How to point at incompatible pointer type in C?

Question

In the book Learn C The Hard Way at excercise 15 there is suggestion to break program by pointing integer pointer at array of strings and using C cast to force it. How can I do it?

Answer 1

Here is a small example. the result depends on the endianness of your system and the size of int. I would expect the first or fourth character to change to the next character in the alphabet.

#include<stdio.h>
int main(void) {
    char string[100] = "Somestring";
    int *p;

    /* Let p point to the string */
    p = (int*)string;

    /* modify a value */
    (*p)++;

    /* Let's see if any character got changed */ 
    printf("%s", string);

    return 0;
}

It should be pointed out that not all casts are safe and that the result could be implementation defined or undefined. This example is actually undefined, since int could have stricter alignment constraints than char.

When writing portable code you need to take great care when using casts.

The code above could break on any system where sizeof(int) is greater than the string length regardless of alignment issues. In this case, where the string has size 100, we wouldn't expect that to happen in a long while. Had the string been 4-7 bytes it could happen sooner. The jump from 32- to 64-bit pointers broke a lot of old code that assumed that pointers and int were the same size.

Edit:

Is there an easy fix to the alignment problem? What if we could somehow make sure that the string starts in an address that is also suitable for an int. Fortunately, that is easy. The memory allocation function malloc is guaranteed to return memory aligned at an address that is suitable for any type.

So, instead of

char string[100] = "Somestring";

we can use

char *string = malloc(100);
strcpy(string, "Somestring");

The subsequent cast is now safe alignment-wise and is portable to systems where int is smaller than 100.

Note that malloc is declared in stdlib.h, so we should add the following at the top of our code file:

#include<stdlib.h>

Answer 2

That's simply an abusive way of casting.

// setup the pointers to the start of the arrays
int *cur_age = ages;
char **cur_name = names;

What the author of that link meant by "to break program by pointing integer pointer at array of strings and using C cast to force it." He meant that you can write something like this int *cur_age = (int *)names; That is to cast a pointer to pointer to char to a pointer to int. You can do that in C, which allows you to cast from one type of pointer to another type of pointer; but be warned you need to know what you are doing.

Here the author wanted to show how to break a program by pointing a pointer to a wrong type. His example, however, is probably making you more confused rather than helping you to understand pointers.

Answer 3

To cast, use the cast operator: (type)expression. For example, to cast an expression of type double to int:

(int)sqrt(2);

In your specific case, cast names to int* (the type of cur_age) to break the program:

cur_age = (int*)names;

Answer 4

To point incompatible pointer in c you only need to cast it to void.

//array of string declaration
char aStr[50][50];
Int *pint;

//do whatever you need with string array

pint = (*int)(*void)aStr;

I'm writing this from my cell phone.

if you increment your pointer past the allocated memory, you might end up in your program stack and change value to it.