4.4. Assignment Operators

The left-hand operand of an assignment operator must be a modifiable lvalue. For example, given

int i = 0, j = 0, k = 0; // initializations, not assignment const int ci = i; // initialization, not assignment

1024 = k; // error: literals are rvalues i + j = k; // error: arithmetic expressions are rvalues ci = k; // error: ci is a const (nonmodifiable) lvalue

The result of an assignment is its left-hand operand, which is an lvalue. The type of the result is the type of the left-hand operand. If the types of the left and right operands differ, the right-hand operand is converted to the type of the left:

k = 0; // result: type int, value 0 k = 3.14159; // result: type int, value 3

Under the new standard, we can use a braced initializer list (§ 2.2.1, p. 43) on the right-hand side:

k = {3.14}; // error: narrowing conversion vector<int> vi; // initially empty vi = {0,1,2,3,4,5,6,7,8,9}; // vi now has ten elements, values 0 through 9

If the left-hand operand is of a built-in type, the initializer list may contain at most one value, and that value must not require a narrowing conversion (§ 2.2.1, p. 43).

For class types, what happens depends on the details of the class. In the case of vector, the vector template defines its own version of an assignment operator that can take an initializer list. This operator replaces the elements of the left-hand side with the elements in the list on the right-hand side.

Regardless of the type of the left-hand operand, the initializer list may be empty. In this case, the compiler generates a value-initialized (§ 3.3.1, p. 98) temporary and assigns that value to the left-hand operand.

Assignment Is Right Associative

Unlike the other binary operators, assignment is right associative:

int ival, jval; ival = jval = 0; // ok: each assigned 0

Because assignment is right associative, the right-most assignment, jval = 0, is the right-hand operand of the left-most assignment operator. Because assignment returns its left-hand operand, the result of the right-most assignment (i.e., jval) is assigned to ival.

Each object in a multiple assignment must have the same type as its right-hand neighbor or a type to which that neighbor can be converted (§ 4.11, p. 159):

int ival, *pval; // ival is an int; pval is a pointer to int ival = pval = 0; // error: cannot assign the value of a pointer to an int string s1, s2; s1 = s2 = "OK"; // string literal "OK" converted to string

The first assignment is illegal because ival and pval have different types and there is no conversion from the type of pval (int*) to the type of ival (int). It is illegal even though zero is a value that can be assigned to either object.

On the other hand, the second assignment is fine. The string literal is converted to string, and that string is assigned to s2. The result of that assignment is s2, which has the same type as s1.

Assignment Has Low Precedence

Assignments often occur in conditions. Because assignment has relatively low precedence, we usually must parenthesize the assignment for the condition to work properly. To see why assignment in a condition is useful, consider the following loop. We want to call a function until it returns a desired value—say, 42:

// a verbose and therefore more error-prone way to write this loop int i = get_value(); // get the first value while (i != 42) { // do something ... i = get_value(); // get remaining values }

Here we start by calling get_value followed by a loop whose condition uses the value returned from that call. The last statement in this loop makes another call to get_value, and the loop repeats. We can write this code more directly as

int i; // a better way to write our loop---what the condition does is now clearer while ((i = get_value()) != 42) { // do something ... }

The condition now more clearly expresses our intent: We want to continue until get_value returns 42. The condition executes by assigning the result returned by get_value to i and then comparing the result of that assignment with 42.

Without the parentheses, the operands to != would be the value returned from get_value and 42. The true or false result of that test would be assigned to i—clearly not what we intended!

Note

Because assignment has lower precedence than the relational operators, parentheses are usually needed around assignments in conditions.

Beware of Confusing Equality and Assignment Operators

The fact that we can use assignment in a condition can have surprising effects:

if (i = j)

The condition in this if assigns the value of j to i and then tests the result of the assignment. If j is nonzero, the condition will be true. The author of this code almost surely intended to test whether i and j have the same value:

if (i == j)

Bugs of this sort are notoriously difficult to find. Some, but not all, compilers are kind enough to warn about code such as this example.

Compound Assignment Operators

We often apply an operator to an object and then assign the result to that same object. As an example, consider the sum program from § 1.4.2 (p. 13):

int sum = 0; // sum values from 1 through 10 inclusive for (int val = 1; val <= 10; ++val) sum += val; // equivalent to sum = sum + val

This kind of operation is common not just for addition but for the other arithmetic operators and the bitwise operators, which we cover in § 4.8 (p. 152). There are compound assignments for each of these operators:

+= -= *= /= %= // arithmetic operators <<= >>= &= ^= |= // bitwise operators; see § 4.8 (p. 152)

Each compound operator is essentially equivalent to

a = a op b;

with the exception that, when we use the compound assignment, the left-hand operand is evaluated only once. If we use an ordinary assignment, that operand is evaluated twice: once in the expression on the right-hand side and again as the operand on the left hand. In many, perhaps most, contexts this difference is immaterial aside from possible performance consequences.

Exercises Section 4.4

Exercise 4.13: What are the values of i and d after each assignment?

int i; double d;

(a) d = i = 3.5;

(b) i = d = 3.5;

Exercise 4.14: Explain what happens in each of the if tests:

if (42 = i) // ... if (i = 42) // ...

Exercise 4.15: The following assignment is illegal. Why? How would you correct it?

double dval; int ival; int *pi; dval = ival = pi = 0;

Exercise 4.16: Although the following are legal, they probably do not behave as the programmer expects. Why? Rewrite the expressions as you think they should be.

(a) if (p = getPtr() != 0)

(b) if (i = 1024)