increement and addition in c

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#include <stdio.h>
int main(int argc, char** argv)
{
   int x = 3;
   printf("%d ";, x++ + ++x);
   return 1;
}</stdio.h>

Please, I need to know why the output will be 8 since I think it has to be 7?

Posted 29-Nov-10 3:16am

metyouba

Updated 29-Nov-10 3:30am

JF2015

v2

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2 solutions

Solution 2

thanks for ur reply Mamun
but i want to ask something
i know that x++ here won't increment 3 during printf but after it and save as x=4 then comes the addition operation this means till now we have 3+ then comes the ++x and this increment x firstly so x now will be 5 beacause it was 4 and increased by 1
finally we have 3+5=8

is this what u mean
thanks :)

Posted 29-Nov-10 3:38am

metyouba

Comments

Abdul Quader Mamun 29-Nov-10 9:59am

NO, NO, 3+ 5= 8 its wrong, It is 4+4 =8. So How,

printf("%d ";, x++ + ++x); // in the line

First x++ that 3+1= 4.
Now x holds 4 in memory OK.
x++ + ++x,
4 + 4 =8 then
so,

printf("%d ";, x++ + ++x);

Will display 8.

then the following statement execute but it will not for output.

++x=3

Thanks,
Mamun

fjdiewornncalwe 29-Nov-10 10:23am

Think of the syntax as being (x++) + (++x) and it will make sense. Order or precedence makes this 4+4 and not 3+4

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Abdul Quader Mamun · Accepted Answer · 2010-11-29T03:21:00

Solution 1

Than you for your question,

C

int x = 3; //now three
printf("%d ";, x++ + ++x); // in the line
x++ means 3 + 1 = 4
++x this statement will decrease 1 from 4 but after addition.
so, x++ + ++x will be 8.

Thanks,
Mamun

Posted 29-Nov-10 3:21am

Abdul Quader Mamun