C - How to input and insert a string into an array of strings?

Question

I want to insert string to the array until I type "ok". Why I am getting just "ok" and original array at the output?

int main(void)
{
    char b[20];
    char* str[10] = { "1","2" };
    int i = 2;
    while (1) {

        gets(b);
        if (strcmp(b, "ok") == 0) break;
        str[i] = b;
        i++;
    }

    for (int j = 0; j < i; j++)
        printf("%s ", str[j]);
    return 0;
}

`char* str[10]` is an array of *pointers*. You are storing a *pointer* to the input buffer. They all point to the *same* buffer. Therefore they all point to the most recent input data. — Weather Vane, Aug 30 '19 at 19:29
Please read [Why is the gets function so dangerous that it should not be used?](https://stackoverflow.com/questions/1694036/why-is-the-gets-function-so-dangerous-that-it-should-not-be-used) — Weather Vane, Aug 30 '19 at 19:37
Code like this suggests you need to spend more time understanding C strings and the implications of pointers vs. arrays. It's a conceptual block you *need* to have before you can be effective in C. There's many resources, both video and print, which can help you here. — tadman, Aug 30 '19 at 19:45
@tadman - I used to 'love the challenge' of coding strings in C. I know C is a great language and worth learning and there are historical reasons for the way its strings work. Also now that I've had java and c# for so long, I see the truth about c strings - they suck! — FastAl, Aug 30 '19 at 20:17
@FastAl C was from an era where 100KB of memory was a luxury, so everything is super simple and a lot of the responsibility falls to the programmer. It's still relevant in some parts of the embedded world, but it's *still* super difficult to get your C code 100% correct. A lot of open-source C projects spend innumerable hours dealing with C string issues, especially the never-ending risk of *buffer overflow bugs*. — tadman, Aug 30 '19 at 20:20

score 0 · Answer 1 · answered Aug 30 '19 at 19:31

0

They all point to b, which gets overwritten in each iteration.

answered Aug 30 '19 at 19:31

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Nathan Xabedi · Accepted Answer · 2019-08-30T20:09:58.577

0

You need to allocate a string on each iteration:

int main(void)
{
    char* b;
    char* str[10] = { "1","2" };
    int i = 2;
    while (1) {
        b = malloc(20);
        gets(b);
        if (strcmp(b, "ok") == 0) break;
        str[i] = b;
        i++;
    }

    for (int j = 0; j < i; j++)
        printf("%s ", str[j]);

    // free allocated strings
    while (i > 2)
        free(str[--i]);

    return 0;
}

edited Aug 30 '19 at 20:09

answered Aug 30 '19 at 19:37

Nathan Xabedi

1,097
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`while (i > 2) free(str[--i]);` – Weather Vane Aug 30 '19 at 19:40
@Weather Vane (y) – Nathan Xabedi Aug 30 '19 at 19:46
Nathan Xabedi, Curious, why the unneeded `(char*)` cast in `b = (char*) malloc(20);`? – chux - Reinstate Monica Aug 30 '19 at 19:59

Bob Jarvis - Слава Україні · Answer 3 · 2019-08-30T23:32:16.130

0

You need to make a copy of the input string, then save a pointer to the copy of the input string in your array. Something like:

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

int main(void)
  {
  char b[20];
  char *str[10] = { "1","2" };
  int i = 2;
  char *p;
  size_t lenb;

  for(i = 2 ; i < 10 ; ++i)
    {
    fgets(b, sizeof(b), stdin);

    lenb = strlen(b);

    if(lenb > 0 && *(b+lenb-1) == '\n')
      {
      *(b+lenb-1) = '\0';  /* overwrite the trailing \n */
      lenb = strlen(b);
      }

    if (strcmp(b, "ok") == 0)
      break;

    p = malloc(lenb+1);
    strcpy(p, b);

    str[i] = p;
    }

  for (int j = 0; j < i; j++)
    printf("%s\n", str[j]);

  return 0;
  }

edited Aug 30 '19 at 23:32

answered Aug 30 '19 at 19:41

Bob Jarvis - Слава Україні

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It would be better to read the user input into the allocated array. Copying is then unnecessary. – Nathan Xabedi Aug 30 '19 at 19:50
Note that `if(*(b+strlen(b)-1) == '\n')` allows a hacker exploit of causing UB by having the first letter entered as `'\0'`. Consider `b[strcspn(b, "\n")] = 0;` instead. – chux - Reinstate Monica Aug 30 '19 at 20:02

S.S. Anne · Answer 4 · 2019-08-30T20:30:07.583

You need to create a copy of the string when you assign it:

str[i] = strdup(b);

You also may consider using fgets instead of gets; however, you will need to remove the newline:

size_t size;
fgets(str, 20, stdin);
size = strlen(str);
if(str[size-1] == '\n')
    str[size-1] = '\0';

Also, print a newline at the end of your program, so it won't interfere with the shell:

putchar('\n');

Full code:

int main(void)
{
    char b[20];
    char* str[10] = { "1","2" };
    int i = 2;
    while (1) {
        size_t size;
        fgets(str, 20, stdin);
        size = strlen(str);
        if(str[size-1] == '\n')
             str[size-1] = '\0';
        if (strcmp(b, "ok") == 0)
            break;
        str[i] = strdup(b);
        i++;
    }

    for (int j = 0; j < i; j++)
        printf("%s ", str[j]);
    putchar('\n');
    return 0;
}

C - How to input and insert a string into an array of strings?

4 Answers4