为什么(以及什么时候)需要在 sizeof 之后使用括号？

以下内容无法编译:

typedef int arr[10];
int main(void) {
return sizeof arr;
}


sizeof.c:3: error: expected expression before ‘arr’

但如果我改成

sizeof(arr);

一切都很好，怎么了？

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最佳答案

According to 6.5.3, there are two forms for sizeof as the following:

sizeof unary-expression
sizeof ( type-name )

Since arr in your code is a type-name, it has to be parenthesized.

小开

That's the way the language is specified, type names must be parenthesized here.

Suppose the grammar looked like this:

sizeof unary-expression sizeof type-name

Now, e.g. the following expression would be ambiguous:

sizeof int * + 0

It could be either sizeof(int *) + 0 or sizeof(int) * +0. This ambiguity doesn't arise for unary expressions, as an asterisk appended to an expression isn't an expression (but for some type names, appending one, is again a type name).

Something had to be specified here and requiring type-names to be parenthesized is a way to solve the ambiguity.

小开

I think it's because you have typedef. If you remove it, it should compile.

Example from wikipedia:

/* the following code fragment illustrates the use of sizeof
* with variables and expressions (no parentheses needed),
* and with type names (parentheses needed)
*/
char c;
printf("%zu,%zu\n", sizeof c, sizeof (int));

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