1.9 — Introduction to expressions

By Alex on June 6th, 2007 | last modified by nascardriver on January 25th, 2020

Expressions

Consider the following series of statements:

1 2 3 4

int x{ 2 };             // initialize variable x with value 2 int y{ 2 + 3 };         // initialize variable y with value 5 int z{ (2 * 3) + 4 };   // initialize variable z with value 10 int w{ y };             // initialize variable w with value 5 (the current value of variable y)

Each of these statements defines a new variable and initializes it with a value. Note that the initializers shown above make use of a variety of different constructs: literals, variables, and operators. Somehow, C++ is converting each of these literals, variables, and operators into a single value that can be used as the initialization value for the variable.

What do all of these have in common? They make use of an expression.

An expression is a combination of literals, variables, operators, and explicit function calls (not shown above) that produce a single output value. When an expression is executed, each of the terms in the expression is evaluated until a single value remains (this process is called evaluation). That single value is the result of the expression.

Here are some examples of different kinds of expressions, with comments indicating how they evaluate:

1 2 3 4 5 6

2               // 2 is a literal that evaluates to value 2 "Hello world!"  // "Hello world!" is a literal that evaluates to text "Hello world!" x               // x is a variable that evaluates to the value of x 2 + 3           // 2 + 3 uses operator + to evaluate to value 5 x = 2 + 3       // 2 + 3 evaluates to value 5, which is then assigned to variable x std::cout << x  // x evaluates to the value of x, which is then printed to the console

As you can see, literals evaluate to their own values. Variables evaluate to the value of the variable. We haven’t covered function calls yet, but in the context of an expression, function calls evaluate whatever value the function returns. And operators let us combine multiple values together to produce a new value.

Note that expressions do not end in a semicolon, and cannot be compiled by themselves. For example, if you were to try compiling the expression x = 5, your compiler would complain (probably about a missing semicolon). Rather, expressions are always evaluated as part of statements.

For example, take this statement:

1	int x{ 2 + 3 }; // 2 + 3 is an expression that has no semicolon -- the semicolon is at the end of the statement containing the expression

If you were to break this statement down into its syntax, it would look like this:

1	type identifier { expression };

Type could be any valid type (we chose int). Identifier could be any valid name (we chose x). And expression could be any valid expression (we chose 2 + 3, which uses 2 literals and an operator).

Key insight

Wherever you can use a single value in C++, you can use an expression instead, and the compiler will resolve the expression down to a single value.

Expression statements

Certain expressions (like x = 5) are useful by themselves. However, we mentioned above that expressions must be part of a statement, so how can we use these expressions by themselves?

Fortunately, we can convert any expression into an equivalent statement (called an expression statement). An expression statement is a statement that consists of an expression followed by a semicolon. When the statement is executed, the expression will be evaluated (and the result of the expression will be discarded).

Thus, we can take any expression (such as x = 5), and turn it into an expression statement (such as x = 5;) that will compile.

Note that we can make expression statements that compile but are meaningless/useless (e.g. 2 * 3;). This expression evaluates to 6, and then the value 6 is discarded.

Rule

Values calculated in an expression are discarded at the end of the expression.

Quiz time

Question #1

What is the difference between a statement and an expression?

Show Solution

Statements are used when we want the program to perform an action. Expressions are used when we want the program to calculate a value.

Question #2

Indicate whether each of the following lines are statements that do not contain expressions, statements that contain expressions, or are expression statements.

a)

1

int x;

Show Solution

Statement does not contain an expression (this is just a variable definition)

b)

1	int x = 5;

Show Solution

Statement contains an expression (The right hand side of the equals side is an expression containing a single value, but the rest of the statement isn’t an expression)

c)

1

x = 5;

Show Solution

Expression statement (the entire statement is an expression, terminated by a semicolon)

d) Extra credit:

1	std::cout << x; // Hint: operator<< is a binary operator.

Show Solution

If operator<< is a binary operator, then std::cout must be the left-hand operand, and x must be the right-hand operand. Since that’s the entire statement, this must be an expression statement.

Question #3

Determine what values the following program outputs. Do not compile this program. Just work through it line by line in your head.

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#include <iostream>   int main() { std::cout << 2 + 3 << '\n'; int x{ 6 }; int y{ x - 2 }; std::cout << y << '\n';   int z{ 0 }; z = x; std::cout << z - x << '\n';   return 0; }

Show Solution

5
4
0

1.10 -- Developing your first program

Index

1.8 -- Introduction to literals and operators

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138 comments to 1.9 — Introduction to expressions

• Values calculated in an expression

  May 3, 2020 at 3:26 pm · Reply

  "Values calculated in an expression are discarded at the end of the expression."
  Are those values stored temporally in a memory or in a register?

  • nascardriver

    May 4, 2020 at 8:43 am · Reply

    That's not standardized.

    • Jesse

      May 12, 2020 at 7:33 pm · Reply

      Hello Nescar,

      I have to say this tutorial is the best of all the programming tutorials I've ever seen. Thank you!
          
      For this one:

      #include <iostream>

      int main()
      {
          int x{ 6 };
          int y{ x - 2 };
          std::cout << y << '\n';
          return 0;
      }

      6 and x-2 are both expressions, the values after evaluation are assigned to x and y respectively. Why the values calculated in an expression are discarded? Does it mean like these values are no longer stored in the memory? Would you please elaborate on it?

      Thanks in advance.

    • Values calculated in an expression

      May 19, 2020 at 4:12 am · Reply

      Thank you

• Wherever you can use a single value

  May 3, 2020 at 3:22 pm · Reply

  "Wherever you can use a single value in C++, you can use an expression instead, and the compiler will resolve the expression down to a single value."

  What if we had a function call returns a reference type (code in the below), so in that case we can't use a single value for the whole function call but we can use an expression, so isn't this in contrast to what you stated above? on the other hand, if we have a regular function call, then what you stated was correct, but how about this example?

  1 2	int& sumTwoNum(int x,int y) {...};

  1	sumTwoNum(1,2)=10;

  • Alex

    May 18, 2020 at 1:48 pm · Reply

    I said wherever you can use a single value, you can use an expression instead. I did not say all expressions can be replaced by single values.

    • komurugov

      May 24, 2020 at 9:45 am · Reply

      You said that "the compiler will resolve the expression down to a single value".
      It may be true in cases like

      1	cout << 2 + 2;

      (There will be just 4 in compiled code, without any addition).
      But in most cases like

      1	cout << foo(x) + foo(y);

      the compiler cannot resolve the expression. There'll be the same function calls and operation of addition in compiled code. Resolving to a single value is performing in runtime in this case, not in compile time.

• Alek

  May 2, 2020 at 4:25 am · Reply

  hello,in question #2 what you explain is rational,but it also performs and action which is printing the value of x?so how can it just be a expression statement ?(since the result is shown and not discarded).
  clear me if I've got something wrong thx.

• Shakib

  April 29, 2020 at 1:17 pm · Reply

  Can these be called "Expression statements"?
  - Please explain the reason.

  1 2	x{55}; x{1+2};

  • Alex

    May 1, 2020 at 7:25 pm · Reply

    They look like malformed statements to me.

    Are these intended to be definitions (in which case, x needs a type specifier before it)?

    • Shakib

      May 3, 2020 at 12:57 pm · Reply

      Sir, please forget my previous question. I am asking you again-
      1. In a program, does this x need a type specifier before using it?

      1	x = 5;//Expression statement

      2. From Question 3: at line 12, why can't I write- z{x}; ?

      1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

      #include <iostream> int main() {     std::cout << 2 + 3 << '\n';          int x{ 6 };     int y{ x - 2 };     std::cout << y << '\n';     int z{ 0 };     z = x;     std::cout << z - x << '\n';     return 0; }

      3. Which kinds of initializations and assignments can be used in Expression statements?