2014-09-25 56 views
0

我正在努力讓一個結構指向另一個依賴於傳入命令行的參數,問題是,結構我看起來正在指向所需的初始化結構但是,當我在函數調用後打印它們的地址時,在main中,如果播放器是A,則它看起來沒有改變(在輸出之後):指向一個結構到另一個? (不是永久性的)

Before initialise: 0x7f8a88403990, 0x7f8a884039f0, 0x7f8a88403a50, 0x7f8a88403ab0, 0x7f8a88403b10 
After initialise: 0x7f8a884039f0, 0x7f8a884039f0, 0x7f8a88403a50, 0x7f8a88403ab0, 0x7f8a88403b10 
After parse args: 0x7f8a88403990, 0x7f8a884039f0, 0x7f8a88403a50, 0x7f8a88403ab0, 0x7f8a88403b10 

int main (int argc, char *argv[]) { 

Player *me = NULL, *playerA = NULL; 
Player *playerB = NULL, *playerC = NULL, *playerD = NULL; 

me = malloc(sizeof(*me)); 
playerA = malloc(sizeof(*playerA)); 
playerB = malloc(sizeof(*playerB)); 
playerC = malloc(sizeof(*playerC)); 
playerD = malloc(sizeof(*playerD)); 

parse_args(me, playerA, playerB, playerC, playerD, argv); 

//should be pointing to the same memory location 
printf("After parse args: %p, %p, %p, %p, %p\n", me, playerA, playerB, playerC, playerD); 

} 

void parse_args(Player *me, Player *a, Player *b, Player *c, Player *d, 
      char *argv[]) { 
initialise_game(*tempChar, tempNum, me, a, b, c, d); 

} 

void initialise_game(char playerID, int numPlayers, Player *me, Player *a, 
       Player *b, Player *c, Player *d) { 
printf("Before initialise: %p, %p, %p, %p, %p\n", me, a, b, c, d); 

switch((int)playerID) { 
    case 'A': 
     me = a; 
     break; 
    case 'B': 
     me = b; 
     break; 
    case 'C': 
     if (numPlayers < 3) { 
      exit_prog(EXIT_PLAYERID); 
     } 
     me = c; 
     break; 
    case 'D': 
     if (numPlayers < 4) { 
      exit_prog(EXIT_PLAYERID); 
     } 
     me = d; 
     break; 
} 

printf("After initialise: %p, %p, %p, %p, %p\n", me, a, b, c, d); 

} 

回答

1

指針通過值傳遞。這意味着你不能在函數體內修改它們()。看到這個StackOverflow的question/answer

我想你想做的事是這樣的:

void parse_args(Player **me, Player *a, Player *b, Player *c, Player *d, char *argv[]) { 
    initialise_game(*tempChar, tempNum, me, a, b, c, d); 
} 

void initialise_game(char playerID, int numPlayers, Player **me, Player *a, 
      Player *b, Player *c, Player *d) 
{ 
    printf("Before initialise: %p, %p, %p, %p, %p\n", me, a, b, c, d); 
    switch((int)playerID) { 
    case 'A': 
     *me = a; 
     break; 
    case 'B': 
     *me = b; 
    break; 
    case 'C': 
     if (numPlayers < 3) { 
      exit_prog(EXIT_PLAYERID); 
     } 
     *me = c; 
     break; 
    case 'D': 
     if (numPlayers < 4) { 
      exit_prog(EXIT_PLAYERID); 
     } 
     *me = d; 
     break; 
} 

您對parse_args通話將需要一個指針傳遞給一個指針對我來說:

parse_args(&me, playerA, playerB, playerC, playerD, argv); 
+0

謝謝你提供了一個明確的答案,我現在明白了這項工作是如何的(對不起,我的頭腦有些棘手) – user3603183 2014-09-25 22:50:25

+0

通過價值傳遞指針並嘗試修改它們是棘手的 - 它捕捉到許多新的C程序員未料到。我鏈接到的Stackoverflow問題可以幫助你非常。但是如果你想修改一個函數中的指針,你需要傳遞它的地址,然後在函數中去引用來完成賦值 – 2014-09-25 22:54:08

0

您需要使用指向函數參數結構的指針(例如,Player **x)。使用指向結構的指針只能修改結構對象,而不能修改指針對象。記住所有的函數參數都是通過C中的值傳遞的,如果你修改函數中的指針參數,你只是修改指針的一個副本。

+0

哪些功能會我需要那些參數?我剛剛嘗試了它自己的parse_args,它自己的initialise_game,並且兩者在一起,但他們似乎並沒有工作...... – user3603183 2014-09-25 22:33:09

+0

@ user3603183所有這些,但你必須正確地將參數傳遞給函數並正確解引用函數中的指針。在你的編譯器中啓用警告,如果有的話修復它們。 – ouah 2014-09-25 22:35:34

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