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I was doing some testing for a program of mine and was wondering why the program was crashing when entering my function. Don't mind the logic of the program, since i was still in the phase of making sure i understood how to use my tools at hand. 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

/* Constants */
#define HEX_CAPITAL_LETTERS_BEGIN 65
#define HEX_CAPITAL_LETTERS_END 90
#define HEX_NUMBERS_BEGIN 48
#define HEX_NUMBERS_END 57
#define EXIT_SUCCES 0

/* Prototypes */
void codeToField(char *charArray, int i, int hexFloor, int hexCeil, char *outputArray);

/* Main Function */
int main(void) {
  char *code, warehouse, product, qualifiers;
  int i = 0;

  printf("Enter a MMOC product code: ");
  scanf("%s", &code);

  codeToField(code, i, HEX_CAPITAL_LETTERS_BEGIN, HEX_CAPITAL_LETTERS_END, &warehouse);
  codeToField(code, i , HEX_NUMBERS_BEGIN, HEX_NUMBERS_END, &product);

  strcpy(&qualifiers, code + i);

  printf("\n\nWarehouse: %s\nProduct: %s\nQualifiers: %s\n", &warehouse, &product, &qualifiers);

  return EXIT_SUCCES;
}

void codeToField(char *charArray, int i, int hexFloor, int hexCeil, char *outputArray) {
  int j = 0;
  while (charArray[i] >= hexFloor && charArray[i] <= hexCeil) {
    outputArray[j] = charArray[i];
    i++;
    j++;
  }
}

Thanks in advance.

Mike Lund Andersen
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  • Why use numbers like `65` and `90` instead of `'A'`, `'Z'`, `'0'`, etc.? – trent Nov 02 '17 at 01:30
  • Why are you trying to `strcpy` into a single character? – M.M Nov 02 '17 at 01:37
  • @M.M as I said, don't mind anything but the problem at hand. Was still trying to find out what i wanted and really just tested things out. Everything put the problem was commented out before posting haha. – Mike Lund Andersen Nov 02 '17 at 14:13

3 Answers3

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First, this does not do what you want:

char *code, warehouse, product, qualifiers;

The only pointer is code, the other are just a single char. You're printing them as strings with printf, and you use warehouse and product as outputArray in your function. They need to be pointers (or arrays) too:

char *code, *warehouse, *product, *qualifiers;

Then you need memory. The pointers are still uninitialized, so any reading from them is undefined behavior.

You can either allocate memory with automatic storage duration (on the stack) or dynamically (on the heap).

Stack:

char codestr[100];
code = codestr;

but then, you could also just have declared code as

char code[100];

to avoid to have two variables.

If you want to allocate the memory dynamically, you would use malloc:

code = malloc(100);

Don't forget to free the memory again:

free(code);

warehouse, product, qualifiers all need memory too. Some of the array sizes could be deduced form the defined constants.

alain
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The reason is because code has no memory allocated to it. It is an uninitialized pointer. Try this instead:

// ...
char myString[16], *code, warehouse, product, qualifiers;
code = &myString[0];
int i = 0;

printf("Enter a MMOC product code: ");
scanf("%15s", code);
// ...
vicatcu
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  • You should use a length limiter in the scanf format string to avoid buffer overflows – M.M Nov 02 '17 at 01:23
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  char *code, warehouse, product, qualifiers;
  int i = 0;

  printf("Enter a MMOC product code: ");
  scanf("%s", &code);

Here, code is an uninitialized pointer to memory so once you do the scanf call, your program is hosed.

I think you want something more like

char code[100];
printf ("Enter a MMOC product code: ");
scanf ("%s", code);
QuestionC
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