How is the following code evaluated in C?

Question

#include<stdio.h> 

int main()
{ 
  int const SIZE=5; 
  int expr; 
  double value[SIZE]={2.0,4.0,6.0,8.0,10.0}; 
  expr=1|2|3|4; 

  printf("%f",value[expr]); 
  return 0; 
}

How is the above code evaluated? Especially the statement:

expr=1|2|3|4;

Answer 1

It's a bitwise OR.

This line:

expr=1|2|3|4; 

is like:

expr = b0001 | b0010 | b0011 | b0100;

So:

0001
0010
0011
0100
----
0111

expr = b0111;

expr = 7;

Answer 2

Especially the statement: expr=1|2|3|4;

The | operator calculates the bitwise or of its operands:

    1 = 00000001b
    2 = 00000010b
    3 = 00000011b
    4 = 00000100b
    -------------
Result: 00000111b = 7

So, you will finally observe undefined behaviour, since you are accessing an array element which does not exist:

expr=1|2|3|4;               // evaluates to 7
printf("%f", value[expr]);  // accesses array element at index 7 which is out of bounds

Answer 3

In C (and many other programming languages), | is a bitwise operator OR.

Convert decimal to binary 1 = 0001 2 = 0010 3 = 0011 4 = 0100

Hence, 1|2|3|4 equivalents to 0001 OR 0010 OR 0011 OR 0100 = 0111 ( = 7 in decimal)

So expr = 7. In this case, accessing value[expr] turns out unexpected result.