c - 如何使用指针初始化和打印结构并将其提供给函数？

Question

我想用指针打印并初始化一个结构并将其提供给另一个函数。

我的问题是：我怎样才能用变量名“注册”来初始化结构并将其打印出来以测试并查看我的值和地址？我的程序目标基本上是模拟 CPU。因此，我有 3 个用于订单、堆栈和计算器的指针（这是我们老师给的任务）。但是我对结构和指针不太好

#include<stdio.h>
#include<stdint.h>
#include<stdbool.h>

struct reg {
  unsigned char pc; // Befehlszeiger
  unsigned char sp; // Stapelzeiger
  unsigned char fa; // Flags + Akkumulator
};

bool cpu(struct reg *registers, unsigned char data[128], uint16_t cmd[256]);


int main(){
unsigned char data[128];
uint16_t cmd[256];
cmd[127] = 12;
data[127] = 'D';
data[128] = 'Z';
struct reg *registers = { '1', '2', '3'};
printf("The number before function: %d\n", registers->pc);
cpu(registers, data, cmd);
return 0;
}

bool cpu(struct reg *registers, unsigned char data[128], uint16_t cmd[256])
{
printf("The number in function: %d\n", registers->pc);
return 0;
}

因为程序说：

|20|warning: initialization makes pointer from integer without a cast [-Wint-conversion]|
20|note: (near initialization for 'registers')|
20|warning: excess elements in scalar initializer|
|20|note: (near initialization for 'registers')|
c|20|warning: excess elements in scalar initializer|
c|20|note: (near initialization for 'registers')|

标签： cpointersstruct

Answer 1

为了使它工作重写代码如下：

struct reg registers = { '1', '2', '3'};  // "registers" isn't a pointer
printf("The number before function: %d\n", registers.pc); // "." instance of a struct
cpu(&registers, data, cmd);// pass the address of "registers"