrakash said:
1 decade ago
Has anyone compile the code? Do it and you will find '3' is the answer.
Avinash said:
1 decade ago
Answer = 4 is appropriate among the given options.
To say accurately the answer should be "undefined".
i=i++;// the behavoiur of this expresiion is undefined.
In an expression never use value of a variable if the value of the variable is getting changed..
Similarly,
i=i++*i++;//undefined b/c trying to modify twice with in an expression.
i=i*i++;//undefined.
To say accurately the answer should be "undefined".
i=i++;// the behavoiur of this expresiion is undefined.
In an expression never use value of a variable if the value of the variable is getting changed..
Similarly,
i=i++*i++;//undefined b/c trying to modify twice with in an expression.
i=i*i++;//undefined.
Himanshu said:
1 decade ago
Actually precedence of post increment operator (++) is greater than = operator.
So, i=i++; first increment occurs and then assignment occurs.
So, i=i++; first increment occurs and then assignment occurs.
Siraj said:
1 decade ago
#include<stdio.h>
int main()
{
int i=3,j;
j = i++;
printf("%d %d\n", i,j);
return 0;
}
What will be the output now?
int main()
{
int i=3,j;
j = i++;
printf("%d %d\n", i,j);
return 0;
}
What will be the output now?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers