005 # puzzles on c

First of all, this program will not compile... WHY? Let me explain... See the statement: x = 10 && y = 6 The precedence of ‘&&’ is more than assignment… so it would become like “x = (10 && y) = 6” Now, “10 && y” is actually a constant.. so it would become like “x = (const number) = 6; Here… “const number = 6” is an error known as ‘L-value’ error. I hope this is clear now… Again, let me try to correct you, may be u want to write your program in this way:- void main() { int x=10,y=5,z; z= (x=10 && (y=6))?100:200; printf("\n z=%d ",z); } Now .. It will compile and result would come ‘z = 100’ because x = 10 is a true condition and y=6 is a assignment of 6 in y which becomes true. So … z = true ? 100 : 200,,,, very clear that the value of z would become 100 here.

Hi dholey, output is simple. tell me whether i am right. output: z=100 Reason: 1. As x=10 so x=10 condition is true. 2. y=6 is a nonzero value even it is true. 3. So condition of ternary operator is true. Hence 100 is returned and assigned to z 4. finally we get x=10 and y=6 and z=100

hi dholey me back aftr a long time u nd tmahes.... doin extremely well. i dont know much abt programming but these discussions doing a great job for beginners

sorry dholey i am not a student of C language i don't know its output please tell me its out put .. can you teach me C language? send me private message

If the program is executed as given above it will lead to an error since it is y=6 and it should be y==6. So the correct program should be int main() { int x=10,y=5,z; z= (x=10 && y==6)?100:200; // changed y=6 to y==6 printf("\n z=%d ",z); return 0; } The above program would output z=200 since the condition y==6 is false so 200 would be stored to z.

z=200