Can anyone explain the conclusion of the outputs ?

Question

int main()
{
   int x=5,*p;
   p=&x;
   *p=x++ + *p;
   printf ("%d %d",x,*p);
}
We have this program
now
here *p= 5++ + (*p) -> *p= 5 + 5 =10   
so x location is 10
and then the postfix x++= 10+1=11
so output is 11 11

int main()
{
   int x=5,*p;
   p=&x;
   *p=x++ + 5;
   printf ("%d %d",x,*p);
}
Now in the same question we add a constant 5 , Here x is not incrementing later .
*p= x++ + 5;
*p= 5+5=10
output 10 10  , when x is not taking the postfix increment ???

Answer 1

Hii @Rajib,

Postfix increment/decrement has higher precedence than unary operators.

source : https://stackoverflow.com/questions/24579014/does-postfix-operator-really-has-a-higher-precedence-than-prefix

So in your 1^st Program first x++ executed and then  *p which takes incremented value(6)

so *p=(x++) + *p ==> 5+6 ==> 11.

in second program simple addition happens

*p=(x++)+5 ==>5+5 ==>10;

 

Comments

Thanks Sir

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So here there is a pointer p pointing to data type x

*p=x++; will not be a problem ?

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@Rajib Datta Roy it is undifinede...