What is the difference between int* and char* in context of my load function?

Question

Please take a look at the codes (both correct and incorrect) first-

incorrect code-

    char* buffer = malloc(*counter);
    char* ch = malloc(sizeof(char));
    for (*ch=fgetc(file);*c != EOF;*ch=fgetc(file))
    {
        buffer[*i] = *ch;
        *i = *i + 1;
    }
    *content = buffer;

Then I changed

char* ch = malloc(sizeof(char));

to-

int* ch = malloc(sizeof(int));

And everything worked fine (of course only the static data, but that is not a point i want to make)

and then to-

int* ch = malloc(sizeof(char));

and still everything alright.

1. What difference it makes to assign int* or char* to "ch".
2. I specified int* ch and size of an Int in malloc (in first correction) but my buffer elements are still of type char and of size of a char. Then how the code still works?

Answer 1

Because of internal casting in fgetc, this char* ch = malloc(sizeof(char)); can return a "false eof", and almost always does on binary (image) files. Some char in the image files is being cast to the int -1. EOF is defined (in stdio.h) as -1. The loop ends prematurely. When you declare ch as an int (instead of char), it should work. This article may help clarify the point.