C Programming - Control Instructions - Discussion

Here the ternary operator is used so that it will need the left-hand side value like in the form of ternary syntax.

y==1?n=0:n=1; <=> (y==1?n=0:n)=1;

Here, within parenthesis is an expression so, we can't assign value in an expression.

Expression? True(1):false(0)
In false condition it's always value is 0.
We don't modify it,

If can't assign the value n=1 in false condition.

The syntax y == 1 ? n=0 : n=1; should be written as y==1?0:1;.

Yes, @Sandya is right when run these code in GCC it also gives "NO" as a output and if you assign n=1 in if than it gives "YES" as a output so before If there is no bug.

In grouping is required otherwise error as Lvalue required.

y == ? (n=0):(n=1);

Here you will not get error friends.

#include<stdio.h>
int main()
{
int n = 0, y = 1;
y == 1 ? n=0 : n=1;
if(n)
printf("Yes\n");
else
printf("No\n");
return 0;
}

Compile and run the following code in Turbo C or in GCC.
The conditional operator ?: has following form.
(condition)?true part : false part.

Thus, the value of n will be 0 not 1.

Thank you kiran.

In turbo c this prog gives error: "L-value required".

While in dev c++ its executes successfully and gives the o/p as: No

When the compiler reaches the second n in
y == 1 ? n = 0 : n = 1;
It is convinced that the conditional expression is completely scanned. This expression evaluates to an R-value after which the compiler encounters an assignment operator (=). So essentially the compiler sees the above statement as
(y == 1 ? n = 0 : n) = 1;.
And the expression inside the brackets evaluates to an R-value. Since we can not place an R-value before the assignment operator (=), the compiler generates the error message "L-value required".