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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

MathBoy
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    Read about post-increment and pre-increment. – devnull Feb 20 '14 at 02:27
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    `x < y` is false, so `y++` is selected. `y` is pushed onto the stack for `printf` before `y` is incremented with the `++` since it's a post-increment. *Voila!* A 2! :) – lurker Feb 20 '14 at 02:28

5 Answers5

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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!

keshlam
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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.

PomfCaster
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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
elixenide
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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.

Karthik Kalyanasundaram
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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.

happyWinner
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