報價

Java Language Specification, 15.7 Evaluation Order：

The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right.

The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.

If the operator is a compound-assignment operator (07001), then evaluation of the left-hand operand includes both remembering the variable that the left-hand operand denotes and fetching and saving that variable’s value for use in the implied binary operation.

所以，基本上，我會在評估右邊之前記住我左邊的舊值。

記住，操作數的評估順序和操作符的優先級是兩個不同的東西。

在{大括號}中顯示評估順序，一步一步保存值：

int i = 0;

i = i += (++i + (i += 2 + --i) - ++i); // i = 0

i{0} = i += (++i + (i += 2 + --i) - ++i); // i = 0

i{0} = i{0} += (++i + (i += 2 + --i) - ++i); // i = 0

i{0} = i{0} += (1 + (i += 2 + --i) - ++i); // i = 1

i{0} = i{0} += (1 + (i{1} += 2 + --i) - ++i); // i = 1

i{0} = i{0} += (1 + (i{1} += 2 + 0 ) - ++i); // i = 0

i{0} = i{0} += (1 + (i{1} += 2 ) - ++i); // i = 0

i{0} = i{0} += (1 + 3 - ++i); // i = 3

i{0} = i{0} += (4 - ++i); // i = 3

i{0} = i{0} += (4 - 4 ); // i = 4

i{0} = i{0} += 0 ; // i = 4

i{0} = 0 ; // i = 0

0 ; // i = 0

跟進編輯問題

如果我們命名初始值I和常數N：

int i = I;

i = i += (++i + (i += N + --i) - ++i);

那么我們可以看到這些值是：

i{I} = i{I} += ((I+1) + (i{I+1} += N + I) - ((I+1+N+I)+1));

i{I} = i{I} += (I + 1 + (I + 1 + N + I) - (I + 1 + N + I + 1));

i{I} = i{I} += (I + 1 + I + 1 + N + I - I - 1 - N - I - 1);

i{I} = i{I} += I;

i{I} = I + I;

i = 2 * I;

總結

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