Inaccurate size of string while dynamic memory allocation

Question

As per the image, you can see I have inserted whole lot of text but the total length by strlen is only showing 8 (k=8).

I wanted to print full text inserted into one string a line but it is stopping in 8th character.

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

int main() {
    
    int i;
    char *s;
    s = malloc(1024 * sizeof(char));
    scanf("%[^\n]", s);
    s = realloc(s, strlen(s) + 1);
    int k=sizeof(strlen(s));
     for(i=0;i<sizeof(strlen(s)+1);i++)
    {
        if(*(s+i)!=' ')
            printf("%c",*(s+i));
        else
            printf("\n");
      fflush(stdin);
    }
    free(s);
    return 0;
}`

```int k=sizeof(strlen(s))``` ----> What are you trying to do with this statement? — Harith, Jan 08 '23 at 11:12
`sizeof(strlen(s))` is the same as `sizeof(size_t)` -- it doesn't actually call strlen, just gives the size of its return type. — Chris Dodd, Jan 08 '23 at 11:17
You don't actually use the incorrect `int k = sizeof(strlen(s));`. I suggest `int len = strlen(s);` move it up and use it instead of repeatedly calling `strlen(s)`. — Weather Vane, Jan 08 '23 at 11:24
What is the purpose of `realloc` ? It wont change `strlen(s)` — Support Ukraine, Jan 08 '23 at 11:32
The length of the string is `int k=strlen(s);` : `sizeof` is for types only. — QuentinUK, Jan 08 '23 at 11:46
Yeah, I m stupid enough to end up there and I was confused for hours. This is actually a hacker rank question so Extention and realloc was there already ; why realloc was there I didn't get it either. Thank you very much it was actually the case. — Ghimire Suraj, Jan 08 '23 at 12:07

score 2 · Accepted Answer · answered Jan 08 '23 at 12:20

First, the sizeof operator returns the size of the type in bytes, not the number of elements in the array. To get the number of elements in s, you should use strlen(s) instead.

Second, the scanf function will stop reading input when it encounters a newline character, so if you enter a string that is longer than 8 characters, it will only read the first 8 characters. To read a string that may contain spaces, you can use fgets instead:

fgets(s, 1024, stdin);

Finally, it's not necessary to call realloc after calling scanf or fgets, because the size of s is already 1024 characters, which is more than enough to hold the input string.

With these changes, your code should work as expected. Here's the modified version:

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

int main() {
    int i;
    char *s;
    s = malloc(1024 * sizeof(char));
    fgets(s, 1024, stdin);
    int k = strlen(s);
    for(i = 0; i < k; i++) {
        if(*(s+i) != ' ') {
            printf("%c", *(s+i));
        } else {
            printf("\n");
        }
    }
    free(s);
    return 0;
}

score 1 · Answer 2 · answered Jan 08 '23 at 11:41

1

sizeof(strlen(s)+1) is giving you the size of the size_t type (return type of strlen function) not the length of the string
i should have type size_t
fflush in the input stream is not needed and is strongly implementation-defined. Read:Is this the proper way to flush the C input stream?
You use realloc incorrectly. If it fails it will result in a memory leak. This realloc is not needed anyway. But you need to save the result in the temporary variable to not loose reference to the allocated memory.

char *tmp = realloc(s, strlen(s) + 1);
if(tmp) s = tmp
else { /* handle error */}

answered Jan 08 '23 at 11:41

0___________

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Thank you very very much. It is really helpful but can you please explain me about size_t in this – Ghimire Suraj Jan 08 '23 at 12:08
1

@GhimireSuraj See [What is `size_t` in C?](https://stackoverflow.com/questions/2550774/what-is-size-t-in-c) – pzaenger Jan 08 '23 at 12:26
1

`size_t` is the correct type to store the sizes. instead `int i;` => `size_t i;` – 0___________ Jan 08 '23 at 12:26

Inaccurate size of string while dynamic memory allocation

2 Answers2