[Tutorial] Introduction to programming using C (VII)

[m0skit0]

Here is a rewrite of Ex4. It encrypts, decrypts, and prints at each stage. It's a lot more compact than my first attempt and hopefully, you'll like it better.

Cool, but we didn't see loops yet, so it's not valid Also as Freddy points out, you don't free your mallocs

char *str=(char*)malloc(5+1); //I can't initialize as a const literal because I can't modify the string that way

Anyone said strlen()?

char *str=(char*)malloc(5+1); //I can't initialize as a const literal because I can't modify the string that way
strcpy(str,"Snail");

Why such complications? Isn't char* str = "Snail" easier?

Can you please explain why? I've seen that (int*) typecast convention with malloc() used in several books including K&R.

Memory leaks means you're not freeing the reserved memory. It has nothing to do with casting

I know this is unorthodox code, but it is perfectly valid code

Yup, my bad, didn't look closer

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[Wdingdong]

Yeah!
Correct!

Yeay Though I'll be asking more doubt on this.

Repeat after me: always free your memory,always free your memory...

Always free your memory, always free your memory, always free your memory....

Still, I should probably avoid using this weird method since it confuses and frightens people.

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[Wdingdong]

Till now I was thinking that pointers are abstract identities which points to some memory locations. But, now I came to know that pointers are also variables which stores address of another variable. Is it right? Also, can there be a pointer to point memory location of another pointer? Something like this:

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int x;
int* a;
int** b;
a = &x;
b = a;

[JJS]

pointers are (also) variables which stores address of another variable

This is exactly what a pointer variable is. Edit: It can also point to functions and const values, not only "variables" in the strictest sense.

Also, can there be a pointer to point memory location of another pointer?

Of course. But in your example you probably mean the last line to read

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b = &a;

because otherwise dereferencing it twice to get the value that is ultimately pointed to you would dereference x as a pointer which is wrong.

[m0skit0]

now I came to know that pointers are also variables which stores address of another variable. Is it right?

As JJS points out, yes, you're correct. Pointers are variables that hold a memory address (this not necesssarily points to a variable, not even to data, as JJS also points out). This is why you're able to do operations with them and even print them using printf() like any other variable type.

Also, can there be a pointer to point memory location of another pointer?

Yes, and just to clarify: and a pointer to another pointer that points to another pointer, and a pointer to another pointer that points to another pointer that points to another pointer... And so on. For example, an array of character strings would be a pointer to a pointer, e.g.:

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char* lotsofstrings[20];
char** decaying = lotsofstrings;

[ChemDog]

Using only pointers: declare an array of signed integers and assing it a size of 50 integers.
Same as the last exercise but also assign positions 5, 6, and 8, and then print them.
[spoiler]

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#include <stdio.h>

int main()
{
	int array1[50];
	array1[4] = 8;
	array1[5] = 20;
	array1[7] = 30;
	printf("%d, %d, %d\n", array1[4], array1[5], array1[7]);
	
	return 0;
}

[/spoiler]
Using only pointers: given a random string, print from the 5th character onwards.
[spoiler]

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

int main()
{
	char* s1 = "Random string.";
	char* s2 = s1 + 4;
	
	printf("%s\n", s2);

	return 0;
}

[/spoiler]
Given 2 strings of length 5, encrypt the first one using the second one using an XOR operation.
I'm having a hard time with this one. The only way I can think to do it would be: to get the ASCII values of both strings, then convert them to binary, and then do the XOR opperation. This seems beyond what we know though, so I'm guessing I just am thinking about it wrong.

[m0skit0]

Using only pointers: declare an array of signed integers and assing it a size of 50 integers.
Same as the last exercise but also assign positions 5, 6, and 8, and then print them.

You're using arrays and not pointers in your answer

Using only pointers: given a random string, print from the 5th character onwards.

All right

The only way I can think to do it would be: to get the ASCII values of both strings, then convert them to binary, and then do the XOR opperation. This seems beyond what we know though, so I'm guessing I just am thinking about it wrong.

You're thinking about it the right way, but you just believe it's complicated because you're confusing ASCII values, binary and characters. This 3 things are exactly the same one

[ChemDog]

In C, arrays can "decay" into pointers and viceversa, which means that arrays are equivalent to pointers. Thus for example char[] is the equivalent of char*.

This part has me a little confused . Here is my fix for the first two:
[spoiler]

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

int main()
{
	int* a = (int*) malloc(50*sizeof(int));
	a[4] = 8;                                           // Are these pointers decayed into arrays or viceversa?
	a[5] = 20;
	a[7] = 30;
	printf("%d, %d, %d\n", b[4], a[5], a[7]);
	free(a);
	
	return 0;
}

[/spoiler]

Given 2 strings of length 5, encrypt the first one using the second one using an XOR operation.
[spoiler]

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

int main()
{	
	char* s1 = "buffy";
	char* s2 = "point";
	int a = (int)s1;
	int b = (int)s2;
	int c = a ^ b;
	s1 = (char*) c;
	printf("%08X\n", s1);
	return 0;
}

[/spoiler]
This was by far the hardest practice excercise yet

[m0skit0]

This part has me a little confused . Here is my fix for the first two

You still using arrays. You aren't supposed to use array's syntax. Just work with a variable as a pointer.

Given 2 strings of length 5, encrypt the first one using the second one using an XOR operation.

Why are you converting to char to int and then int to char? This is weird because (in 32-bit machines) char is 1 byte and int is 4 bytes. By converting to int you're using 4 times the memory you should use with no real reason. Just use char right away. char is a number. Don't ever forget that for a computer, everything is a number. Also watch out with casting because in C you can almost always cast anything to anything, and this can be a source of very hard to track bugs. Use casting only when you're very sure of what you're doing.

[ChemDog]

Alright no more arrays:
[spoiler]

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

int main()
{
	int *a = (int*) malloc(50*sizeof(int));
	*(a+5) = 1;
	*(a+6) = 2;
	*(a+8) = 3;
	printf("%d, %d, %d\n", *(a+5), *(a+6), *(a+8));
	free(a);
	
	return 0;
}

[/spoiler]

Why are you converting to char to int and then int to char?

Well I started trying to do something like "int a = s1" but once I casted char to int, it didn't "click" that all of those conversions were unneccesary. So here is without all of the converting:
[spoiler]

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

int main()
{	
	char* s1 = "buffy";
	char* s2 = "point";
	s1 = (int)s1 ^ (int)s2;
	printf("%08X\n", (unsigned int)s1);
	return 0;
}	

[/spoiler]
I think this is an ok time to use casting but if it is not I will try and figure something else out.