Array of Char in C

Question

I'm trying to create a function that get a string, taking char by char:

char * GetString (int dim) // dim is the size of the string
{
    char c, *S;
    int i = 0;

    S = (char*)malloc(dim*(sizeof(char));

    while ((c=getchar() != '\n') && (i < dim) && (c != EOF))
    {
        S[i] = c;
        i++;
    }

    S[i] = '/0';

    return S;
}

The problem is when i try to use a "printf" in this function, trying to see if the input was taken correctly, it shows me strange characters, not the ones i inserted. I don't know what i'm missing.

Thanks very much.

[**DO NOT** cast the return value of `malloc()`](https://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc/605858#605858) — The Paramagnetic Croissant, Mar 07 '15 at 10:05
`c` should be an `int` and don't cast the result of `malloc`. — Spikatrix, Mar 07 '15 at 10:06
Unrelated to the problem, but `sizeof(char)` is 1 by definition, so multiplying by it is pointless. — Arkku, Mar 07 '15 at 10:15
Do you compile on a 64bit system **and** miss to `#include` `stdlib.h`? — alk, Mar 07 '15 at 10:18
CodeBlocks with GCC, no got 0 warnings. I think the problem is when i do: S[i] = c, because if i print the value of c = getchar() che char is taken correctly, but if i do S[i] = c or *(S) = c and try to print S, it shows me strange characters. — Mauri, Mar 07 '15 at 10:28

Jonathon Reinhart · Answer 1 · 2015-03-07T10:09:44.227

6

This should be a backslash:

S[i] = '\0';

Make sure your compiler warnings are turned on and you're paying attention to them. '/0' is a non-standard multi-character integer literal.

Backslash zero '\0' is the C string NUL-terminator you're after.

Also, in the case where your input is larger than dim, you are overrunning your buffer, writing one byte more than you allocated. I would suggest allocating dim+1 bytes.

Finally, there's no point in multiplying by sizeof(char) - it is 1 by definition.

P.S. Now is an excellent time to learn how to use a debugger. If you're using GNU tools: gcc -Wall -Werror main.c then gdb a.out, and type r to run.

edited Mar 07 '15 at 10:09

answered Mar 07 '15 at 10:04

Jonathon Reinhart

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5

also he's not allocating space for that character. – The Paramagnetic Croissant Mar 07 '15 at 10:04
Thanks very much, i edited but nothing, same output. – Mauri Mar 07 '15 at 10:08
1

For completeness: `S[i] = 0;` would do also. – alk Mar 07 '15 at 10:22

Vlad from Moscow · Answer 2 · 2015-03-07T10:26:09.237

3

This '/0' is a character constant which value is implementation defined. So you have to replace it with escape character '\0'.

Also you have to allocate dim + 1 bytes if you want to append exactly dim characters with the terminating zero.

The function can look like

char * GetString( int dim )
{
    char *s;
    int i;

    s = ( char* )malloc( ( dim + 1 ) *( sizeof( char ) );

    i = 0;
    while ( i < dim && ( ( c = getchar() ) != '\n') && ( c != EOF ) )
    {
        s[i++] = c;
    }

    s[i] = '\0';

    return s;
}

edited Mar 07 '15 at 10:26

answered Mar 07 '15 at 10:10

Vlad from Moscow

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I tryed with your code but nothing, the results is the same.. for example, i insert "aaa" and it prints three shi**y smiley ! – Mauri Mar 07 '15 at 10:15
@Mauri I do not see what code you are running. The code snippet I showed is vlaid. – Vlad from Moscow Mar 07 '15 at 10:17
"*vlaid*" nice typo! ;-) – alk Mar 07 '15 at 10:20
I'm using your snippet but got the same problem. – Mauri Mar 07 '15 at 10:20
@ Mauri See my updated post and try the function one more by copy and paste it. There was a typo. – Vlad from Moscow Mar 07 '15 at 10:27
IT WORKS, what changes? – Mauri Mar 07 '15 at 10:32
THE fu***** parenthesis in c != getchar()... Thanks very much mate. – Mauri Mar 07 '15 at 10:37
@Mauri , Please stop using bad language!! That attitude might get you suspended or banned. – Spikatrix Mar 07 '15 at 10:51

score 2 · Answer 3 · answered Mar 07 '15 at 10:38

2

Thanks to Vlads from Moscow for the right snippets. The problem was here:

while (c = getchar() != '\n')

It's wrong, the right way:

while ((c=getchar()) != '\n')

answered Mar 07 '15 at 10:38

Mauri

41
7

If Vlad's answer was useful to you, vote it up. You may want to comment on the answer itself. – pmg Mar 07 '15 at 11:59
I did it, i couldn't first because i had a low level. – Mauri Mar 07 '15 at 12:18
Not enough rep to comment is not an excuse to post a comment as an answer. – Lightness Races in Orbit Mar 08 '15 at 03:08
Man are you serious? i've found the right way, haven't i to tell others? bah.. you think more for a reputation of a true problem solved.. good logic man – Mauri Mar 21 '15 at 08:56

score 1 · Answer 4 · answered Mar 07 '15 at 11:49

The code had an amazingly simple problem: you missed parentheses around c=getchar(). Try ((c=getchar()) != '\n') instead of (c=getchar() != '\n') The code has other problems also, like using /0 instead of \0

Here is the rectified code:

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

char * GetString (int dim) // dim is the size of the string
{
    char c, *S;
    int i = 0;

    S = (char*)malloc(dim*(sizeof(char)));

    while (((c=getchar()) != '\n') && (i < dim) && (c != EOF))
    {
        S[i] = c;
        i++;
    }

    S[i] = '\0';

    return S;
}

int main()
{
    char *s;

    s = GetString(10);
    printf("%s\n", s);
    free(s);

    return 0;
}

Array of Char in C

4 Answers4