C. Passing pointers to be modified causing segmentation faults

2
votes

I am making a decimal to binary converter in C for a class. I want to pass a char array to my function as well as the the decimal as an int. ie void DtoB(int decimal, char *array); DtoB will do the math and modify array to be the binary value. Ideally by giving it int values. Main() will just scanf(decimal), DtoB(decimal,array), and printf(array).

Here is what I have. This just returns a segmentation fault

1 #include <stdio.h>
2 #include <math.h>
3
4 void DecToBin(unsigned int, char *binary);
5
6 int main()
7 {
8         unsigned int dec;
9         char *binary[];
10         while(1) {
11                 scanf("%d", &dec);
12                 DecToBin(dec,*binary);
13                 printf("In binary is ");
14                 printf("%s\n",binary);
15         }
16 }
17 void DecToBin(unsigned int dec, char *binary)
18 {
19         int counter=0;
20         while(dec) {
21                 binary[counter]=dec%2;
22                 dec/=2;
23                 counter++;
24         }
25 }

I want it done like this since this seems like the best way to be able to do 32 bit integers, while keeping the array at the minimum size. Sorry if I killed formatting. Any help is appreciated.

2
cpointerssegmentation-fault
char *binary[33]; declares an array-of-pointers -- none of which are allocated. DecToBin(dec,*binary) is the same as DecToBin(dec, binary[0]). – David C. Rankin
As currently written, char *binary[]; won't compile; you can't define an array in the body of a function without a non-zero size (GCC may permit a zero size, but that's a compiler extension). You actually want char binary[33]; and then pass binary (not *binary) to your function. – Jonathan Leffler
when writing/posting code, do no use tabs for indenting. Because each wordprocessor/editor has the tab stops/tab widths set differently Suggest using 4 spaces for each indent level as that is wide enough to be visible even with variable width fonts and still allows many levels of indenting across the page – user3629249
this line: DecToBin(dec,*binary); is dereferencing the array/pointer binary[] Suggest: DecToBin(dec,binary);. This is because in C, an array name (in most instances) degrades to the address of the first byte of the array. – user3629249
in general, the code should always check the returned value (not the parameter value) from scanf() to assure the operation was successful. In this case, the returned value should be 1. – user3629249

2 Answers

1
votes
char *binary[33]

binary is array of pointers. So each element in it is a pointer.

The segmentation fault is because you are not initializing your array and trying to use it.

You are dereferencing a pointer which is not pointing to any valid memory location.

You need to allocate memory to the members of the array before using them

1
votes

incorporating all the comments, incorporating error checking, etc. the posted code becomes :

#include <stdio.h>
#include <stdlib.h>  // exit(), EXIT_FAILURE
#include <string.h>  // memset()


// prototypes
void DecToBin(unsigned int, char *binary);

int main()
{
        unsigned int dec;
        char binary[sizeof(int)*8 +1];


        while(1)
        {
                if( 1 != scanf("%u", &dec) )
                { // then scanf failed
                    perror( "scanf for decimal value failed" );
                    exit( EXIT_FAILURE );
                }

                // implied else, scanf successful

                DecToBin(dec, binary);
                printf("In binary is ");
                printf("%s\n",binary);
        }
}


void DecToBin(unsigned int dec, char *binary)
{
        size_t counter= sizeof(int)*8;

        memset( binary, ' ', counter );
        binary[ counter ] = '\0'; // terminate string
        counter--;

        // do...while allows for dec being 0
        do
        {
                binary[counter]= (char)((dec%2)+ 0x30);
                dec /= 2;
                counter--;
        }  while(dec);
}

which still has the shortcoming that the user is left with a blank screen and a blinking cursor. I.E. the code should be prompting the user, by requesting the input value.