错误：从“void *”类型分配类型“值”时出现不兼容的类型？

Question

I get this error when I'm trying to malloc some memory for my struct of ints.

typedef struct _values {
random ints;
} values;

I have tried the lines below but my compiler doesn't like it. How do I fix the error?

values v;
v = malloc(sizeof(values));

Answer 1

You forgot to add the asterisk (*) after the values and before the v to mark it as a pointer: values *v;

The way you set it now, the v (without the asterisk) is defined as a stack variable and would be allocated on the stack and discarded once the function ends. It's type will be simply values . malloc is used to allocate memory on the heap and returns a pointer to the memory. Sine the function has no way of knowing the type it returns it as a void * type - Which gives you your error - you're attempting to assign a void * type into a struct type, which the compiler can't do, nor can the compiler find a legitimate cast that could resolve the problem.

Answer 2

You don't need to malloc memory in this case; the line values v; already allocated the memory ("on the stack"). It will automatically be freed when you leave the current scope.

If you really wanted to allocate the memory "on the heap", so that it persists past the current scope, you need v to be a pointer; that is:

values *v;
v = malloc(sizeof(values));

Remember to free it when you're finished.

Answer 3

values v;

this is an instance of the structure that's allocated on the stack.

v = malloc(sizeof(values));

malloc returns a void* pointer and compiler won't allow you to assign a pointer to an instance.

you need to declare a pointer and then assign the malloc return pointer to it.

something of the sort

values *v = NULL;

v = malloc(sizeof(values));