Code explanation

[sidrules1984]

int a=10,b;
b=(a>=5)?b=100:b=200;
printf("%d",b)

why does the output shows as 200? Plz explain

[Narue]

That code shouldn't compile at all, much less output the wrong value. Try this program and see if you get the same erroneous output:

#include <stdio.h>

int main(void)
{
    int a = 10, b;

    b = (a >= 5) ? 100 : 200;

    printf("%d\n", b);

    return 0;
}

[sidrules1984]

this program does compile. but i dont understand what's wrong with the code. Your program gives 100

[Narue]

this program does compile

What compiler are you using? It's either broken or you're relying on an extension.

but i dont understand what's wrong with the code.

It's a syntax error, and your compiler is incorrect in allowing it. But in allowing it, the expression is being parsed like so:

b = ((a >= 5) ? (b = 100) : b) = 200;

a is greater than 5, so b is set to 100, then the ternary expression evaluates to b (the result of b = 100 ), which is then assigned a value of 200. Thus the end result of the expression is that b is set to 200.

However, the ternary operator does not result in an lvalue, so assigning 200 to it is illegal. That's why your compiler is doing the wrong thing.