2013-10-07 41 views
-2

我創建了二維數組來模擬緩存。對於每個緩存行,我使用struct來定義它。當我想初始化cahce時,使用malloc時出現錯誤。我在代碼中標記了錯誤的地方。 謝謝!使用malloc分配內存失敗

typedef struct { 
    int valid; 
    int tag; 
    int lruIndex; 
} Line; 

Line** initCache(int s, int E){ 

    int i, j; 
    //int setSize = sizeof(Line *); 
    //int lineSize = sizeof(Line); 
    /* allocate memory to cache */ 
    //printf("%d %d\n", setSize, lineSize); 
    Line** cache = NULL; 
    cache = (Line **)malloc((1 << s) * sizeof(Line *)); //set 

    //check for memory error 
    if (!cache) 
    { 
     printf("%s\n", "allocate memory failed 1111111"); 
     exit(-1); 
    } 

    for (i = 0; i < (1 << s); i++){ 
     cache[i] = (Line *)malloc(E * sizeof(Line)); <<<<<<< i think here something is wrong, cache[i] returns NULL and then print "allocate memory failed 22222222" 

     //check for memory error 
     if (cache[i]) 
     { 
      printf("%s\n", "allocate memory failed 22222222"); 
      exit(-1); 
     } 

     for(j = 0; j < E; j++){ 
      cache[i][j].valid = 0; //initial value of valid 
      cache[i][j].lruIndex = j; //initial value of lruIndex 0 ~ E-1 
     } 
    } 

    return cache; 
} 
+0

[不要轉換'malloc()']的返回值(http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc/605858#605858) – 2013-10-07 19:36:48

+1

'使用malloc時出錯了嗎?這是什麼東西' – exexzian

+0

's'和'E'的值是什麼?什麼平臺? –

回答

1
if (cache[i]) 
    { 
     printf("%s\n", "allocate memory failed 22222222"); 
     exit(-1); 
    } 

退出時cache[i]是!= NULL,拿什麼當內存分配退出。

若要正確的方式改變條件:

(cache[i]==NULL)

這將評估爲TRUE,當malloc的失敗。

1
malloc((1 << s) * sizeof(Line *) 

可以的malloc(0 * 4)呢!

也可以用完內存。