segment fault after parsing unsigned char to int pointer

Question

I'm facing a problem when trying to read a binary file of fixed size. The code below returns segmentation Fault just before closing the file. What I want to achieve it is to return an int pointer back to the main function.

The file it s a raw grayscale image, which has values from 0 to 255, that's why I'm using unsigned char.

How can I convert and assign from unsigned char to *int properly?

Any kind of help will be welcome!

void readBinaryFile(char *filename, int *in){

    FILE *file;
    long length;
    unsigned char *imagen;

    int c;

    file = fopen(filename, "rb");
    fseek(file,0,SEEK_END);
    length = ftell(file);

    imagen = (unsigned char *) malloc(length);
    fseek(file, 0, SEEK_SET);
    fread(imagen, length,  1 , file);

    int cont;
    //c+4: file contains values from 0 to 255 
    for(c=0,cont=0;c<length;c=c+4,cont++){
       in[cont] = (unsigned char) imagen[c];         
    }


    for(cont=0;cont<length/4;cont++){
       printf("%d",(int) in[cont]);
    }

    fclose(archivo);
    free(imagen_buffer);
}

void main(int argc, char **argv){

    int *in;           
    int fixed_size = 784;
    in = (int *) malloc((fixed_size)*(fixed_size));
    readBinaryFile("test.raw", in);

    int c;

    for(c=0;c<((fixed_size)*(fixed_size));c++){
        printf("%d", (unsigned char) in[c]);
    }           
}

alk · Answer 1 · 2018-09-23T13:33:21.687

2

This

int *in;

makes in point to an int.

This

 in = (int *) malloc((fixed_size)*(fixed_size));

allocates to in 784*784=614656 bytes.

You need 784*784 ints.
An int needs sizeof (int) bytes, which ~~by definition is~~ very well might be and most often is larger then 1.

From 1. and 2. above it can be deduces that 784*784 int needs more then 784*784 bytes.

So change

in = (int *) malloc((fixed_size)*(fixed_size));

to be

in = (int *) malloc((fixed_size)*(fixed_size) * sizeof (int));

or even nicer and safer with less noise do

in = malloc(fixed_size*fixed_size * sizeof *in);

because

in C there is no need to cast void* (which malloc() returns).
the parentheses around fixed_size are useless.
using sizeof *in returns the same as sizeof (int), but would "survive" if you changed int * in to become for example unsigned * in.

Also while reading you need to make sure to not read more then 784*784 char, as else the memory allocated for the int will be overflowed, which in turn invoked the infamous Undefined Behaviour.

edited Sep 23 '18 at 13:33

answered Sep 23 '18 at 09:56

alk

69,737
10
105
255

dude, I was stuck in this like all night, THANKS ! – Charles Sep 23 '18 at 10:02
The size of `int` is not more than one by definition. It would have to be a derived property, as there is no such definition. However, by extensive discussion elsewhere on Stack Overflow, it may be possible for a conforming implementation to have `int` and `char` be the same size. (`EOF` is a troublesome issue for that.) – Eric Postpischil Sep 23 '18 at 11:47
@EricPostpischil "*possible ... to have `int` and `char` be the same size*" Really? I memorise the C Standard defining `CHAR_MAX` to be 127 or 255 as maximum and `INT_MAX` to be 32767 as minimum. Mind sharing the link to the discussion you mention? – alk Sep 23 '18 at 13:06
@alk: The standard does not limit `CHAR_MAX` to being 127 or 255. There is discussion in [this question](https://stackoverflow.com/questions/3860943/can-sizeofint-ever-be-1-on-a-hosted-implementation). – Eric Postpischil Sep 23 '18 at 13:25
@EricPostpischil Oh, yes I see: Only the minimum values are defined by the Standard. My bad. Thank you for the correction. – alk Sep 23 '18 at 13:31

segment fault after parsing unsigned char to int pointer

1 Answers1