C 多维数组中奇怪的数组语法
我知道这是真的：
x[4] == 4[x]
多维数组的等效项是什么？以下内容是否属实？
x[4][3] == 3[x[4]] == 3[4[x]]
I've known that this is true:
x[4] == 4[x]
What is the equivalent for multi-dimensional arrays? Is the following true?
x[4][3] == 3[x[4]] == 3[4[x]]
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x[y]定义为*(x + (y))x[y][z]将变为*( *(x + (y)) + z)x[y[z]]将变为*(x + (*(y + (z))))< /code>x[4][3]将变为*(*(x + (4)) + 3)将变为*(*(x + 4) + 3)3[x[4]]将变为*(3 + (*(x + (4))))将变为*(* (x + 4) + 3)3[4[x]]将变为*(3 + (*(4 + (x))))将变为*(*(x + 4 ) + 3)这意味着它们都是等价的。
x[y]is defined as*(x + (y))x[y][z]would become*(*(x + (y)) + z)x[y[z]]would become*(x + (*(y + (z))))x[4][3]would become*(*(x + (4)) + 3)would become*(*(x + 4) + 3)3[x[4]]would become*(3 + (*(x + (4))))would become*(*(x + 4) + 3)3[4[x]]would become*(3 + (*(4 + (x))))would become*(*(x + 4) + 3)Which means they are all equivalent.
是的。在每种情况下，x 都是一个数组，它会衰减为指针，然后对其执行指针算术。
Yes. In each case
xis an array which decays to a pointer and then has pointer arithmetic performed on it.