Question

我有一个

无法将参数'1'的'gpio **'转换为'gpio *'为'void timer_hard（gpio *）'

此测试代码的

错误：

#include <iostream>

struct gpio{
  short const pin;
  short must;
  short flag;
}handle_out{0,0,0},handle_in{1,0,0},lock{2,0,0},mouth{3,0,0},ring_hard{4,0,0},ring_soft{5,0,0};

gpio *gpio_0[]={&handle_out,&handle_in,&lock,&mouth};

void timer_hard(gpio *array){
  std::cout << sizeof(gpio_0) << '\n';
  std::cout << sizeof(array) << '\n';
  for (int i = 0; i<sizeof(array)/sizeof(array[0]);i++){
    std::cout << array[i].pin << '\n';
  };
  std::cout << '\n';
  for (int i = 0; i<sizeof(array)/sizeof(array[0]);i++){
    std::cout << array[i].flag << '\n';
  };
  lock.flag = 1;
  std::cout << '\n';
  for (int i = 0; i<sizeof(array)/sizeof(array[0]);i++){
    std::cout << array[i].flag << '\n';
  };
}

main(){
  timer_hard(gpio_0);
}

请告诉我我做错了什么

Answer 1

gpio *gpio_0[]是一个二维数组，这意味着它的类型为gpio **。

参数'1'无法将'gpio **'转换为'gpio *'

1 个答案: