Answer E14

The first sum:  6.500000
The second sum: 4.500000
The third sum:  5.000000

Here is the program, again:

#include <stdio.h>

/* Puzzle E14 -- assignment */
int main()
{
  int a = 3;
  double b = 1.5;
  double sum;
  
  sum = a+b+2.0;
  printf("The first sum:  %f\n", sum );   
  
  printf("The second sum: %f\n", sum = a+b );   
  
  sum = sum+0.5;
  printf("The third sum:  %f\n", sum );   
  
  return 0;
}

Examine this statement:

sum = a+b+2.0;

The assignment operator = and the variable sum are part of an expression. The assignment operator has low precedence, so it is performed last. It saves the value of the subexpression a+b+2.0 in the variable sum.

The entire assignment expression has a value and a type, which is the value and type stored in the variable.

Another expression is part of the second printf:

sum = a+b

It computes a value a+b, stores that value in sum. The value of the entire expression is a parameter to the printf The entire assignment expression has a value and a type, the same as the value stored in the variable. That is the value that the printf uses.

printf("The second sum: %f\n", sum = a+b );   

The third expression is:

sum = sum+0.5

This adds 0.5 to the value currently in sum and puts the result back into sum.

The second printf works, but is best avoided since the code is unclear.