lvalue required as left operand of assignment

buaaspy

typedef unsigned int uint32;
typedef struct node* node_t;
struct node {
  const void* key;
  node_t next;
} __attribute__((aligned(sizeof(uint32))));

我本意是想让next指针的最低位代表一个mark标记位的，于是定义了下面的宏
#define cast(type, expr) ((type)(expr))
#define is_marked_reference(p) (cast(uint32, (p)) & 0x1)
#define get_unmarked_reference(p) cast(node_t, cast(uint32, (p)) &= ~0x3)
但是这个get_unmarked_reference在使用的时候会报lvalue required as left operand of assignment的错误，这是肿么回事儿呢

linux_c_py_php

typedef unsigned int uint32;
typedef struct node* node_t;
struct node {
  const void* key;
  node_t next;
} __attribute__((aligned(sizeof(uint32))));


((node_t)(((uint32)((p))) &= ~0x3));

uint32转了p之后就是左值了, 怎么可以&=.

buaaspy

回复 2# linux_c_py_php
可是如果我不做一个类型强制转换的话，编译器又会报错


   

linux_c_py_php

哥, 好好学学C语言吧,

int n = 2;

(double)n = 5;

你见过这样的?

buaaspy

回复 4# linux_c_py_php
多谢关注哈，现在问题已经解决了