C - 通过引用传递struct

Question

我试图在C中通过引用传递结构,以便我可以修改函数内的值.这是我到目前为止的代码,但它会产生一些警告和一个错误.

main.c中

#include <stdio.h>
#include "myfunctions.h"
#include "structures.h"

int main(int argc, char const *argv[] {

    struct MyStruct data;

    data.value = 6;

    printf("Before change: %d\n", data.value);

    changeData(data);

    printf("After change: %d\n", data.value);

}

myfunctions.c

#include "structures.h"

void changeData(MyStruct data) {
    data.value = 7;
}

myfunctions.h

#ifndef MyStruct
#define MyStruct

void changeData(MyStruct data);

#endif

structures.h

typedef struct {

    int value;

} MyStruct;

产生的错误

In file included from main.c:2:0:
myfunctions.h:4:1: warning: parameter names (without types) in function declaration
 void changeData(MyStruct data);
 ^
In file included from main.c:3:0:
structures.h:5:1: warning: unnamed struct/union that defines no instances
 } MyStruct;
 ^
main.c: In function ‘main’:
main.c:9:5: error: ‘data’ undeclared (first use in this function)
     data.value = 6;
     ^
main.c:9:5: note: each undeclared identifier is reported only once for each function it appears in

Answer 1

这都是由于

#define MyStruct

使用此行,您已定义MyStruct为扩展为空的宏.也就是说,你已经有效地删除MyStruct了以下代码中出现的所有内容,这就是为什么编译器在看到类似内容时会如此困惑

typedef struct {

    int value;

} ;

要么

void changeData( data);

要解决此问题,请使用

#ifndef MYFUNCTIONS_H_
#define MYFUNCTIONS_H_

代替.(这就是我们ALL_UPPERCASE为宏使用名称的原因:避免使用普通标识符进行意外名称冲突.)