Segmentation Fault in C program that read a text file

Question

I´m writing a small C program that reads a text file and the prints it´s contents on screen. I do get the prints, but at the end of the printing I get a segmentation fault. Also, if I uncomment the atoi function I only get the first print and then a segmentation fault.Can anyone tell me what my error is?

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/mman.h>
#include <sys/shm.h>
#include <sys/stat.h> /* For mode constants */
#include <sys/types.h>
#include <fcntl.h> /* For O_* constants */


int sudoku[9][9];

int main(int argc, char *argv[]){

    if(argc!=2){
        printf("error en numero de parametros, el directorio de archivo de sudoku.\n");
        return -1;
    }
    char *directorio = argv[1];
    //printf("el directorio es: %s \n", directorio);
    int fp;
    fp=open(directorio,O_RDWR);
    if(fp<0){
        printf("error al leer el archivo");
        return -1;
    }   

    char *data = mmap(0, 4096, PROT_READ, MAP_SHARED, fp,0);

    if (data == -1) {
        perror("mmap");
        exit(1);
    }
    int i=0;
    for(i=0; i<81 ; i++){
        printf("data : %c \n",data[i]);
        //int val =atoi(data[i]);
    }

    if ((munmap(data,4096)) != 0) perror("!mumap failed");
    fclose(fp);


}

atoi takes a character array (string), in your case you want to use single characters. This will most likely crash since there isn't a null in the end, unless the sunspots are favorable. In any case you won't get what you want (a single character's numeric value) — Sami Kuhmonen
– Sami Kuhmonen, Commented Aug 23, 2015 at 4:47
As to the crash without atoi, is your file surely at least 81 bytes long? What does the debugger say what's wrong? — Sami Kuhmonen
– Sami Kuhmonen, Commented Aug 23, 2015 at 4:51
My file just contains this string 534678912672195348198342567859761423426853791713924856961537284287419635345286179 it is exactly 81 characters long. I don´t really know how to get the debugger going. I´m totally new to C — Pablo Estrada
– Pablo Estrada, Commented Aug 23, 2015 at 4:54
Honestly, it sounds to me like you would be better off with data[i] - '0' rather than atoi anyway, assuming you fixed all the other items wrong in the posted code. — WhozCraig
– WhozCraig, Commented Aug 23, 2015 at 5:29
Strongly suggest compiling with all warnings enabled. For gcc, at a minimum use: '-Wall -Wextra -pedantic'. The posted code does not cleanly compile. There are several warnings raised by the compiler. Several of those warnings are very serious. For instance, passing a 'int' to the fclose()' function. The parameter has to be the returned 'FILE *' from a call to fopen(), not the 'int' from a call to open() — user3629249
– user3629249, Commented Aug 23, 2015 at 5:29

Sunny · Accepted Answer · 2015-08-23 04:50:10Z

5

if I uncomment the atoi function I only get the first print and then a segmentation fault.

The reason for that is the atoi takes argument as char *, its prototype is

int atoi(const char *nptr);

When a char is passed as argument the ascii value of the char is treated as an address, the value will be (0-255). You dont have permission to access these locations and hence segmentation fault.

answered Aug 23, 2015 at 4:50

Sunny

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Comments

cgriff · Accepted Answer · 2015-08-24 06:16:55Z

The reason that you have experience a segfault is because you have used fclose() to close a file that you have opened with open().

int fclose(FILE *stream) is the function prototype for fclose() and int close(int fildes) is the prototype for close().

What this means is that fclose() is expecting a FILE pointer but is receiving the argument fp which you've declared as an int, this is what is causing your segfault. The program is trying to access the memory address location of the value of fp which it doesn't have the correct permissions to access, thus the segfault.

I've compiled and ran your program replacing fclose() with close() and I no longer had any segfault issues.

Regarding your atoi() call, this as has been explained by other posters, is due to atoi(const char *str) requiring a char pointer as its argument. What atoi() is receiving is not a char pointer but a char variable therefore it doesn't get terminated by the '\0' character, that is the cause of your segfault with the atoi() function.

Since you are operating on one character at a time it makes more sense to use int val = data[i] - '0'. Chars in C are integer types and the C standards do not guarantee that the character representation of '0' will be the same across all implimentations, for example the number zero on my machine is 48, '1' is 49 and 2 is 50. The C standards do however state that no matter the representations of numbers as chars, that for the characters '0' to '9', they will all be a value of 1 higher than the number before it. Therefore doing a small bit of char math you can work out that '2' - '0' = 2 as '2' is 50 (on this system) and '0' is 48.

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