Accelerated C++ Solution to Exercise 0-1

Exercise 0-1

What does the following statement do?

3 + 4;

Solution

The statement yields a result of integer (int) type, with the value of 7. As 3 + 4 = 7.

In Detail

  • The first operand, 3, is of type integer (int). The result is therefore of type int.
  • The addition operator, +, takes in the second operand (4) and add to the first operand (3). The second operand must be of the same (int) type for the addition operator to work.
  • The semi-colon at the end discards the result.
  • Note that the result is not expected to show in the console output window (e.g. a command window) as it is of type int. Only result of type std::ostream (standard output stream) is displayed in the console output window. For instant, if we submit the following program…
#include <iostream>
int main()
{
   3 + 4;
   return 0; 
}

We should expect to see the following output. i.e. the result of 7 does not get output to the console window.

Process returned 0 (0x0) execution time : 0.359 s
 Press any key to continue.

To get the result to display the int result of 7, we use the std::cout (standard console output) facility, which is of type std::ostream (standard output stream).

#include <iostream>
int main()
{
   std::cout << 3 + 4;
   return 0; 
}

This now output the result to the console window.

7
Process returned 0 (0x0) execution time : 0.248 s
Press any key to continue.

Just a one-line statement and we have covered lots of ground here!

Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

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