C语言 从“void*”到“node*”的无效转换[-fpermissive]
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invalid conversion from 'void*' to 'node*' [-fpermissive]
提问by STAR S
I have a C program that produces an error:
我有一个产生错误的 C 程序：
invalid conversion from 'void*' to 'node*' [-fpermissive]
Here's my code:
这是我的代码：
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
struct node
{
int data;
struct node* next;
};
struct node* onetwothree();
int main()
{
struct node* ptr;
ptr = onetwothree();
return 0;
}
struct node* onetwothree()
{
struct node* head;
struct node* temp;
head = malloc(sizeof(struct node));
temp = head;
for(int i=1; i<=3; i++)
{
temp->data = i;
if(i<3)
temp=temp->next;
else
temp->next = NULL;
}
return head;
}
What am I doing wrong?
我究竟做错了什么？
回答by hmjd
In C, a void*is implicity convertible to a T*where Tis any type. From section 6.3.2.3 Pointersof the C99 standard:
在 C 中， avoid*可以隐式转换为 a T*whereT是任何类型。来自C99 标准的第6.3.2.3节指针：
A pointer to void may be converted to or from a pointer to any incomplete or object type. A pointer to any incomplete or object type may be converted to a pointer to void and back again; the result shall compare equal to the original pointer.
指向 void 的指针可以与指向任何不完整或对象类型的指针相互转换。指向任何不完整或对象类型的指针可以转换为指向 void 的指针，然后再返回；结果应与原始指针相等。
malloc()returns a void*and is assignable without casting to head, a struct node*. This is not true in C++, so I suspect a C++ compiler is being used to compile this C code.
malloc()返回 avoid*并且无需强制转换为head, a即可赋值struct node*。这在 C++ 中并非如此，因此我怀疑正在使用 C++ 编译器来编译此 C 代码。
For example:
例如：
#include <stdlib.h>
int main()
{
int* i = malloc(sizeof(*i));
return 0;
}
when compiled with:
编译时：
gcc -Wall -Werror -pedantic -std=c99 -pthread main.c -o main
gcc -Wall -Werror -pedantic -std=c99 -pthread main.c -o main
emits no errors. When compiled with:
不发出错误。编译时：
g++ -Wall -Werror -pedantic -std=c++11 -pthread main.cpp -o main
g++ -Wall -Werror -pedantic -std=c++11 -pthread main.cpp -o main
emits:
发出：
main.cpp: In function 'int main()': main.cpp:5:31: error: invalid conversion from 'void*' to 'int*' [-fpermissive]
main.cpp：在函数“int main()”中：main.cpp:5:31：错误：从“void*”到“int*”的无效转换[-fpermissive]
Additionally, the onetwothree()function is not allocating memory correctly. It allocates one struct nodeonly:
此外，该onetwothree()函数未正确分配内存。它只分配一个struct node：
head = malloc(sizeof(struct node));
and then, eventually, dereferences head->next->nextwhich is undefined behaviour. An individual malloc()is required for every struct node.
Remember to free()what was malloc()d.
然后，最终，取消引用head->next->next这是未定义的行为。一个人malloc()需要每一个struct node。记住free()什么是malloc()d。
回答by Alex Erny
You're having this warning/error because you are using malloc(which returns a void*)to initialize a structure of type node*without doing an explicit cast.
您收到此警告/错误是因为您正在使用malloc（返回 a void*）来初始化类型结构node*而不进行显式转换。
To get rid of this error you could change your code this way :
要摆脱此错误，您可以通过以下方式更改代码：
head = (struct node *)malloc(sizeof(struct node));
or you could as well add the "-fpermissive" flag to your compiler which will then ignore these errors.
或者您也可以将“-fpermissive”标志添加到您的编译器，然后将忽略这些错误。
EDIT: But yeah, i didn't think about the fact that this should not happen in a C compiler anyways
编辑：但是，是的，我没有想过这不应该发生在 C 编译器中的事实
