Thread: Evaluation of arithmetic expression that involves increment operators

Here is a simple code to evaulate an arithmetic expression :

a=(++a)+(++a)+(a++)+(a++) where a is an integer . Taking an initial value of a=10,we have,

a= (++10)+(++a)+(a++)+(a++)
= 11+(++11)+(a++)+(a++)
= 11+12 +(12++)+(a++)
= 11+12+12+(13++)
= 11+12+12+13
a= 48

A simple Code executed using Turbo C++3.0 on Windows OS gives an output of 50 for the same.

CODE:

Code:

#include<stdio.h>
void main()
{
int a=10;
a=(++a)+(++a)+(a++)+(a++);

printf("%d",a);

}

As already said when this code is executed the value of a is printed as 50.

Thus i am most certain that i have gone wrong in evaluating the expression myself as i have got 48 as the answer. Can anybody identify this error and tell me how this expression gets evaulated?

Thanks in advance.

you shouldnt try to work out this problem because your code is just poor and you should never have anything at all resembling that line in a real program.

why?

for the very reason that you are confused about this problem. you should only do one thing at a time. it makes your code clear. eg

Code:

x=y++;

that is terrible code. does it mean?

Code:

x=y;
y++;

or

Code:

y++;
x=y;

the answer? who cares. we could compile and run it and find out but that isnt the problem. if anyone should ever ask that question about your code, then the code is ambigious and should be fixed so that it is absolutely clear what the program isa meant to be doing.

(btw, if you are dying to know, your arithmetic is wrong. go re read what ++a and a++ do. its a little different to what you thikn they do, which is clear after looking at your arithmetic)

Originally Posted by abc1000

As already said when this code is executed the value of a is printed as 50.


Thanks in advance.

The ONLY answer, is, the result is undefined. Period. End of story.
50 is not right. 48 is not right. 32.8765 is not right. It is undefined. Trying to figure out an actual number is a 100% waste of time.

You should never post/pre increment or decrement a variable within the same code statement. This will surely lead to undefined behavior.

Originally Posted by Brain_Child

you shouldnt try to work out this problem because your code is just poor and you should never have anything at all resembling that line in a real program.

why?

for the very reason that you are confused about this problem. you should only do one thing at a time. it makes your code clear. eg

Code:

x=y++;

that is terrible code. does it mean?

Code:

x=y;
y++;

or

Code:

y++;
x=y;

the answer? who cares. we could compile and run it and find out but that isnt the problem. if anyone should ever ask that question about your code, then the code is ambigious and should be fixed so that it is absolutely clear what the program isa meant to be doing.

(btw, if you are dying to know, your arithmetic is wrong. go re read what ++a and a++ do. its a little different to what you thikn they do, which is clear after looking at your arithmetic)

You're wrong. The OP code is wrong, but your example is not. The code:

Code:

x=y++;

is perfectly fine and it means:

Code:

x = y;
++y;

There's no uncertainty or implementation defined behaviour about it.