Why does each element of my array automatically turn into the last element?

Question

char *string[n];    
for (int i = 0; i < n; i++) {
    char string[60];
    fscanf(fptr, "%s%*[^\n]", string);
    strings[i] = string;
    printf("%s\n", strings[i]);
}
puts("Ended");
for (int i = 0; i < n; i++) {
printf("%s\n", strings[i]);

This is part of my code in C ^. strings[] contains pointers to the first word of each line in the file. So if my file contains:

ABC 123
DEF 123
GHI 123

I get the following output:

ABC
DEF 
GHI
Ended
GHI
GHI
GHI

Which means there's nothing wrong with the initialisation. Then what is going on?

`strings[i] = string;` stores a pointer to an array that goes out of scope. Your `strings` array is full of *dangling pointers* — UnholySheep, May 29 '20 at 18:23
You need to store pointers to memory that outlives the scope of the first `for` loop, e.g.: allocated using `malloc` — UnholySheep, May 29 '20 at 18:32

Henry Le Berre · Accepted Answer · 2020-05-29T18:55:06.000

3

When you wrote :

strings[i]=string

You were not copying the original string into the array because the “string” variable only contains a pointer (memory address) of the stack allocated string. When you go into the next iteration of the for loop, the stack allocated “string” variable gets freed/thrown away.

So your “strings” array contains pointers to unallocated memory. You should use “strcpy” instead. If the size of your “strings” array is not constant, use “malloc” then “strcpy”.

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

int main()
{
    FILE *fptr = fopen("test.txt", "r");
    char strings[10][60];
    int n = 3;
    for (int i = 0; i < n; i++)
    {
        char string[60];
        fscanf(fptr, "%s%*[^\n]", string);
        strcpy(strings[i], string);
        printf("%s\n", strings[i]);
    }
    puts("Ended");
    for (int i = 0; i < n; i++)
    {
         printf("%s\n", strings[i]);
    }
}

Or, simpler:

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

int main()
{
    FILE *fptr = fopen("test.txt", "r");
    char strings[10][60];
    int n = 3;
    for (int i = 0; i < n; i++)
    {
        fscanf(fptr, "%s%*[^\n]", strings[i]);
        printf("%s\n", strings[i]);
    }
    puts("Ended");
    for (int i = 0; i < n; i++)
    {
        printf("%s\n", strings[i]);
    }
}

edited May 29 '20 at 18:55

answered May 29 '20 at 18:31

Henry Le Berre

910
9
18

Thanks for the answer! Could you explain why "If the size of your “strings” array is not constant, use “malloc” then “strcpy"? – Archer May 29 '20 at 18:35
Yes sorry if I was unclear, I was talking about the size/length of each string. For instance, if you know you will have 10 strings and that each string is fewer than 60 characters, you can initialize the “strings” array in the following manner : “char strings[10][60]” – Henry Le Berre May 29 '20 at 18:39
`char string[60];` `fscanf(fptr, "%s%*[^\n]",string);` `strcpy(strings{i], string);` – Archer May 29 '20 at 18:42
How does your “strings” array look? – Henry Le Berre May 29 '20 at 18:44
You could also use fscanf on strings[i] instead of creating the “string” variable – Henry Le Berre May 29 '20 at 18:46
The program doesnt execute succesfully. There's a segmentation fault at the strcpy line – Archer May 29 '20 at 18:46
it works for me, I'm posting my full source code of my working fix – Henry Le Berre May 29 '20 at 18:53
Please consider accepting my answer if it solved your issue in order to make show to other with the same issue that it is a viable solution. Thanks! – Henry Le Berre May 29 '20 at 18:57
@HenryLeBerre consider if `n > 3` (for example `n=10`) – Hitokiri May 29 '20 at 19:07
I'm looking into it – Henry Le Berre May 29 '20 at 19:31
I works perfectly for me as long as the file has enough lines (where n <= lineCount) – Henry Le Berre May 29 '20 at 19:33
Is your file long enough and is your array large enough (char strings[n][60]) – Henry Le Berre May 29 '20 at 19:38
1

If would be safer to check if `fscanf(fptr, "%s%*[^\n]", strings[i]);` returns `1` and break from the loop if not, adjusting the number of strings read. – chqrlie May 29 '20 at 20:21
You are right but I didn’t say anything since it is not the question but he should still add it – Henry Le Berre May 29 '20 at 20:31
You have used a 2D array, what's the harm in using a 1D array the way I have done? – Archer May 30 '20 at 12:02
Your code would be slower since you would have use malloc, but you you have the flexibility of having each string be any length! Did my solution work out for you? – Henry Le Berre May 30 '20 at 12:06
Yes, it works. Why do we have to malloc with single array and not with 2D array? – Archer May 30 '20 at 12:19
I'm glad to hear that! The so called "1D" array is a list of pointers that point to the place in memory of the string they represent. You have to use malloc to allocate the area in memory where each string is stored. However, with the "2D" array, you specify the maximum size of each string so you don't have to dynamically allocate them. – Henry Le Berre May 30 '20 at 12:22
So 2D array is stored in heap by default? – Archer May 30 '20 at 12:29
It is actually stored on the stack (in order to store on the stack, you have to know at compile-time how much memory you need) – Henry Le Berre May 30 '20 at 12:29
I need to pass strings to another function and access all its strings, how can I do that? – Archer May 30 '20 at 13:02
Hi! https://stackoverflow.com/questions/30999951/passing-array-of-strings-to-functions-c should help you out! – Henry Le Berre May 30 '20 at 15:28

Why does each element of my array automatically turn into the last element?

1 Answers1