什么是“无效(* old_sigint_handler)(INT)”? [英] What is 'void (*old_sigint_handler)(int)'?
问题描述
这是一个语法问题。我穿越线路传来:
无效(* old_sigint_handler)(INT);
和我不知道它在做什么。这似乎是三种类型的不带变量名的串联。我想AP preciate澄清!
无效(* old_sigint_handler)(INT);
这个定义 old_sigint_handler
是一个指针,它接受一个 INT
并返回无效
,即没有价值。围绕括号old_sigint_handler
是必要的其他位置如下:
无效* old_sigint_handler(INT);
声明的函数 old_sigint_handler
这需要一个 INT
,并返回一个指向无效
类型。这是因为在 C
的precedence规则。括号裹紧indentifier old_sigint_handler
比 *
使其成为一个功能,而不是一个指针的函数。阅读此精神上解析复杂的C声明 - 。顺时针/螺旋规则
This is a syntax question. I came across the line:
void (*old_sigint_handler)(int);
And I have no idea what it is doing. It seems like the concatenation of three types with no variable name. I would appreciate clarification!
void (*old_sigint_handler)(int);
This defines old_sigint_handler
to be a pointer to a function which takes an int
and returns void
, i.e, no value. The parentheses around old_sigint_handler
are necessary here else the following:
void *old_sigint_handler(int);
declares a function old_sigint_handler
which takes an int
and returns a pointer to void
type. This is because of the precedence rules in C
. Parentheses bind tightly to the indentifier old_sigint_handler
than the *
making it a function rather than a pointer to a function. Read this to mentally parse complex C declaration - Clockwise/Spiral Rule.
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