当非静态函数声明跟在静态函数声明之后会发生什么? [英] What happens when non-static function declaration follows static function declaration?

问题描述

以下编译:

static int foo() { return 1; }
int foo();

但是，它将始终编译吗?在这种情况下，行为是否定义明确?当非静态原型遵循静态声明时，这意味着什么?

But, will it always compile? Is the behavior in this case well defined? And what does it means when a non-static prototype follows a static declaration?

推荐答案

是的，它将进行编译并且行为已明确定义.由于foo在int foo(); ¹ 之前声明为static，因此foo具有内部链接.

Yes it will compile and behavior is well defined. Since foo is declared static earlier than int foo();¹, foo has internal linkage.

对于在可见该标识符的先前声明的范围内用存储类说明符extern声明的标识符， ³¹⁾ (如果该先前声明指定了内部或外部)关联，则后面的声明中标识符的链接与先前声明中指定的链接相同. [...]

For an identifier declared with the storage-class specifier extern in a scope in which a prior declaration of that identifier is visible,³¹⁾ if the prior declaration specifies internal or external linkage, the linkage of the identifier at the later declaration is the same as the linkage specified at the prior declaration. [...]

和脚注指出:

31)按照6.2.1的规定，后面的声明可能会隐藏前面的声明.

31) As specified in 6.2.1, the later declaration might hide the prior declaration.

---

_{1 .如果未指定存储类别，则假定该函数具有外部链接.标准说:如果函数标识符的声明没有存储类说明符，则其链接 完全确定是否与存储类说明符extern-6.2.2(p5)声明的一样.}

---

_{1. If no storage class is specified , the function is assumed to have external linkage. Standard says: If the declaration of an identifier for a function has no storage-class specifier, its linkage is determined exactly as if it were declared with the storage-class specifier extern -- 6.2.2 (p5).}

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