Having Run time check failure #2-stack around the variable "numbers"

This is a discussion on Having Run time check failure #2-stack around the variable "numbers" within the C Programming forums, part of the General Programming Boards category; Having Run time check failure #2-stack around the variable "numbers" Need help thank you my input file is below 5 ...

  1. #1
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    Having Run time check failure #2-stack around the variable "numbers"

    Having Run time check failure #2-stack around the variable "numbers"
    Need help thank you
    my input file is below

    5
    8
    10



    below is my code

    Code:
    ..
    
    #include "stdafx.h"
    #include <ctype.h>
    #include <string.h>
    #include <stdlib.h>
    #include <stdio.h>
    
    int _tmain(int argc, _TCHAR* argv[])
    {
    	FILE *test;
    	int j=0;
    	int numbers[]={0};
    	int record[]={0};
    	int i=0;
    	char *array1;
     	if((test=fopen("Input1.txt","r"))==NULL)
    	{
    		printf("File could not be opened\n");
    	}
    	else
    	{
    		array1 = (char*)malloc(1*sizeof (char));
    		if((test=fopen("Input1.txt","r"))==NULL)
    		{
    			printf("File could not be opened\n");
    		}
    		else
    		{
    			while(fgets(array1,(sizeof array1)-1,test)!=NULL) 
    			{
    				numbers[i]=atoi(array1);
    				i++;
    			}
    			for(i=0;i<sizeof(array1)-1;i++)
    			{
    				printf("%d\n",numbers[i]);
    			}
    		}
    	fclose (test);
    	}
    	system("pause");
    	return 0;
    	free(array1);
    }

  2. #2
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    In the following snippet:
    Code:
    int numbers[]={0};
        int record[]={0};
    What do you think the size of these arrays will be?
    Also:
    Code:
    array1 = (char*)malloc(1*sizeof (char));
    How many characters are you allocating for this array? Do you think you are actually allocating enough memory for a C-string.

    Jim
    Last edited by jimblumberg; 07-04-2012 at 08:10 PM.

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    I want the size of array depending on the file line. so i write like this so that the size of array is vary.

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    You must supply a size for the array. Also in order to use atoi() your array1 array must contain at least two characters one space for a single digit, and one character for the end of string character.

    Jim

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    Thanks Jim. I solve the stack error. but can u further explain for the atoi()? not sure what you mean by it. Thank you very much.

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    And Jim, if i really dont know the size of array, what i should do? as my text file will be very long. this is just make me easy to code and debug only

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    Any help?

  8. #8
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    The following:
    Code:
    array1 = (char*)malloc(1*sizeof (char));
    Allocates space for exactly 1 character. In C strings are arrays of char terminated with the end of string character '\0'. The atoi() function works with C strings. Therefore allocating exactly one character for your read buffer does not allow room for a character plus this end of string character. Since this is just a buffer for reading one line of your text file I would skip the malloc() call altogether and just use a fixed sized C string for your buffer. Maybe a string with a length of BUFSIZ which is a value defined in stdio.h. You should also not cast the return value of malloc in a C program and since sizeof(char) is one you don't actually need the multiplication either.
    Code:
     char buffer[BUFSIZ]; // Declare a buffer to use for extracting a line from a file.
    char *array1 = malloc(1024);  // Declare and allocate 1024 bytes to the array array1.
    If you don't know how many numbers will be in your file you will need to use the malloc/calloc/realloc to allocate the memory for your array of numbers.

    Jim

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    Jim, sorry for troublesome you again. I go to website study about buffer[BUFSIZ], but i cannot read it into a variable and store it. and for the

    char *array1 = malloc(1024); // Declare and allocate 1024 bytes to the array array1.
    the malloc is void and how it going to assigned to char.

  10. #10
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by zhengkoon8
    the malloc is void and how it going to assigned to char.
    The return type of malloc is void*, not void. There is an implicit conversion from void* to char* (and indeed to any other pointer to object type).
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  11. #11
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    ok. if i am using realloc, how this function be used?

  12. #12
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by zhengkoon8 View Post
    ok. if i am using realloc, how this function be used?
    Probably very badly, if the track record of posters on this forum is anything to go by.

    How about finding out the length of the file first, then allocating the whole amount using just malloc, then reading the file into that?
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    Code:
        return 0;
        free(array1);
    }
    The return statement will exit the function/program and the call to free will not happen. The free call must be before the return statement.
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