[英]C struct variables allocation?
I'm new to C so please correct any mistakes I have. 我是C的新手所以请纠正我的错误。
Here's some code that's sort of like the code I have 这里的代码有点像我的代码
//typdef stuff for apple, *apple_t here
apple_t get() {
apple a;
a.num = 0;
apple_t ap = &a;
printf("set num to %d\n", ap->num);
return ap;
}
// ap above is placed into read(ap)
void read(apple_t ap) {
printf("num: %d\n", ap->num);
}
Why is it that for the "set" print ap->num == 0, but when I do the printf in read function I get some junk number like -1218550893? 为什么对于“set”打印ap-> num == 0,但是当我在read函数中执行printf时,我会得到一些像-1218550893这样的垃圾编号? What's going on?
这是怎么回事? Is the integer set being freed?
是否释放了整数集? What is C doing?
C做什么? And how do you fix this?
你是如何解决这个问题的?
You are returning the address of a local variable. 您正在返回本地变量的地址。
In this case the variable a
is a local variable. 在这种情况下,变量
a
是局部变量。 It is lost after the function ends. 功能结束后丢失。
There are two options to fix this: 有两种方法可以解决这个问题:
malloc()
. malloc()
为它分配内存。 But you must be sure to free()
it later. free()
它。 You are returning a local variable, that is not available after the function has returned. 您将返回一个局部变量,该函数在函数返回后不可用。
C supports returning structs, so no need for a pointers at all: C支持返回结构,因此根本不需要指针:
apple_t get() {
apple_t a;
a.num = 0;
return a;
}
The following code will copy the result, not returning a local variable. 以下代码将复制结果,而不是返回局部变量。
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