c - 如何正确使用宏函数？

Question

我有一个包含数字的数组。三种数字(float、double 等)，当我搜索特定数字时，它应该返回/显示该数字在数组中的位置。

所以我的想法是，我可以将 -3、1.25 这样的数字放入数组中，而无需专门定义数组类型，因为我需要接收任何类型的数字。

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#define finder(find_num, array,type,format)\
void find_num(type array[], type ,int size)\
{\
    int location = 0, i;\
    for (i = 0; i < size; i++)\
    {\
        location++;\
        if (array[i] == num)\
            break;\
    }\
    if (location != arry[0]\
        printf("Location of the number is at %d place\n", result);\
    else\
        printf("Location of number not found\n");\
}

finder(funcfind, int array[], int %d\t, int %d\t);
finder(funcfind, int array[] ,double %f, int %d\t)

int main()
{
    int array[] = { 5,6,7,88,12,3,20 }, result = 0, num = 0, size = 0;
    size = sizeof(array);
    scanf("%d", &num);
    funcfind(array, num, size);
}

我需要在这个练习中使用宏，但我陷入了 finder 函数和 funcfind 的困境，这是我大部分错误的来源。

**更新:

#include <stdio.h>
#include <stdlib.h>
#define finder(find_num, array ,type ,format)\
void find_num(type array[], type num,int size)\
{\
    int location = 0, i;\
    for (i = 0; i < size; i++)\
    {\
        location++;\
        if (array[i] == num)\
            break;\
    }\
    if (location != array[0])\
        printf("Location of the number is at %d place\n", location);\
    else\
        printf("Location of number not found\n");\
}

finder(funcfind, array, double, int %d\t);
finder(funcfind, array, int, int %d\t);
finder(funcfind, array, float, int %d\t);

int main()
{
    int array[] = { 5,6,7,-3,1.5,3,20 }, num = 0, size = 0;
    size = sizeof(array) / sizeof(array[0]);
    scanf("%d", &num);
    funcfind(array, num, size);
}

它可以工作，但某些数字(例如 -3 和 1.5)不起作用

Answer 1

如果我通过预处理(gcc -E)，结果是:

void funcfind(int %d\t int array[][], int %d\t ,int size){ int location = 0, i; for (i = 0; i < size; i++) { location++; if (int array[][i] == num) break; } if (location != arry[0] printf("Location of the number is at %d place\n", result); else printf("Location of number not found\n");};
void funcfind(double %f int array[][], double %f ,int size){ int location = 0, i; for (i = 0; i < size; i++) { location++; if (int array[][i] == num) break; } if (location != arry[0] printf("Location of the number is at %d place\n", result); else printf("Location of number not found\n");}

类型是 int %d\t所以第一个参数是错误的，为什么int %d而不是int v例如？

因为很难猜测你想要什么，所以我让你继续解决问题

无论如何在宏中-void find_num(type array[], type ,int size)很奇怪，也许一定是void find_num(type array[], type num,int size)和电话void find_num(type array[], type ,int size);和finder(funcfind, array ,double, int %d\t)-if (location != arry[0]\必须是if (location != array[0])\

给出:

void funcfind(int array[], int num ,int size){ int location = 0, i; for (i = 0; i < size; i++) { location++; if (array[i] == num) break; } if (location != array[0]) printf("Location of the number is at %d place\n", result); else printf("Location of number not found\n");};
void funcfind(double array[], double num ,int size){ int location = 0, i; for (i = 0; i < size; i++) { location++; if (array[i] == num) break; } if (location != array[0]) printf("Location of the number is at %d place\n", result); else printf("Location of number not found\n");}

如果我缩进第一个函数:

void funcfind(int array[], int num ,int size){
  int location = 0, i;
  for (i = 0; i < size; i++) {
    location++;
    if (array[i] == num)
      break;
  }
  if (location != array[0])
    printf("Location of the number is at %d place\n", result); 
  else
    printf("Location of number not found\n");
};

所以有一些错误，例如结果未知等，但我认为你可以轻松解决