Not being printed out when using fgets

Question

I'm learning about file input/output in C. But there is something I don't understand.
I was about to print the simple sentence in the [words.txt] using fgets, but it doesn't work.

[words.txt] as below:

This speech caused a remarkable sensation among the party.

And my code is :

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

int main()
{
    char *buffer = malloc(100);
    memset(buffer, 0, 100);

    FILE *fp = fopen("words.txt", "r");

    fgets(buffer, sizeof(buffer), fp);

    printf("%s\n", buffer);

    fclose(fp);

    free(buffer);

    return 0;
}

The output is "This sp"

I thought there was no problem with my code, but the output result was unexpected.

I mean, why isn't the whole string printing out?
Also, what is the reason for this output string?
(There seems to be no indication that this code will have 7 character,that is, "This sp")

Side note: The line `memset(buffer, 0, 100);` is redundant, as `fgets` will write a terminating null character into `buffer`. — Andreas Wenzel, Nov 28 '21 at 19:17

Scouarn. · Answer 1 · 2021-11-28T19:25:48.803

Your buffer is declared as a char* and not a fixed size array so sizeof(buffer) is 8, hence the 7 characters (plus a '\0') being read and printed. sizeof doesn't know how many bytes you asked to malloc anyway. I think this is the most confusing thing when you start learning about pointers and realizing arrays work like pointers but not quite.

Solution :

fgets(buffer, 100, fp);

Or

char buffer[100];
fgets(buffer, sizeof(buffer) fp);

Andreas Wenzel · Accepted Answer · 2021-11-28T19:41:04.930

Since you declared buffer as char *buffer, it is not an array. Instead, it is simply a pointer.

The expression sizeof buffer will evaluate to the size of the pointer itself, not the size of the object that the pointer is pointing to. Since pointers typically have a size of 32 or 64 bits, the expression sizeof buffer will evaluate to either 4 or 8. In your case, it seems to be 8.

If you had declared buffer instead like this

char buffer[100];

then buffer would not be a pointer, but an array, and the expression sizeof buffer would have evaluated to 100.

In your case, there is nothing wrong with using malloc, instead of an array. However, if you use malloc, you must remember the size of the allocated memory yourself. You cannot use the sizeof operator for this.

Instead of writing the number 100 in several places in your code, you can use a preprocessor macro for this:

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

#define ALLOCATION_SIZE 100

int main()
{
    char *buffer = malloc( ALLOCATION_SIZE );
    memset(buffer, 0, ALLOCATION_SIZE );

    FILE *fp = fopen("words.txt", "r");

    fgets(buffer, ALLOCATION_SIZE, fp);

    printf("%s\n", buffer);

    fclose(fp);

    free(buffer);

    return 0;
}

This has the advantage that if you ever decide to change the number 100 to some other number, you only have to change the number in one place in your code, instead of several places.

ohh thanks a lot !!!! Btw, is there any way to know the size of the allocated memory while executing program without using that way or remembering? — y J., Nov 28 '21 at 19:41
@yJ.: That is not possible in ISO C (i.e. the common standard). However, individual operating systems may provide a way to do this. See [this question](https://stackoverflow.com/q/1281686/12149471) for further information. But I do not recommend doing this, except maybe as an academic exercise. It is generally easier to simply remember the size of the allocation, for example by storing it into a variable. — Andreas Wenzel, Nov 28 '21 at 19:44

Not being printed out when using fgets

2 Answers2