c - 带数组的 If 语句。如何判断数组整数是否为0？

Question

所以我的程序被设置为满足捐赠请求，我试图制作一个 if 语句，表示如果请求大于捐赠且捐赠为 0，则打印该请求无法满足。那么如何检查数组中的整数是否为0呢？程序中的案例3就是我的问题所在。谢谢

#include <stdio.h>

int main() {
    int choice, i, number, type;
    int temp_donations[5] = {0};
    int donations[5] = {0};
    int request[5] = {0};
    char TYPES[5][20] = {"Protein", "Dairy", "Grains", "Vegetables", "Fruits"};

    printf("Welcome to the Food Bank Management Program!\n");

    //Give and ask for the user's choice
    printf("What would you like to do?\n");
    printf("\t1. Enter a Donation\n");
    printf("\t2. Enter a Request\n");
    printf("\t3. Fulfill Request\n");
    printf("\t4. Print status report\n");
    printf("\t5. Exit\n");
    scanf("%d", &choice);
    printf("\n");

    //Print if choice is greater than 5 or less than 1
    if(choice > 5 || choice < 1)
        printf("Sorry, that was not a valid input.\n\n");

    while (choice != 5) {
        switch (choice) {
            case 1:
                //ask user the type of food they would like to donate
                printf("\nWhat donation type would you like to enter?\n");
                number = 0;
                for(i=0; i<5; i++){
                    printf("\t%d. %s\n",number, TYPES[i]);
                    number += 1;
                }
                //user input for food type and amount to donate
                scanf("%d", &type);
                printf("How much would you like to donate? ");
                scanf("%d", &donations[type]);
                printf("Donation Added!\n\n");
                break;
            case 2:
                //ask user the type of food they would like to request
                printf("\nWhat would you like to request?\n");
                number = 0;
                for(i=0; i<5; i++){
                    printf("\t%d. %s\n",number, TYPES[i]);
                    number += 1;
                }
                //user input for request and amount requested
                scanf("%d", &type);
                printf("How much would you like to donate? ");
                scanf("%d", &request[type]);
                printf("Donation Added!\n\n");
                break;
            case 3:
                //go through foods and fulfill the requests if possible
                for(i=0; i<5; i++){
                    if (request[i] > donations[i] && //I'm not sure what to put in here)
                        printf("%s requests cannot be fulfilled.\n", TYPES[i]);
                    else if (request[i] > donations[i]){
                        printf("%s requests will be partially fulfilled.\n", TYPES[i]);
                        temp_donations[i] = donations[i];
                        donations[i] -= donations[i];
                        request[i] -= temp_donations[i];
                    }
                    else {
                        donations[i] -= request[i];
                        request[i] -= request[i];
                    }
                }
                printf("\n");
                break;
            case 4:
                //print table of current donations and requests
                for(i=0; i<5; i++){
                    printf("\t%-10s:    Donations: %-2d    Requests: %-2d\n", TYPES[i], donations[i], request[i]);
                }
                printf("\n");
                break;
        }
        //reask for user's choice
        printf("What would you like to do?\n");
        printf("\t1. Enter a Donation\n");
        printf("\t2. Enter a Request\n");
        printf("\t3. Fulfill Request\n");
        printf("\t4. Print status report\n");
        printf("\t5. Exit\n");
        scanf("%d", &choice);
        printf("\n");

        if(choice > 5 || choice < 1)
            printf("Sorry, that was not a valid input.\n\n");
    }

    printf("Thank you for running our system!\n");

    return 0;
}

Answer 1

你想要这个吗？

if (request[i] > donations[i] && donations[i] == 0)