Why Does Assigning a Map Value in C Produce Unexpected Results?

Evaluation Order in C Assignment Statements

When assigning a value to a map in C , the order of evaluation can lead to counterintuitive results. Consider the following code:

map<int, int> mp;<br>printf("%d ", mp.size());<br>mp[10]=mp.size();<br>printf("%dn", mp[10]);<br>

This code prints 0 and then 1, whereas one might expect 0 and 0. This is because:

1. Right-to-Left Evaluation: Assignment operators in C associate from right to left, meaning that mp[10] is evaluated before mp.size().
2. Temporary Reference: The left-hand side of the assignment mp[10] creates a temporary reference to the underlying map element. This element is initially uninitialized, so mp.size() initially returns 0.
3. Value Modification: The assignment mp[10]=mp.size() sets the value of mp[10] to the current size of the map, which is 0.
4. Reference Life Extension: The assignment temporarily extends the lifetime of the temporary reference created by mp[10].
5. Final Evaluation: mp[10] is now pointing to the modified element, so printf("%d", mp[10]); prints the updated value of 1.

Unspecified Behavior

This particular behavior is unspecified in the C standard. However, a recent proposal (N4228) seeks to clarify the order of evaluation in such cases.

Section [expr.ass]p1 of the revised proposal states that:

"The right operand is sequenced before the left operand."

This means that in the above example, mp.size() would be evaluated before mp[10], resulting in the expected output of 0 and 0.

Update

It is important to note that in C 17, this behavior has been specified in the standard, as per revision 3 of proposal p0145. The right operand is now explicitly sequenced before the left operand in assignment statements.

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