Learned C here's my version of Hello,World!

#include <stdio.h>

int main() {

printf("Thank you @drenal couldn't do it without ya!");
return 0;
}

Comments

G
Or there is this variant:
Code: 
#include <stdio.h>
#include <string.h>
int main()
{
    const char thanks[] = "Thank you @drenal couldn't do it without ya!";
    for (int i = 0; i <= (strlen(thanks) - 1); i++)
        putchar(thanks[i]);
    return 0;
}
I would recommend purchasing a book. In my experience, free web-based tutorials are a bad way to go because the creators aren't making money off of it (usually), so they have no motivation. When an author knows they are going to be making money, they want to make it the best they can and as appealing as possible for the most income.
 
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@blujay I don't know about C. but you use "char" as a type of array? Isn't char supposed to be used with that, chars? Lol
 
G
There is no 'string' variable. char is the identifier for character, which in turn will make strings. Now, I could assign each individual character of the array by saying this:
Code: 
const char thanks[] = {'T', 'h', 'a', 'n', 'k', ' ', 'y', 'o', 'u', [etc]};
However that is unreasonable.
 
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I see! That was the problem. I didn't knew that. I used string name[] (or whatever type I need) to create arrays. Now it makes sense. Thank you!
 
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G
By defining type string as a char* this code also works:
Code: 
#include <stdio.h>
#include <string.h>
typedef char* string;
int main()
{
    const string thanks = "Thank you @drenal couldn't do it without ya!";
    for (int i = 0; i <= (strlen(thanks) - 1); i++)
        putchar(thanks[i]);
    return 0;
}
However it still just shows that a string is a character array (char* name is the same as char name[])
 
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Now, if you haven't created an array (at least not using []), how is the FOR supposed to run through the supposed array you tell him to run (even if it doesn't exist)?
And making var types looks great lol
 
G
In C, the following two statements function the exact same:
Code: 
putchar(thanks[i]);

and 

putchar(*(thanks + i));
That just shows that pointers and arrays are the exact same thing, just arrays make things more legible.

Defining an array as const char* name and const char name[] end up the exact same. It declares name to be a pointer to a character value.
 
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Ookay. I think I understand. Then, when would you use a "pointer" (that if I understood correctly, they were represented with a [type]*) instead of an array and viceversa? I mean, they should have a situational difference, or they have not?

Thank you for your answers, by the way.
 
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I am not sure if that was sarcasm or something, but if it wasn't, everything that has been showed here looks "difficult" to understand if you have never tried to program, but in fact, they are quite easy -and very logical too-.

I'm not a pro programmer nor anything similar, in fact, I have studied 1 year of Java and now I'm learning PHP and JS. The three lenguages are kind of similar in some aspects, for example, the three of them, use the (almost) same loop structure (FOR, IF, WHILE...). It's not that hard if you have time and try to understanding.

Now, that being said, I don't know many things about programming, and of course, there are MANY that are hard to understand and use, and that's why programmers -in many cases- need more credits and kudos than they get.

That's my humble opinion.
 

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