Virtual Gate Test Series: Programming & DS - Output Of The Program

Question

What will be the output$?$

int main()
{
  int varl = 35,*var2,*var3;
  var2 = &var1; //suppose the address of var1 is 1006
  var3 = var2;
  *var2++ = *var3++;
  var1++;
  printf("var1 = %d var2 = %d var3 = %d ",var1,var2,var3);
  return 0;
}

1. 36  1010  1010
2. 38  1006  1006
3. 37  1006  1010
4. 38  1010  1006

Comments

Read operator precedence in C, only then you can attempt this question. Here confusing statement is *var2++=*var3++ --> here 3 operators are there i.e '=' , '++' , '*' --> read who has high precedence and solve it accordingly.

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Wasn't able to find it on net. Is you have some link plz share

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http://www.difranco.net/compsci/C_Operator_Precedence_Table.htm

Answer 1

Address of variables var2 and var3 are different.

after main() {

line no 3. var2 and var3, both pointing to same address, i,e &var1.

line no 4.  *var2++ = *var3++

*a++ can be written as:

*a;

a++;

here both * and ++ are unary operator precedence same but associaitvity right to left.

so (*(var2++)) = (*(var3++)), assignment has the least precedence among them (right to left), but increment is post, so variables' value (which is &var1) will increment after evaluation of expression.

It is just assigning 35 (*var3) to *var2 (which is var1), which was already 35, and they are incrementing var2 and var3 by 1×size_of_data_type

line no 5. var1++ ;  just increment 35 to 36.

Answer should be 36, var2, var3

= 36, &var1, &var1

Whatever the address of var1, it will be printed and both are equal.

Comments

I've ran the code on ide, address of var2 and 3 is same.

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yes u are right. It is the address of var1, so it will be equal.

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Addresses of var2 var3 are different but they both contain the same value i.e., the address of var1

Answer 2

* and ++ have same precedence so to evaluate them you to go with associativity  
Associativity is     e.g:   if you same precedence operators on both sides of your variable like  +y+ or *ptr++ then

Answer 3

* and ++ operators have same precedence so you have to go with associativity which is right to left for them

Associativity : if your operand is surrounded(right side and left side ) with same precedence operators like *ptr++  then to whom the ptr should be associated with is called associativity.

so in this case it is right to left and its evaluated as *(ptr++)  
ptr++ = address inside ptr + step size increment(integer size increment 4B or 2B)

Find all precedence and associativity here
https://www.programiz.com/c-programming/precedence-associativity-operators

Answer 4

both * and ++ hav the same precendence 

http://www.difranco.net/compsci/C_Operator_Precedence_Table.htm