将带有typedef的struct数组传递给函数

Question

我需要C编程方面的帮助.我有以下情况:

struct Product {
    int code;
    char *name;
    char *spec;
    int quantity;
    float price;
};

typedef struct Product products[8];
products product = {
    {100, "Mouse", "Ottico", 10, 8.30},
    {101, "Tastiera", "Wireless", 6, 15.50},
    {102, "Monitor", "LCD", 3, 150.25},
    {103, "Webcam", "USB", 12, 12.00},
    {104, "Stampante", "A Inchiostro", 6, 100.00},
    {105, "Scanner", "Alta Risoluzione", 9, 70.50},
    {106, "Router", "300 Mbps", 10, 80.30},
    {107, "Lettore Mp3", "10 GB", 16, 100.00}
    };

请忽略上面使用的意大利语.

我想将名为"product"的结构数组传递给函数.例如,如果我想做类似的事情

product[1].name = "Computer"

但是在一个函数里面,我应该怎么做呢？我想知道如何从main()调用该函数以及如何在头文件中编写原型.

在此先感谢您的帮助.

编辑

我给你这个测试程序.这个不起作用,甚至没有主要功能调用.它只是不编译.

#include <stdio.h>
#include <stdlib.h>

void test(Card *card);

int main()
{
    struct Card {
        char *type;
        char *name;
    };

    typedef struct Card cards[2];
    cards card = {{"Hi", "Hi"}, {"Foo", "Foo"}};
    return 0;
}

void test(Card *card) {
    printf("%s", card[1].type);
}

Answer 1

这里:

void foo(struct Product *bla)
{
    bla[1].name = "Computer";
}

或使用您的类型别名

void foo(products bla)
{
    bla[1].name = "Computer";
}

然后像这样调用函数:

foo(product);