C / C++ / MFC

int i = 10;
int result;
result = i++;  // result equals 10, i equals 11
result = ++i;  // result equals 12, i equals 12
result = i--;  // result equals 12, i equals 11
result = --i;  // result equals 10, i equals 10

// note do not use expressions such as
result = ++i + i++; // results undefined

Richard MacCutchan wrote:

but it is still worth avoiding expressions of that sort.

Marco Bertschi wrote:

my comment was intended to be an addition to yours.

sukanta kumar mangal wrote:

As written the last line gives output undefine i think its wrong .it will give correct output as 23.
If u had checked then u already got the result.
here i will explan u how it's work..

i = 10;
i = i + 1; // pre-increment of the first argument
first_arg = i;
second_arg = i;
i = i + 1; // post-increment of the second argument
result = first_arg + second_arg; // 11 + 11

i = 10;
second_arg = i;
i = i + 1; // post-increment of second argument
i = i + 1; // pre-increment of first argument
first_arg = 1;
result = first_arg + second_arg; // 12 + 10

int i = 0;

i++;

++i;

i = 10;
x = ++i + 5;

i = 10;
i = i + 1;
x = i + 5;

i = 10;
x = --i + 5;

i = 10;
i = i - 1;
x = i + 5;

i = 10;
x = i++ + 5;

i = 10;
x = i + 5;
i = i + 1;

i = 10;
x = i-- + 5;

i = 10;
x = i + 5;
i = i - 1;

i = 10;
x = ++i + i++;

i = 10;
i = i + 1;
x = i + i;
i = i + 1;

i = 10;
i1 = i;
i = i + 1;
x = i1 + i;
i = i + 1;

i = 10;
i1 = i;
i = i + 1;
x = i + i1;
i = i + 1;