c - 链接列表不兼容的指针类型

Question

我不明白为什么将“root”分配给等于“trieu”会是一个不兼容的指针:/

#include <stdio.h>
#include <stdlib.h>

struct uniform {
    char size;
    int number;
    struct uniform *ext;
};

struct uniform *trieu;
struct unifrom *root;

int main(void) {
    trieu = malloc(sizeof(struct uniform));
    root = trieu;
...
    trieu = root;

当我用 gcc 编译它时，它给我:

program.c: In function ‘main’:
program.c:15:7: warning: assignment from incompatible pointer type
  root = trieu;
   ^
program.c:57:8: warning: assignment from incompatible pointer type
  trieu = root;

它之前在我制作的另一个程序中有效:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct ll {
    char store;
    struct ll *ext;
};

struct ll *trieu;
struct ll *root;

int main(int argc, char* argv[]) {
    trieu = malloc(sizeof(struct ll));
    root = trieu;
...

Answer 1

你打错了。使用 struct uniform *root; 而不是 struct unifrom *root;。