将静态分配的数组的地址获取到结构内部的指针中

Question

The simplified version of my code is the following:

struct args {
    int p;
    int *A;
};

typedef struct sort_args sort_args;

void func(int A[], int p) 
{
    args new_struct = {&A, p};
    args *args_ptr = &new_struct;
}

I'm trying to convert a statically (I think that's the term) allocated array into a pointer, but the compiler keeps throwing these warnings:

warning: initialization makes integer from pointer without a cast [enabled by default] args new_struct = {&A, p, r};

warning: (near initialization for 'new_struct.p') [enabled by default] warning: initialization makes pointer from integer without a cast [enabled by default] warning: (near initialization for 'new_struct.A') [enabled by default]

What am I doing wrong?

Answer 1

You got the parameters backwards.

args new_struct = {&A, p};

=>

args new_struct = {p, A};

You need to initialize the members of a struct in exactly the same order as they appear in the structs declaration, or you need to use named syntax like this:

args new_struct = { .A = A, .p = p };

But this is usually only used to improve code clarity with larger structs who have more members.