c - 分配int存储？

Question

我是 C 语言的初学者。

main() {    
   int *a-ptr = (int *)malloc(int); 
   *a-ptr = 5;
   printf(“%d”, *a-ptr);
}

问题是:这能保证打印 5 吗？

答案是:不，有两个原因:

• 变量名称中不能使用“-”
• “您没有分配 int 存储空间”

我不明白第二点。存储不是用这一行分配的吗？
int *a-ptr = (int *)malloc(int);

Answer 1

    main() {    // wrong. Should return int
int main() {    // better

int *a-ptr =  //wrong. no dashes in variable names
int *a_ptr =  // better, use underscores if you want to have multiparted names

(int *)malloc(int); // wrong. Don't typecast the return of malloc(), also it takes a 
                    // size, not a type
malloc(sizeof(int)); // better, you want enought memory for the sizeof 1 int 

所以更好的代码版本是:

int main() {    
   int *a_ptr = malloc(sizeof(int)); 
   *a_ptr = 5;
   printf("%d", *a_ptr);
   free(a_ptr); // When you're done using memory allocated with malloc, free it
   return 0;
}