[英]setting variable to the return type of a function
I can't seem to figure out why this will not work, I am passing the 'aHouse' variable a function which returns a House.我似乎无法弄清楚为什么这不起作用,我将“aHouse”变量传递给 function ,它返回一个房子。 I am new to C so am still trying to get my head around a few things.
我是 C 的新手,所以我仍在努力解决一些问题。
#include <stdio.h>
typedef struct house {
int id;
char *name;
} House;
House getHouse()
{
House *myHouse = NULL;
char c = getchar();
myHouse->id = 0;
myHouse->name = c; /*only single char for house name*/
return *myHouse
}
int main()
{
House *aHouse = NULL;
aHouse = getHouse();
}
First: You are using a NULL pointer and assigning values to it in the 'getHouse' function.首先:您正在使用 NULL 指针并在“getHouse”function 中为其分配值。 This is undefined behaviour and should give an access violation.
这是未定义的行为,应该给出访问冲突。
Also, you are returning a House object by value from getHouse and trying to assign to a pointer type.此外,您将通过 getHouse 的值返回 House object 并尝试分配给指针类型。 A pointer and a value are two different things.
指针和值是两个不同的东西。
You don't need pointers here at all unless you want to allocate your Houses dynamically on the heap.除非您想在堆上动态分配房屋,否则您根本不需要指针。
House getHouse()
{
House myHouse;
char c = getchar();
myHouse.id = 0;
myHouse.name = c; /*only single char for house name*/
return myHouse
}
int main()
{
House aHouse;
aHouse = getHouse();
}
EDIT: for the sake of efficiency, you could implement it like this though:编辑:为了提高效率,你可以像这样实现它:
void getHouse(House* h)
{
char c = getchar();
h->id = 0;
h->name = c; /*only single char for house name*/
}
int main()
{
House aHouse;
getHouse(&aHouse);
}
EDIT again: Also in the House structure, since the name can only be one char, don't use a char* for name, just use a char.再次编辑:同样在 House 结构中,由于名称只能是一个字符,因此不要使用 char* 作为名称,只需使用 char。
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