Question

explain output of

int x = 4;
x = x + x + x++;
printf("%d\n",x);
x = 4;
printf("%d\n",x + x + x++);

output of above code is
13
12

Answer 1

x++ means:
first read the current value of x and then increment its value.

so in the statement:
x = x + x + x++

what happens is as follows:
x = 4 + 4 + 4 = 12

then the "post increment" on x is applied (i.e. x++) resulting in x getting the value 13

In the second statement:
printf("%d\n",x + x + x++);

the same thing happens, except this time the result of "x + x + x++" is printed out first and then the current value of x is incremented - which in this case would mean x would end up with a value of 5.

Answer 2

it doesn't work on
#define p(x) printf("%d\n",x);
int x = 4;
x = x + x++ + x;
p(x);
x = 4;
p(x + x++ + x);


same output
13
12

Answer 3

what do you mean by it doesn't work?

Answer 4

according to ur explanation first + is evaluated then increment but
post increment has higher precedence than normal addition and assignment operator so it should be evaluated first

Answer 5

post increment always occurs AFTER the value has been read and processed

Answer 6

thanks i get it.

Answer 7

yw :)