函数指针变量声明中有错字，但代码可以编译 [英] typo in function pointer variable declaration, but code compiles

问题描述

回答有关函数指针的问题，OP这样做是为了声明一个函数指针，该函数指针带有1个整数参数且不返回任何内容:

While answering a question about function pointers, OP did that to declare a function pointer which takes 1 integer argument and returns nothing:

void *(intr_handlerptr)(int);  // wrong but compiles!!
intr_handlerptr = intr_handler;  // compiler error: cannot assign to this weird thing (lvalue required as left operand of assignment)

正确的声明是

void (*intr_handlerptr)(int);  // correct

有趣的是，错误是在分配给该函数指针时发生的，而不是在声明它时发生的(使用gcc 7.3.1测试)

What's funny is that the error occurs when assigning to this function pointer, not when declaring it (tested with gcc 7.3.1)

那第一行是做什么的?

推荐答案

表达式void *(intr_handlerptr)(int);读为"intr_handlerptr是一个接受int并返回指向未命名对象的指针(空指针)的函数."函数名称周围的括号没有用，并且被丢弃了.这是一个原型.

The expression void *(intr_handlerptr)(int); reads as "intr_handlerptr is a function taking an int and and returning a pointer to an unnamed thing (a void pointer). So the parenthesis around the function name have no use and are discarded. This is a prototype.

表达式void (*intr_handlerptr)(int);读为"intr_handlerptr是指向一个带int且不返回任何内容的函数的指针.必须使用括号将名称与指针类型相关联.这是一个变量.

The expression void (*intr_handlerptr)(int); reads as "intr_handlerptr is a pointer to a function taking an int and returning nothing. The parenthesis are necessary to associate the name with the pointer type. This is a variable.

分配给第一个时，编译器会抱怨您试图分配给"nothing"，因为左侧符号不是变量.实际上，编译器在其符号表中仅知道该名称的原型.当然，第二个赋值语句是正确的，因为要分配给变量.

When assigning to the first, the compiler complains that you try to assign to "nothing" because the left-hand symbol is not a variable. Actually, the compiler only knows a prototype of that name in its symbol table. Of course, the second assignment statement is correct because you are assigning to a variable.

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