Insert element into dynamic char array in C programming

Question

I was having some problem when trying to add element to dynamic char array in C programming. Here is the expected output:

How many characters do you want to input: 5
Input the string:datas
The string is: datas
Do you want to 1-insert or 2-remove or 3-quit?: 1
What is the character you want to insert: a
Resulting string: adata

I already did those user input part in the main function and here is the code in main where I take in the string input, size and pass them to insert():

printf("How many characters do you want to input: ");
scanf("%d", &n);
str = malloc(n + 1);
printf("Input the string class: ");
scanf("%s", str);

case '1':
    printf("What is the character you want to insert: ");
    scanf(" %c", &input);
    insert(str, input, n);
    break;

And the part where my insert():

void insert(char *str, char input, int n) {
int i;
size_t space = 1;
for (i = 0; i < n; i++) {
    str[i] = (char)(input + i);
    space++;                       
    str = realloc(str, space); 
    if (i > 2) {
        break;
    }
}

for (i = 0; i < n; i++) {
    printf("%c", str[i]);
}
}

However, when I tried to print out the string from insert(), let's say I entered 'a' to append to the first element of dynamic array with a size of 5, the result that I am getting is abcd=

I referenced from the stackoverflow thread and I not sure how to fix this. Thanks in advance.

Sorry but do you mean checking the input passing as parameter from main to insert()? I checked it and it was correct. Just that for example when I entered 'b', the output became bcde=. Then let's say q, the outout became qrst= etc etc — QWERTY
– QWERTY, Commented Oct 10, 2015 at 10:19
The function scanf returns a value - how successful it is. This should be checked to ensure that the relevant variable has been "filled in". People make typos and perhaps give them an opportunity to re-enter the data — Ed Heal
– Ed Heal, Commented Oct 10, 2015 at 10:21
(please read the manual page for scanf - linux.die.net/man/3/scanf — Ed Heal
– Ed Heal, Commented Oct 10, 2015 at 10:22
Why are you continuously storing input+i into the str buffer? Since input is a of course you then get b (a+1), c (a+2), etc. And why continuously realloc? You know exactly how long the final string needs to be. Just malloc once and be done with it. — kaylum
– kaylum, Commented Oct 10, 2015 at 10:22

Ed Heal · Accepted Answer · 2015-10-10 10:46:04Z

1

Here is the code - with the contract that the caller does the free bit! The caller calls it with insert(&str, input, n)

void insert(char **str, char input, int n) {

char* temp = *str;
int i;

*str = realloc(*str, n + 2); /* realloc first */

if(!*str) /* realloc failed */
{
    fputs("realloc failed", stderr);
    free(temp); /* Free the previously malloc-ed memory */
    exit(-1); /* Exit the program */
}

for (i = n; i >= 0; i--) {
    (*str)[i + 1] = (*str)[i]; /* Move all characters up */ 
}

(*str)[0] = input; /* Insert the new character */

printf("%s", *str); /* Print the new string */
}

Sorry about the formatting. That is left to the reader. I have not checked the algorithm but this does not leak memory

answered Oct 10, 2015 at 10:46

Ed Heal

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4 Comments

QWERTY Over a year ago

Thanks a lot! It works and thanks for the explanation too!

QWERTY Over a year ago

But short question, when you realloc the *str, the n is increase by 2. So can I assume that 1 space is for the new input whereas the other is for the '\0'?

Ed Heal Over a year ago

One assumes that the null character is there already. We just need to shift it. So the +2 could be +1

QWERTY Over a year ago

I see I see. Thanks a lot for the helps!

Spikatrix · Accepted Answer · 2015-10-10 10:45:39Z

1

You can use

void insert(char **str, char input, int n) {

    char* temp = *str;
    int i;

    *str = realloc(*str, n + 2); /* realloc first */

    if(!(*str)) /* realloc failed */
    {
        fputs("realloc failed", stderr);
        free(temp); /* Free the previously malloc-ed memory */
        exit(-1); /* Exit the program */
    }

    for (i = n; i >= 0; i--) {
        (*str)[i + 1] = (*str)[i]; /* Move all characters up */ 
    }

    **str = input; /* Insert the new character */

    printf("%s", *str); /* Print the new string */
}

And pass str by reference using

insert(&str, input, n); /* Note the '&' */

edited Oct 10, 2015 at 10:45

answered Oct 10, 2015 at 10:24

Spikatrix

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8 Comments

Ed Heal Over a year ago

This has a memory leak

Spikatrix Over a year ago

Yes. Fixed it and also a bug. Thanks.

Ed Heal Over a year ago

Not quite fixed - realloc can return a different address that is thrown away

fuz Over a year ago

@CoolGuy No. realloc() cannot magically update all pointers to str.

Spikatrix Over a year ago

@FUZxxl, Ed Heal, Fixed. Thanks!

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