Question

我正在制作一个关于动物和其他功能的结构来控制它。当来自其他文件的所有代码都在一个文件中时，它运行良好，但是当我决定将代码设置为其他文件时，我收到以下错误：

d:/programms/mingw64/bin/../lib/gcc/x86_64-w64-mingw32/11.0.1/../../../../x86_64-w64-mingw32/bin/ld.exe: C:\Users\yuriy\AppData\Local\Temp\ccAi1tOR.o:main.c:(.text+0x2a): undefined reference to `new_animal' 
d:/programms/mingw64/bin/../lib/gcc/x86_64-w64-mingw32/11.0.1/../../../../x86_64-w64-mingw32/bin/ld.exe: C:\Users\yuriy\AppData\Local\Temp\ccAi1tOR.o:main.c:(.text+0x36): undefined reference to `animal_info'
collect2.exe: error: ld returned 1 exit status

main.c

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

int main() {
    animal jack = new_animal("jack", 23, 2.4f);
    animal_info(&jack);
    return 0;
}

animal.h

#pragma once

typedef struct animal {
    char name[25];
    int age;
    float happiness;
} animal;

struct animal new_animal(char name[25], int age, float happiness);

void animal_info(struct animal *an);

animal.c

#include <stdio.h>
#include <stdlib.h>
#include "animal.h"
#include "charfunc.h"

struct animal new_animal(char name[25], int age, float happiness) {
    struct animal *an = malloc(sizeof(struct animal));
    copy_paste(name, an->name);
    an->age = age;
    an->happiness = happiness;
    return *an;
    free(an);
}

void animal_info(struct animal *an) {
    printf("ANIMAL INFO:\n");
    printf("    name -> %s\n", an->name);
    printf("    age -> %d\n", an->age);
    printf("    happiness -> %f\n", an->happiness);
}

charfunc.h

#pragma once
void copy_paste(const char* from, char* to);

charfunc.c

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

void copy_paste(const char* from, char* to) {
    int i = 0;
    for (i; from[i]; i++)
    {
        to[i] = from[i];
    }
    to[i] = 0;
}

对“animal_info”的未定义引用和对“new_animal”的未定义引用

0 个答案: