Reading memory address's in c using user inputted hex values

Question

The aim is to allow the user to input a memory address (in Hex) and print out the contents of the address.

The code is running on the Motorola 68HC11.

So far i am able to read memory addresses using hard coded values:

void displayMemory(){
    char *p = (char*)0x5000;
    printf("Address Ascii: %s\n",p);
}

However i have tried multiple ways of using data input from the user such as:

void displayMemory(char *arr){
    char *p = arr;
    printf("Address Ascii: %s\n",p);
}

^^ where the *arr is the inputted hex value. Such as: 40a

I have also tried hard coding the hex value as a char array like:

void displayMemory(){
    char *p = "0x5000";
    printf("Address Ascii: %s\n",p);
}

This also did not work and gave me a different memory address.

Any ideas would be greatly appreciated, i am probably being very stupid!

UPDATED: how the function is called

char *arr;
arr = malloc(size * sizeof(char));
if(arr){
    strncat(arr, &userInput[3], size);
    displayMemory(arr);
}else{
    return 0;
}

userInput is : dm 40a size is: length of the number part of userInput in this case 3

@EugeneSh. i stated below the second example. The data would come in as for example: 40A — Aaron Nebbs, Feb 01 '18 at 15:18
`*arr` is a single `char`. It can't be a "hex value" (what do you mean by hex value?). — melpomene, Feb 01 '18 at 15:19
Yeah... I can't get myself to answer this. I think you should try a little harder finding the answer of questions this basic on the net yourself. If you try to learn a programming language by only reading answers to questions you asked stackoverflow then that feels more like "cluttering" SO... — Carlo Wood, Feb 01 '18 at 15:21
Your code has undefined behavior: `strncat(arr, ...)` reads from uninitialized storage. Also, buffer overflow: `strncat(arr, ..., 3)` writes up to 4 bytes to `arr`. — melpomene, Feb 01 '18 at 15:31
@melpomene userInput is initialised, it just happens somewhere else. I have a-lot of error trapping that happens before the strncat is called didn't want to just post the whole code. — Aaron Nebbs, Feb 01 '18 at 15:34

score 2 · Accepted Answer · answered Feb 01 '18 at 15:39

You were about to find the solution yourself. If you input an address into a char array, for example with:

char address[16];
printf("Enter address in hexa: ");
scanf("%15", address);

and type 40a into it, address will point to the following character array: { '4', '0', 'a', '\0', ...} what lies after the terminating null is irrelevant. Extactly like in char *p = "0x5000"; p points to the (non modifiable because litteral) character array { '0', 'x', '5', '0', '0', '0', '\0'}`. Assuming ASCII, it gives { 0x34, 0x30, 0x41, 0, ...} in first case, and { 0x30, 0x78, 0x35, 0x30, 0x30, 0x30, 0} in second one. In either case, this exist at an arbitrary address unrelated to the value.

And what works is when the address is the expected value. So you need to read the address as an integer value, and then convert it to a char pointer

int address;
scanf("%i", &address);
char *p = (char *) address;    // NOT &address!

score -1 · Answer 2 · answered Feb 01 '18 at 15:33

First of all, as per my knowledge, it is not possible. I have also ask my professor and he told me that we cannot do that actually you also do not know that which type of data stored at any particular address. so sorry but there is no way for access data by this way.

Reading memory address's in c using user inputted hex values

2 Answers2