Programming in C

Question

#include<stdio.h>
void test(int);
void test1(int *);
int main() { 
    int x = 30, *s = &x;
    test(x++);           //test(30++) ...statement P                         
    test1(s++);        //test(some  address++) ...statement S
    return 0;
}

void test(int x){
    printf("%d\n",x);
}

void test1(int *s){
    printf("%d\n",*s);
}

My doubt is why are we not getting L-value error in both the cases?

The program correctly outputs 

30

31

Comments

when the function is called as test(x++);  won't it be evaluated as //test(30++) ...statement P  ??

---

no,

it is just like

t=x++; and then t is copied to function's local parameter x.

---

why , x is variable and 30 is constant we increment variable value not constant ryt?

x++  = (x=x+1) here indirect addressing is used ryt.

Answer 1

No both will not give lvalue error

Because, in 1st case 30 is an integer value which is inside a variable x and it is incremented, means operation is just post increment

x will replace with integer value, if it is a macro expansion. Then it will give lvalue error

Secondly, s is a pointer here. And pointer increment doesnot give lvalue error. Pointer gives lvalue error, when array value which works like pointer , need to pre or post increment. So, second one will also not give any error.