C：使用字符串数组初始化结构体

Question

我试图执行以下操作，但编译器抱怨括号，但是，我无法找到替代方法。

struct cards { 
    char faces[13][6], suits[4][9]; 
} 

typedef struct cards cards; 

void init_struct(cards *s) { 
    s->suits = {"hearts","spades","clubs","diamonds"}; 
    s->faces = {"ace","two","three","four","five", 
       "six","seven","eight","nine" 
       "ten","jack","queen","king"}; 
} 

我意识到有几个可能的重复线程在那里，但没有一个让我在赛道上。我希望你们中的一个能够:)谢谢

Answer 1

直接初始化语法只能用于初始化，而不能分配。你不能这样做，例如：

char p[2][5]; 
p = {"a", "b"}; //error 

这就是为什么它无法编译。通过串

strcpy(s->suits[0], "hearts"); 
strcpy(s->suits[1], "spades"); 
...etc 

尝试strcpy -ing串，或者初始化一个临时数组，然后将其复制

char suits_tmp[4][9] = {"hearts","spades","clubs","diamonds"}; 
memcpy(s->suits, suits_tmp, 4*9); 

Answer 2

使用const char *内的struct（我假设没有要求修改实际内容西装/面值），并分别初始化它们：

struct cards { 
    const char *suits[4]; 
    const char *faces[13]; 
}; 

typedef struct cards cards; 

void init_struct(cards *s) 
{ 
    s->suits[0] = "hearts"; 
    s->suits[1] = "spades"; 
    s->suits[2] = "clubs"; 
    s->suits[3] = "diamonds"; 
    s->faces[0] = "ace"; 
    s->faces[1] = "two"; 
    s->faces[2] = "three"; 
    s->faces[3] = "four"; 
    s->faces[4] = "five"; 
    s->faces[5] = "six"; 
    s->faces[6] = "seven"; 
    s->faces[7] = "eight"; 
    s->faces[8] = "nine"; 
    s->faces[9] = "ten"; 
    s->faces[10] = "jack"; 
    s->faces[11] = "queen"; 
    s->faces[12] = "king"; 
} 

当然，如果你只是想一次性套卡，这是合理的，那么这将工作：

struct 
{ 
    const char *suits[4]; 
    const char *faces[13]; 
} cards = 
{ 
    {"hearts","spades","clubs","diamonds"}, 
    {"ace","two","three","four","five", 
     "six","seven","eight","nine", 
     "ten","jack","queen","king"} 
}; 

Answer 3

#include <string.h> 
#include <stdio.h> 

struct cards { 
    const char** suits; 
    const char** faces; 
}; 

typedef struct cards cards; 

const char* suits[4] = {"hearts","spades","clubs","diamonds"}; 
const char* faces[13] = {"ace","two","three","four","five", 
       "six","seven","eight","nine" 
       "ten","jack","queen","king"}; 
int main() 
{ 
    cards deck; 
    deck.suits = suits; 
    deck.faces = faces; 
    printf(deck.suits[0]); 
    return 0; 
} 

这工作也是如此。不使用指针。

澄清

我知道我是快速和肮脏的答案，但没有strcpy或memcpy或作业一长串。如果你的计划是为你的游戏使用标准的一副牌，那么无论如何它都会是一组不变的值。如果你的意图是有不同类型的套牌，那么我的答案可能不够。是的，它不具有init_struct功能，但你可以很容易地修改它为你的意图（因为我没有很好地在C和malloc的精通。）

Answer 4

#include <string.h> 

typedef struct cards { 
    char faces[13][6], suits[4][9]; 
} cards; 

cards base_card = { 
    {"ace","two","three","four","five", 
    "six","seven","eight","nine", //forgot "," at nine after 
    "ten","jack","queen","king"}, 
    {"hearts","spades","clubs","diamonds"} 
}; 

void init_struct(cards *s) { 
    memcpy(s, &base_card,sizeof(cards)); 
}