如何访问外部块变量在内部块内的时候有块相同的变量声明？ [英] How to access outer block variable in inner block when inner block have same variable declaration?

问题描述

int main(int argc, char** argv) {
    int i=5;
    {
        int i=7;
        printf("%d\n", i);
    }
    return 0;
}

如果我想访问外 I （ INT I = 5 ）的价值printf的那么就可以怎么做的？

If I want to access outer i (int i=5) value in printf then how it can done?

推荐答案

C99标准的相关部分，第6.2.1节（鉴定网络连接器的作用域）：

The relevant part of the C99 standard, section 6.2.1 (Scopes of identifiers):

的 4 的[...]如果一个标识符指定在同一名称空间两个不同的实体，该范围可能会重叠。如果是这样，一个实体（内部范围）的范围将是其他实体的范围（外部范围）的严格的子集。在内部范围内，该标识符指定的范围内声明的实体;在外部范围声明的实体被隐藏（不可见）内范围之内。

4 [...] If an identifier designates two different entities in the same name space, the scopes might overlap. If so, the scope of one entity (the inner scope) will be a strict subset of the scope of the other entity (the outer scope). Within the inner scope, the identifier designates the entity declared in the inner scope; the entity declared in the outer scope is hidden (and not visible) within the inner scope.

要prevent <一个href=\"http://stackoverflow.com/questions/3882621/how-to-access-outer-block-variable-in-inner-block-when-inner-block-have-same-vari/3884427#3884427\">pmg's从消失回答：您可以通过隐藏之前宣布指向它的指针访问外部块变量发生：

Update

To prevent pmg's answer from disappearing: You can access the outer block variable by declaring a pointer to it before the hiding occurs:

int i = 5;
{
    int *p = &i;
    int i  = 7;
    printf("%d\n", *p); /* prints "5" */
}

当然给这样的隐藏变量是从来不需要总是不好的风格。

Of course giving hiding variables like this is never needed and always bad style.

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