分配結構=結構

分段錯誤我試圖創建一個臨時的'迭代器'結構，它被分配到'列表'的開頭，然後通過檢查iterator->next != NULL遍歷結構列表。我相信這個問題在iterator = start行（35 & 70）。分配結構=結構

該應用程序編譯時沒有任何問題，但是當我應用程序出現分段錯誤時（核心轉儲）。

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

struct record 
{ 
    int    accountno; 
    char    name[25]; 
    char    address[80]; 
    struct record*  next; 
}; 

int  addRecord (struct record **, int, char [], char []); 
void printAllRecords(struct record *); 

int main(int argc, char *argv[]) { 
    struct record ** start; 
    start = NULL; 

    addRecord(start, 1, "Record Name", "Record Address"); 
    printAllRecords(*start); 

    return 0; 
} 

void printAllRecords(struct record * start) 
{ 
    struct record * recordIterator; 

    /* Allocate the required memory and return a pointer to it */ 
    recordIterator = malloc(sizeof(struct record)); 

    /* Start at the beginning */ 
    recordIterator = start; 

    printf("\n\n%10s %20s %20s\n", "accountno", "Name", "Address"); 

    while (recordIterator != NULL) 
    { 
     printf("%10d %20s %20s\n", recordIterator->accountno, recordIterator->name, recordIterator->address); 
     recordIterator = recordIterator->next; 
    } 
} 

int addRecord (struct record ** start, int accountno, char name[], char address[]) 
{ 
    struct record * newRecord; 

    /* Allocate the required memory and return a pointer to it */ 
    newRecord = malloc(sizeof(struct record)); 

    /* Assign values to the new record */ 
    newRecord->accountno = accountno; 
    strcpy(newRecord->name, name); 
    strcpy(newRecord->address, address); 

    if (start == NULL) 
    { 
     start = &newRecord; 
    } 
    else 
    { 
     struct record * recordIterator; 

     /* Allocate the required memory and return a pointer to it */ 
     recordIterator = malloc(sizeof(struct record)); 

     /* Start at the beginning */ 
     recordIterator = *start; 

     while (recordIterator->next != NULL) 
     { 
      recordIterator = recordIterator->next; 
     } 

     recordIterator->next = newRecord; 
    } 

    return 1; 
}

來源

2017-02-13 kneeki

'addRecord'中的'start'是本地副本。函數返回後，這種更改不會被反映出來。 –

爲什麼只是動態分配內存才能立即拋出內存？你的程序泄漏內存......壞 – StoryTeller

沒有理由說明'start'應該是main中的指針指針。將其聲明爲一個普通指針，然後將其地址傳遞給該函數。 – Lundin

你可以聲明start是一個指針在

struct record * start;

然後你就可以通過addRecord(&start, ...)調用該方法。

裏面的方法：

int addRecord (struct record ** start, int accountno, char name[], char address[]) 
{ 
    struct record * newRecord; 

    /* Allocate the required memory and return a pointer to it */ 
    newRecord = malloc(sizeof(struct record)); 

    /* Assign values to the new record */ 
    newRecord->accountno = accountno; 
    strcpy(newRecord->name, name); 
    strcpy(newRecord->address, address); 

    if (*start == NULL) 
    { 
     *start = newRecord; 
    }

當週圍傳遞指針，一個函數中，請記住，你可以永久修改爲佔據地址，而不是地址本身的價值。在修改後的版本中，start的值不會改變（我們無法做到這一點......改變不會反映爲方法返回），但我們正在修改start指向的值。

來源

2017-02-13 08:40:40

這是問題所在。除了在main中分配'struct record ** start'問題（更改爲'struct record * start'）。 – kneeki

此行

addRecord(start, 1, "Record Name", "Record Address");

不會修改start。因此當您撥打printAllRecords時，start仍爲NULL。

來源

2017-02-13 08:31:47

分配結構=結構

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