The post increment operator

[Jayanth1987]

I have a basic doubt...
When we use the post incrment operator the incremnet of the
variable is done after the execution of the statemnt?????
for eg
int i=1;
printf("%d %d",i++,i++);

shud the answer be 1 1
or 1 2

i want to understand the execution pt

[horace1]

I have a basic doubt...
When we use the post incrment operator the incremnet of the
variable is done after the execution of the statemnt?????
for eg
int i=1;
printf("%d %d",i++,i++);

shud the answer be 1 1
or 1 2

i want to understand the execution pt

the problem is that you don't know when the ++ operations are carried out (it is side effect) - they may done immediatly after the operand is used or saved until the end of the statement. For example, gcc and Borland CBuilder V5 gave
2 1

whereas Visual C 5.0 gave
1 1

in all cases the final value of i was 3

in general the advice is do not use a variable more than once in an expression if one (or more) of the references has a ++ or -- attached to it.

[willakawill]

I have a basic doubt...
When we use the post incrment operator the incremnet of the
variable is done after the execution of the statemnt?????
for eg
int i=1;
printf("%d %d",i++,i++);

shud the answer be 1 1
or 1 2

i want to understand the execution pt

Hi. I am getting more and more sensitive to the possibility that these questions are taken from college studies.

It is only necessary to run this code to discover the output. You can run it several times to follow how the output changes. You can insert a pre-increment, ++i, to see how that effects the order of things.

Test it and see. That is the fastest way to learn.

[Sebouh]

I'm surprised Visual C gave 1 1. I think it should be 1 2, cause this statement is broken down when the compiler generates the object file. So it's like saying:
print i : 1
i = i + 1
print i : 2
i = i + 1

[horace1]

I'm surprised Visual C gave 1 1. I think it should be 1 2, cause this statement is broken down when the compiler generates the object file. So it's like saying:
print i : 1
i = i + 1
print i : 2
i = i + 1

try it with different compilers and see what you get but remember the order of evaluation of operands of individual operators and the order in which side effects take place is unspecified in C / C++

see the last line of

http://www.cppreference.com/operator_precedence.html

[willakawill]

try it with different compilers and see what you get but remember the order of evaluation of operands of individual operators and the order in which side effects take place is unspecified in C / C++

see the last line of
http://www.cppreference.com/operator_precedence.html

Right. And vc++ 6 seems to process the parameters for printf in reverse order, i.e. right to left which is another consideration when working out the result of this syntax.