while 循环中的 scanf 不断修改数组的先前值

Question

I'm trying to implement a hash table in C so the idea is simple:

1- the user enters a string 2- the program adds it to the table 3- prints the whole table

but I noticed that the program replaces all previous values with the new one:! THE CODE:

//edit
//initialize array to empty string
void init(char* table[]){
    for (int i=0; i<T_LEN; i++){
        table[i] = "";
    }
}

int hash(char* str){
    int result = 0;
    for (int i=0; i<sizeof(str); i++){
        result += str[i]; 
    }
    return result % T_LEN;
}

void add(char* table[]){
    char* input;
    printf("> ");
    scanf("%s", (char*)&input);
    int index = hash((char*)&input);
    table[index] = (char*)&input;
    return;
}

int main(){
    char* hash_table[T_LEN];
    init(hash_table);
    while(1){
        add(hash_table);
        print_table(hash_table);
    }
    return 0;
}

RESULT:

> abc
1         -------
2         -------
3         -------
4         -------
5         abc
6         -------
7         -------
8         -------
9         -------
10        -------
> esd
1         -------
2         -------
3         -------
4         -------
5         esd
6         -------
7         esd
8         -------
9         -------
10        -------
> ee
1         -------
2         -------
3         ee
4         -------
5         ee
6         -------
7         ee
8         -------
9         -------
10        -------

EDIT: the site says "It looks like your post is mostly code; please add some more details." so here I'm:D

Answer 1

some problems with your code:

1. return; at the end of the function called add is redundant which means that it has no meaning as in all the cases the function will end at this point

---

2. the compiler is giving the warning while(1) in the main as you don't return from the function, so you can do

   int __attribute__((noreturn)) main()

to tell the compiler that the main will never return

---

3. in the function called init , as you are declaring an array of pointers, then it's better to write:

    table[i] = malloc(sizeof(char) * MAX_NAME_LEN); table[i][0] = '\0';

instead of:

 table[i] = "";

to reserve memory in heap for the string not in the read-only memory.

---

4. in the lines:

    scanf("%s", (char*)&input); int index = hash((char*)&input);

the pointer called input is already of type char* , so you don't have to cast it as it's not a better practice to cast everything, also you should reserve a space for the input variable in the heap memory like:

   char* input = malloc(sizeof(char) * MAX_NAME_LEN);  

and so your code becomes:

    char* input = malloc(sizeof(char) * MAX_NAME_LEN);
    printf("> ");
    scanf("%s", input);
    int index = hash(input);
    table[index] = input;

---

5. instead of sizeof() in the line:

    for (int i=0; i<sizeof(str); i++)

I think you should use strlen instead like:

for (int i=0; i<strlen(str); i++)

as sizeof(str) = 4 as sizeof(pointer) = 4 as the pointers has fixed size

---

with all this being said, this is the edited code:

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

#define T_LEN 10
#define MAX_NAME_LEN    20

void print_table(char* table[])
{
    for (int i = 0; i < T_LEN; ++i) {
        printf("%d\t\t%s\n",i, table[i]);
    }
}

//edit
//initialize array to empty string
void init(char* table[]){
    for (int i=0; i<T_LEN; i++){
        table[i] = malloc(sizeof(char) * MAX_NAME_LEN);
        table[i][0] = '\0';
    }
}

int hash(char* str){
    int result = 0;
    for (int i=0; i<strlen(str); i++){
        result += str[i];
    }
    return result % T_LEN;
}

void add(char* table[]){
    char* input = malloc(sizeof(char) * MAX_NAME_LEN);
    printf("> ");
    scanf("%s", input);
    int index = hash(input);
    table[index] = input;
}

int __attribute__((noreturn)) main(){
    char* hash_table[T_LEN];
    init(hash_table);
    while(1){
        add(hash_table);
        print_table(hash_table);
    }
}

and this is some output:

>abc
0
1
2
3
4               abc
5
6
7
8
9
>esd
0
1
2
3
4               abc
5
6               esd
7
8
9
>ee
0
1
2               ee
3
4               abc
5
6               esd
7
8
9
>