C 编程中的 Malloc - 不兼容的指针类型

Question

What is the difference between ptr->Name = (struct rec*)malloc(sizeof(struct rec)); from ptr->Name = malloc(sizeof(struct rec)); Why is it I'm receiving an error whenever I include (struct rec*) on malloc.

struct rec {
 char *Name;
}emp[100];

int main() {

int x;
int i;

struct rec *ptr = NULL;
ptr = emp;

printf("Enter Number of Clients: ");
scanf("%d", &x);
getchar();

for(i=0; i!=x; i++)
{
printf("Enter Name: ");
//I'm receiving an error whenever I add this
ptr->Name = (struct rec*)malloc(sizeof(struct rec));

//Code below is working
ptr->Name = malloc(sizeof(struct rec));

Answer 1

ptr->Name is of type char * .

ptr->Name = (struct rec*)malloc(sizeof(struct rec)) explicitly converts the return value from malloc() to a struct rec * . A struct rec * cannot be implicitly converted to a char * , so the assignment to ptr->Name is invalid.

If there is a preceding #include <stdlib.h> in your code, ptr->Name = malloc(sizeof(struct rec)) works because malloc() return void * , and a void * can be implicitly converted to any pointer type, including to a char * . Without the preceding #include <stdlib.h> (or another header which provides a declaration of malloc() , the conversion is also invalid.

void * is the ONLY pointer type in C that can be converted implicitly to another pointer type. Hence the difference between your two options.

The argument of malloc() is also wrong ie sizeof(struct rec) should not be used to dynamically allocate an array of char , in most circumstances.