在遍历数组并找到匹配的元素后如何停止for循环？

Question

我有一个for循环，该循环遍历struct类型的数组，但是当我将用户输入传递给检索该值的函数时，它将打印10次。我需要main（）来询问用户要打印的地址簿的哪个＃，然后将其传递给将其打印出来的函数，但是该函数会检查数组中是否存在这样的数字，然后再进行打印排除输入的＃的值。这是保持十次打印printAddress（）的代码：

STRUCTURE FOR 
struct addressBook
{ 
int entryNum;
char name[100];
char address[100];
}; 

void printAddress(struct addressBook addresses, int num)
{ 
    int n = 0;
    for(n=0; n < 10; n++)
    {
        if (addresses.checkNum == num)
        { 
            printf("\nBELOW IS ADDRESS # %i DETAILS: \n", num);
            printf("Address Book Number: %i \n", addresses.entryNum);
            printf("Name: %s", addresses.name);
            printf("Address: %s \n", addresses.address);

    }
    }

}


my main function: 

int main()
{ 
// create a addressBook array of at least 10 
struct addressBook addresses[10];

//if statements where user can select to 
//add a name to address book, display a single name or quit the program

//this is if they choose to display a single name using a number 
else if (choice == 2)
    {
    int num; 
    int i = 0;
    printf("Enter book number: ");
    scanf("%i", &num);
    printAddress(addresses[i], num);
    } 

return 0 
}

Answer 1

这是您的代码的改编版，似乎可以正常运行：

#include <stdio.h>

struct addressBook
{
    int entryNum;
    char name[100];
    char address[100];
};

static
void printAddress(int n_addr, struct addressBook addresses[n_addr], int num)
{
    for (int n = 0; n < n_addr; n++)
    {
        if (addresses[n].entryNum == num)
        {
            printf("\nBELOW IS ADDRESS # %i DETAILS:\n", num);
            printf("Address Book Number: %i\n", addresses[n].entryNum);
            printf("Name: %s\n", addresses[n].name);
            printf("Address: %s\n", addresses[n].address);
        }
    }
}

int main(void)
{
    struct addressBook addresses[10] =
    {
        { 1, "John Doe", "1234 Any St, Some Town, CA 94301" },
        { 2, "Mary Fog", "1270 Some St, Any Town, CA 94303" },
        { 3, "Anne Pit", "1240 Any Rd, Some City, CA 94301" },
        { 4, "Bill Zoo", "1252 Some St, Any Town, CA 94303" },
        { 5, "Phil Tin", "1258 Any Rd, Some City, CA 94301" },
    };

    int num;
    printf("Entry number? ");
    scanf("%i", &num);
    printAddress(5, addresses, num);

    return 0;
}

您可以在功能的return;主体中打印后添加break;或if。如果添加return;，则可以在循环之后和函数返回之前添加'printf（“找不到条目％d \ n”，num）;`。

请注意，通常您不希望在换行符前留空格；尾随的空白是草率的。另外，现在名称后面有换行符；如果将名称运行到标签Address:中，则看起来很丑。

鉴于从addr83编译的程序addr83.c，示例运行为：

$ addr83
Entry number? 4

BELOW IS ADDRESS # 4 DETAILS:
Address Book Number: 4
Name: Bill Zoo
Address: 1252 Some St, Any Town, CA 94303
$