"Program received signal SIGSEGV, Segmentation fault." 来自 gdb 调试器的错误消息

Question

while 循环中的某些东西给我这个错误。我不知道要查找什么，因为这个错误似乎很常见，可以找出如何处理我的特定示例

#include <stdlib.h>
#include <stdio.h>
/*Structure declarations*/
struct Data {
    int id_number;
    float ratings;
};
typedef struct Node{
    struct Data player;
    struct Node *next;
}Node;

/*function declaration*/
void push(struct Node *list_head, int unique_id, float grade);
int main(){
    /* Initialize list_head to Null since list is empty at first */
    Node *list_head = NULL;     
    Node *traversalPtr;

    push(list_head, 1, 4.0);
    push(list_head, 2, 3.87);
    push(list_head, 3, 3.60);

    traversalPtr = list_head;
    while(traversalPtr -> next != NULL){
        printf("%d\n",traversalPtr -> player.id_number);
        traversalPtr = traversalPtr -> next;
    }   
}

...function declarations

Answer 1

问题是函数

void push(struct Node *list_head, int unique_id, float grade);

处理 main 中定义的原始指针的副本，因为指针是按值传递的。

你应该像这样声明函数

void push(struct Node **list_head, int unique_id, float grade);

并这样称呼它

push( &list_head, 1, 4.0 );

这是一个如何定义函数的示例（我假设函数将节点附加到它的尾部）。

int push(struct Node **list_head, int unique_id, float grade)
{
    struct Node *node = malloc( sizeof( struct Node ) );
    int success = node != NULL;

    if ( success )
    {
        node->player.id_number = unique_id;
        node->player.ratings   = grade; 
        node->next = NULL;

        while ( *list_head ) list_head = &( *list_head )->next;

        *list_head = node;
    }

    return success; 
}

也是这个循环

traversalPtr = list_head;
while(traversalPtr -> next != NULL){
    printf("%d\n",traversalPtr -> player.id_number);
    traversalPtr = traversalPtr -> next;
}

不正确。它应该看起来像

traversalPtr = list_head;
while(traversalPtr != NULL){
    printf("%d\n",traversalPtr -> player.id_number);
    traversalPtr = traversalPtr -> next;
}

"Program received signal SIGSEGV, Segmentation fault." 来自 gdb 调试器的错误消息

"Program received signal SIGSEGV, Segmentation fault." error message from gdb debugger

c

linked-list

segmentation-fault

singly-linked-list