What brackets in array input mean

Question

为什么在某些动态数组声明中，我们需要将数组名称放在方括号中：

(*allocMat)[count++] = row;

Answer 1

It's about operator precedence, ie which part of the statement is executed first.

Like in simple math. Is x = a + b*c executed like x = (a + b)*c or like x = a + (b*c) ?

So for your code the question is: Is * "stronger" than [] or is it the opposite?

Consider just doing:

*allocMat[count++] = row;

How would you expect that to be executed?

Like A:

(*allocMat)[count++] = row;

or like B:

*(allocMat[count++]) = row;

The answer is that it's executed like B so if you really want A then you need to explicit add the parenthesis.

An example where you would want A is when allocMat is a pointer to an array.

An example where you would want B is when allocMat is an array of pointers.

Answer 2

This is because of operator precedence . The array subscript operator [] has higher priority than the unary dereference operator * . So, unless the parenthesis is used, a statement like

*allocMat[count++] = row;

will be parsed as

* (allocMat[count++]) = row;

which is not desired.

To properly evaluate the statement, we need to first dereference the pointer, and then, index onto it, like

(*allocMat)[count++] = row;

In the above snippet, allocMat is a pointer to an array. So, unless, the dereference is forced on higher priority, the subscripting operator [] , which has higher priority, will be taken into account first and will result in incorrect evaluation.

Answer 3

Allocmat is presumably a pointer to an array.

The parentheses are needed to get the indirection correctly. So (*allocMat)[count++] is the same as allocMat[0][count++] . Would you omit the parentheses, *allocMat[count++] would be equal to allocMat[count++][0] which is completely different. This is because operator precedence - [] binds slightly tighter than * .