task in increment , decrement , printf() , why these are evaluated in this manner in C -

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• why these constructs (using ++) undefined behavior? 12 answers

#include<stdio.h> #include<stdlib.h> int main() { int a=3; //case 1 printf("%d %d %d\n",(--a),(a--),(--a));//output 0 2 0  printf("%d\n",a);//here final value of 'a' //end of case 1 a=3; //case 2  printf("%d\n",(++a)+(a--)-(--a));//output 5 value of 2 printf("%d\n",a);//here final value of 'a' //end of case 2 a=3; //case 3  printf("%d\n",(++a)*(a--)*(--a));//output 48  value of 2 printf("%d\n",a);//here final value of 'a' //end of case 3 //case 4  int i=3,j; i= ++i * i++ * ++i; printf("%d\n",i);//output 81 i=3; j= ++i * i++ * ++i; printf("%d\n",j);//output 80 //end of case 4 return 0; } 

i clear how these outputs coming spent 3 hours in gazing @ , want know in detail why being taken or evaluated in manner.

1. this tricky 1 judge printf evaluates right left first --a pushes 2 , next a-- pushes 2 again post becomes 1 , --a makes first 0 , pushes , guessed output 0 2 2 surprise 0 2 0 final value of a=0 given places used pre increment , decrement . why happens ?

2. this take normal evaluation , printing made 4 in a-- again taken 4 , again in --a taken 3 4+4-3=5 done final value of happens 1 -1 post decrement done in middle becomes 2 again why happening ?

3. this execution same above , takes 4 * 4 * 3 = 48 , final value 2.

4. this not tricky want answer in detail figured first becomes 4 in ++i , @ i++ becomes 4 , @ ++i becomes 5 , 4*4*5 = 80 in case store in , becomes 80 , 1 +1 post increment. can 1 clear decision when store 80 in variable j , saw 80 , final value 6.

so logically judged printing , viewing things in these cases , yet more based in come can go on posting want @ assembly level how happening , why takes post increment separate state or how can different programs can judged generically each cannot keep printing , finding pattern can me in detail here. , worst part executed code in ideone , gives me different output can test @ ideone if interested why these happens please tell me breaking head @ thing 3 hours me professionals.and used gcc compiler.

this undefined behavior modifying variable more 1 in between sequence points in in case within same expression not allowed. can not rely on output of program @ all.

from c99 draft standard 6.5.2:

between previous , next sequence point object shall have stored value modified @ once evaluation of expression. furthermore, prior value shall read determine value stored.

it cites following code examples being undefined:

i = ++i + 1; a[i++] = i;