Any idea why the code snippet below prints 2 and not 3 [duplicate]

Question

This question already has answers here:

Post increment and Pre increment in C (3 answers)

Closed 8 years ago.

I was expecting this code snippet to print 3 as the if condition is false and y++ will execute, but it prints 2, any idea why..Thanks!!

int x = 3;
int y = 2;

printf("%d\n", (x<y) ? x++ : y++);

// This prints 2

Answer 1

It's because y++ returns the value of y then increases it.

Where as if you put ++y it would increase the value of y first then return it.

Answer 2

x++ and y++ are post-increment. That is, they return the current value of the variable, then add one to it.

Pre-increment would be ++x or ++y. Those increment and then return the new value.

Both pre- and post-increment (and -decrement) are useful things when writing loop controls, which is why C supports both.

(Originally, if I remember correctly, C only supported pre-increment and post-decrement, because there happened to be instructions on the machines it was developed on which encapsulated those behaviors. But as C moved to other systems, and as people started noticing that they wanted both pre and post for both, this was generalized.)

Note that this means the c++ language was misnamed. It should have been called ++c -- we want it improved before we use it, not after!

Answer 3

The ++ operators are evaluated last; this is called "post-increment." So, this:

int x = 3;
int y = 2;

printf("%d\n", (x<y) ? x++ : y++);

is equivalent to this:

int x = 3;
int y = 2;

printf("%d\n", (x<y) ? x : y);
y++;

(x++ isn't ever reached because of the ternary condition.) On the other hand, this:

int x = 3;
int y = 2;

printf("%d\n", (x<y) ? ++x : ++y);

would increment y before returning its respective value to printf(), so the logic would be:

printf("%d\n", (3<2) ? 3 : 3); // prints 3

Answer 4

Since you use a post increment y++, value of y will be used first and incremented. That is the printf will be passed the value before increment operation, in your case y is 2 before increment and 2 will be printed.

Answer 5

You should consider x++ and y++ after all the other operations of this line have been completed. So, print y and then increment x and y.