Ç - 冲突的类型错误/ realloc的

Question

初学C语言编程，学习作为大学工作

的一部分的代码如下所述 - 我一直在recieving一个“冲突的类型”错误调用函数get_dog 线23和线32

错误：

task11-1.c:23:35: error: assigning to 'dog' (aka 'struct dog') from incompatible 
task11-1.c:19:30: error: expected ';' after expression 
    new_array = &(*new_array)realloc(array->ptr,array->size*sizeof(new_array)); 
          ^
          ; 
task11-1.c:23:37: warning: implicit declaration of function 'get_dog' is invalid 
     in C99 [-Wimplicit-function-declaration] 
     array->ptr[array->size-1] = get_dog(*new_array); 
            ^
task11-1.c:23:35: error: assigning to 'struct dog' from incompatible type 'int' 
     array->ptr[array->size-1] = get_dog(*new_array); 
           ^~~~~~~~~~~~~~~~~~~~ 
task11-1.c:19:30: warning: ignoring return value of function declared with 
     'warn_unused_result' attribute [-Wunused-result] 
    new_array = &(*new_array)realloc(array->ptr,array->size*sizeof(new_array)); 
          ^~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 
task11-1.c:32:12: error: conflicting types for 'get_dog' 
struct dog get_dog(struct dog_array *array){ 
     ^
task11-1.c:23:37: note: previous implicit declaration is here 
     array->ptr[array->size-1] = get_dog(*new_array); 
            ^
task11-1.c:34:24: error: member reference base type 'int()' is not a structure 
     or union 
    scanf("%s", get_dog.dog_name); 
       ~~~~~~~^~~~~~~~~ 
task11-1.c:36:24: error: member reference base type 'int()' is not a structure 
     or union 
    scanf("%d", get_dog.dog_id); 
       ~~~~~~~^~~~~~~ 
task11-1.c:38:24: error: member reference base type 'int()' is not a structure 
     or union 
    scanf("%d", get_dog.dog_age); 
       ~~~~~~~^~~~~~~~ 
task11-1.c:40:12: error: returning 'int()' from a function with incompatible 
     result type 'struct dog' 
    return get_dog; 

代码：

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

struct dog{ 
    char dog_name[20]; 
    int dog_id; 
    int dog_age; 
}; 

struct dog_array{ 
    int size; 
    struct dog *ptr; 
}; 

void add(struct dog_array *array){ //dog_array *array was intended as a parameter as shown in sample code 
    int *ptr; 
    struct dog *new_array; 
    array -> size++; 
    new_array = &(*new_array)realloc(array->ptr,array->size*sizeof(new_array)); 
    if (new_array) 
    { 
     array->ptr = new_array; 
     array->ptr[array->size-1] = get_dog(*new_array); 
    } 
    else 
    { 
     printf("Out of Memory! Cannot add dog details!\n"); 
     array->size--; 
    } 
} 

struct dog get_dog(struct dog_array *array){ 
    printf("Enter Employee Name: "); 
    scanf("%s", get_dog.dog_name); 
    printf("Enter ID: "); 
    scanf("%d", get_dog.dog_id); 
    printf("Enter Salary: "); 
    scanf("%d", get_dog.dog_age); 

    return get_dog; 
} 

int main(){ 
    int input; 
    struct dog_array array={0, NULL}; 

    printf("Enter in an option:\n"); 
    printf("1. Add to Array\n"); 
    printf("2. Print all Array\n"); 
    printf("3. Exit Program\n"); 
    scanf("%d",&input); 

    switch(input){ 
     case 1: 
      printf("You have selected option 1\n"); 
      add(&array); 
      break; 
     case 2: 
      printf("You have selected option 2\n"); 
      //print_data(dog); 
      break; 
     case 3: 
      printf("You selected to exit, exiting...\n"); 
     return 0; 
    } 
} 

这是任务的香港专业教育学院一直遵循：

Task Requirements

Sample code I followed for add function as given by university

会有人能够纠正我的代码，为什么即时得到冲突类型的错误？ 和get_dog函数，我会在函数中正确调用它，并正确地格式化realloc（）函数吗？

谢谢

UPDATE：插入新代码，采取了评论 - 更多的错误

Answer 1

下面的错误是在提供的代码：

• 功能 '结构狗get_dog（结构dog_array *阵列）' 应该声明在使用之前。
• 对于类型转换和sizeof运算符，请使用type而不是变量。
• “get_dog”变量没有在函数“结构狗get_dog（结构dog_array *阵列）”

以下是校正码，也可以是逻辑上不正确声明。因为这个程序的意图不是很清楚。

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

struct dog{ 
    char dog_name[20]; 
    int dog_id; 
    int dog_age; 
}; 

struct dog_array{ 
    int size; 
    struct dog *ptr; 
}; 

struct dog get_dog(struct dog *array); 

void add(struct dog_array *array){ //dog_array *array was intended as a parameter as shown in sample code 
    //int *ptr; 
    struct dog *new_array; 
    array -> size++; 
    new_array = (struct dog*)realloc(array->ptr,array->size*sizeof(struct dog)); 
    if (new_array) 
    { 
     array->ptr = new_array; 
     array->ptr[array->size-1] = get_dog(new_array); 
    } 
    else 
    { 
     printf("Out of Memory! Cannot add dog details!\n"); 
     array->size--; 
    } 
} 

struct dog get_dog(struct dog *array){ 
    struct dog get_dog= *array; 
    printf("Enter Employee Name: "); 
    scanf("%s", get_dog.dog_name); 
    printf("Enter ID: "); 
    scanf("%d", get_dog.dog_id); 
    printf("Enter Salary: "); 
    scanf("%d", get_dog.dog_age); 

    return get_dog; 
} 

int main(){ 
    int input; 
    struct dog_array array={0, NULL}; 

    printf("Enter in an option:\n"); 
    printf("1. Add to Array\n"); 
    printf("2. Print all Array\n"); 
    printf("3. Exit Program\n"); 
    scanf("%d",&input); 

    switch(input){ 
     case 1: 
      printf("You have selected option 1\n"); 
      add(&array); 
      break; 
     case 2: 
      printf("You have selected option 2\n"); 
      //print_data(dog); 
      break; 
     case 3: 
      printf("You selected to exit, exiting...\n"); 
     return 0; 
    } 
} 

Answer 2

赋值表达式：

array->ptr[array->size-1] = get_dog(); 

是错的 - 在左侧的类型为dog*但右侧是dog。

顺便说一句，在get_dog（）被调用不带参数...