需要左值作为函数指针的赋值左操作数 [英] lvalue required as left operand of assignment for function pointers

问题描述

我正在尝试将函数分配给函数指针，但是出现以下错误:

I am trying to assign a function to a function pointer, but I am getting the following error:

需要左值作为赋值的左操作数.

lvalue required as left operand of assignment.

我的代码如下:

#include <stdio.h>
void intr_handler(int param){
    printf("Hey there!\n");
}

int main(){
    void *(intr_handlerptr)(int);
    intr_handlerptr = intr_handler;
}

我看不到那里是什么问题，因为我为指针"intr_handlerptr"分配了函数"intr_handler"，并且它们具有相同的签名.我想念什么?

I cannot see what is the problem there, since I am assigning to the pointer "intr_handlerptr" the function "intr_handler" and they have the same signature. What am I missing?

推荐答案

我不确定void *(intr_handlerptr)(int);的作用(尽管它可以编译，但是对此单独提出另一个问题现在很有趣，现在是

I'm unsure of what void *(intr_handlerptr)(int); does (it compiles, though, would be interesting to ask another question for this alone which is now done) but this declaration is incorrect. It should be:

void (*intr_handlerptr)(int);

然后您的代码正确编译

关于函数指针的教程: https://www.cprogramming.com/tutorial/function-pointers.html

tutorial on function pointers: https://www.cprogramming.com/tutorial/function-pointers.html

在讨论此语法错误之后，似乎很明显(现在！)

after discussion about this syntax error it seems obvious (now!) that

void *(intr_handlerptr)(int);

与以下相同:

void *intr_handlerptr(int);

因此向前声明一个函数(该函数不存在，因此不会链接，但是由于编译器在尝试为其分配内容时出现错误，因此您看不到它)

so forward declaring a function (which doesn't exist so it would not link, but you can't see that as the compiler issues an error when trying to assign something to it)

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