Question

我想通过C语言制作收银员计划。我使用结构进行记录，但问题是当我尝试按项目detail[i].name的定义名称和价格detail.[i].price显示付款收据进行循环时，我收到此消息

[错误]来自＆＃39; char *＆＃39;的无效转换去＃char;＆＃39; [-fpermissive]

这是我的剧本

#include<stdio.h>
#include <stdlib.h>
    struct item{
        char name[10];
        int price;
        int barcode;
    };
    struct item detail[10]={
        "item1",10,1,
        "item2",20,2,
        "item3",30,3,
        "item4",40,4,
        "item1",50,5,
        "item2",60,6,
        "item3",70,7,
        "item4",80,8,
        "item3",90,9,
        "item4",100,10
    };
    int main(){
        int ibarcode[10];int qty[10];char b[10];int price[10];int ju[10];int tot[10];
        int j,i,k,grand;
        char a;
        printf("Cashier program\n");
        for(j=0;j<10;j++){
            printf("enter barcode : ");scanf("%d",&ibarcode[j]);
            for(i=0;i<10;i++){
                if(ibarcode[j]==detail[i].barcode){
                    printf("item name: %s\n",detail[i].name);
                    printf("price : %d\n",detail[i].price);
                    printf("Quantity : ");scanf("%d",&qty[j]);
                    tot[j]=detail[j].price*qty[j];
                }
               if(ibarcode[j] > 10){
                    printf("Barcode not valid\n");
                    j--;
                    break;
               }
            }
            printf("\nDo you want to buy again? [Y/N] = ");scanf("%s", &a);
            b[j] = detail[i].name;
            ju[j] = detail[i].price;
            if(a=='Y'||a=='y'){
               continue;
            } else {
               break;
            }
        }
        grand = 0; system("cls");
  printf("\n name Kasir = Addzifi Moch G\n");
  printf(" Tanggal    = 03 januari 2017\n");
  printf(" Jam        = 14:05 WIB\n\n");
  printf("+-------------------------------------------------------------------------------------------------+\n");
  printf("| Barcode | name item\t\t\t| price \t\t| Quantity\t| Total   |\n");
  printf("+-------------------------------------------------------------------------------------------------+\n");
 for(k=0; k<=j; k++){
  grand += tot[k];
  printf("|    %d \t  | %s\t                | %d\t\t       | %d\t\t\t| %d |\n", ibarcode[k], b[k], ju[k], qty[k], tot[k]);
 }
 printf("+-------------------------------------------------------------------------------------------------+\n");
 printf("|\t\t\t\t\t\t\t  Total Yang Harus Dibayarkan  =  %d |\n", grand);
 printf("+-------------------------------------------------------------------------------------------------+\n");
}

Answer 1

我敢打赌，就行了

b[j] = detail[i].name;

这是因为b是一个字符数组，name是一个字符串，所以你要分配给b[j]，一个字符串，一个字符串。

使用strcpy：

strcpy(b, detail[i].name);

来自＆＃39; char *＆＃39;的无效转换去＃char;＆＃39; [-fpermissive]

1 个答案: