C语言 '&' 标记前的预期声明说明符或 '...'
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Expected Declaration Specifiers or ‘...’ before ‘&’ token
提问by DotERM
I'm a beginner in C, i just have 13 years, so i'm pretty sure that the error is something very basic.
我是 C 的初学者，我只有 13 年，所以我很确定这个错误是非常基本的。
I was bored in school, i started writing C about a "Game".
我在学校很无聊，我开始写关于“游戏”的 C。
Here is it:
就这个：
#include <stdio.h>
#include <stdlib.h>
typedef struct {
int * id;
int * posx;
int * posy;
char * name;
} entity;
int allocData(&entity) {
entity->id = malloc(sizeof(int));
if(entity->id == NULL) {
return 1;
}
entity->posx = malloc(sizeof(int));
if(entity->posx == NULL) {
return ;
}
entity->posy = malloc(sizeof(int));
if(entity->posy = NULL) {
return 1;
}
entity->name = malloc(sizeof(char) * 129);
if(entity->name == NULL) {
return 1;
}
return 0;
}
int main() {
entity player;
allocData(&player);
player.id = 0;
player.name = "loopbackmain.c:11:15: error: expected declaration specifiers or ‘...' before ‘&' token
int allocData(&entity) {
^
";
printf("ID: %d, Name: %s", player.id, player.name);
return 0;
}
But GCC complains.
但海湾合作委员会抱怨。
int allocData(entity *player);
I can't find the error, and i didn't want to post here, i'm sure is something stupid that i missed.
我找不到错误，我不想在这里发帖，我确定我错过了一些愚蠢的事情。
回答by dinomario10
Your function definition should take pointer as an argument:
您的函数定义应将指针作为参数：
int allocData(entity * p_entity)
回答by artm
Your prototype looks wrong:
你的原型看起来不对：
int allocData(&entity) // <-- what is the type? and why it's address operator here??
int allocData(&entity) // <-- 类型是什么？为什么它是地址运算符？
It should be:
它应该是：
#include <stdio.h>
#include <stdlib.h>
#include <string.h> /* add this for using strcpy() */
typedef struct {
int id;
int posx;
int posy;
char name[129];
} entity;
/* now allocData() won't be needed because entity doesn't have any member to allocate pointer to */
int main(void) {
entity player;
player.id = 0;
/* use strcpy() to copy strings */
/* ##代码## here will be meaningless in this case, so I removed this */
strcpy(player.name, "loopback");
printf("ID: %d, Name: %s", player.id, player.name);
return 0;
}
You should also replace all your entityvariable name in function allocData- remember entityis your typename, not variable.
您还应该替换entity函数中的所有变量名称allocData- 记住entity是您的类型名称，而不是变量。
回答by MikeCAT
You cannot use reference in C.
你不能在 C 中使用引用。
In this case, you should want to use a pointer.
在这种情况下，您应该想要使用指针。
change the lineint allocData(&entity) {to int allocData(entity *entity) {, and the code will compile.
将行更改int allocData(&entity) {为int allocData(entity *entity) {，代码将编译。
Although this change will make the code compile, you will invoke undefined behaviorby passing data having wrong type to printf(): %dexpects for int, but you passed int*. You will also cause memory leak by throwing away the pointer assinged in allocData(). I don't think using pointer unless they are needed isn't good.
尽管此更改将使代码编译，但您将通过将具有错误类型的数据传递给: expects for来调用未定义的行为，但您传递了. 您还将通过丢弃 .assinged 中的指针导致内存泄漏。我不认为使用指针除非需要它们不好。printf()%dintint*allocData()
Your code should be like this:
你的代码应该是这样的：
##代码##