C: pointers, arrays, array of pointers

[Pithikos]

I am really confused about this concept and I have to be done with an exercise for university until tomorrow.
I got an understanding of what pointers and arrays are.
Pointers are "variables" containing the memory adress of some variable.
Arrays are just a bunch of variables that you can access easily by an index number.

Where it gets furstrating is when I have to make a program that takes parameters. An example:


#include <stdio.h>

int main(int argc, char *argv[]) {

int i=0;
printf("first string's adress: %p\n", argv[0]);
printf("number of arguments: %d \n", argc);

printf("first string %s\n", *argv[0]); /*HERE IS THE PROBLEM*/

for (i=0; i<argc; i++){ /*this I saw from somewhere and it works although I don't understand it completely*/
printf("%s\n", *argv);
}
return 0;
}
Isn't it so that "argv[]" is just an array with pointers inside? So to get to the information of the array's first element i should just go with "*argv[0]" as argv[0] holds the address for the first element/parameter. And that element is a string so using "%s" should be right :confused:

Where am I wrong? Please try to explain in simple words. I read a bunch of tutorials and everything but I just got more confused.

[GeneralZod]

argv[0] will be of type char*, which you can use with printf("%s", blahblah).

*argv[0] will be of type char - in fact, it will be the first character of the first argument (argv[0][0] would be another way of writing this :))

This:



for (i=0; i<argc; i++){ /*this I saw from somewhere and it works although I don't understand it completely*/
printf("%s\n", *argv);
}


should be



for (i=0; i<argc; i++){
printf("%s\n", argv[i]);
}


Would it be clearer if you could see where the strings came from?



const char *nice_foodstuffs[] = { "sausage", "bacon", "eggs" };
for (int food_type = 0; food_type < 3; food_type++)
{
printf("Yum - I sure like %s!\n", nice_foodstuffs[food_type]);
};


A minor variation:



const char* third_food_type = "eggs";
const char *nice_foodstuffs[] = { "sausage", "bacon", third_food_type };
for (int food_type = 0; food_type < 3; food_type++)
{
printf("Yum - I sure like %s!\n", nice_foodstuffs[food_type]);
};
printf("But I especially like the third food type: %s\n", third_food_type);


or



const char* third_food_type = "eggs";
const char *nice_foodstuffs[] = { "sausage", "bacon", third_food_type };
for (int food_type = 0; food_type < 3; food_type++)
{
printf("Yum - I sure like %s! It begins with the letter '%c'\n", nice_foodstuffs[food_type], *nice_foodstuffs[food_type]);
};
printf("But I especially like the third food type: %s\n", third_food_type);


or



const char* third_food_type = "eggs";
const char *nice_foodstuffs[] = { "sausage", "bacon", third_food_type };
for (int food_type = 0; food_type < 3; food_type++)
{
printf("Yum - I sure like %s! It begins with the letter '%c'\n", nice_foodstuffs[food_type], nice_foodstuffs[food_type][0]);
};
printf("But I especially like the third food type: %s\n", third_food_type);

[dwhitney67]

And yet another example...


#include <stdio.h>

void function(int argc, char** argv)
{
for (int i = 0; i < argc; ++i)
{
printf("%s\n", *(argv+i)); // here, the pointer is incremented by i, then dereferenced.
}
}

int main(int argc, char** argv)
{
function(argc, argv);

return 0;
}

[pbrane]

argv[0] will be of type char*, which you can use with printf("%s", blahblah).

*argv[0] will be of type char - in fact, it will be the first character of the first argument (argv[0][0] would be another way of writing this :))

This:



for (i=0; i<argc; i++){ /*this I saw from somewhere and it works although I don't understand it completely*/
printf("%s\n", *argv);
}



How does this work? It doesn't work for me:


$ ./argv test two three
./argv
./argv
./argv
./argv

it just prints the first element of argv, the program name, which in this case I called the test program argv

dwhitney67's example works, and I understand why it works.

[r-senior]

No it doesn't work. If you look back at GeneralZod's post, it says that the piece of code you list should be something else, i.e. it should use argv[i].

[dwhitney67]

No it doesn't work. If you look back at GeneralZod's post, it says that the piece of code you list should be something else, i.e. it should use argv[i].

Or use *(argv + i).

Note for the OP:

*argv is the same as *(argv + 0)

[Pithikos]

Sorry, I posted the code after I edited it to test things. So
for (i=0; i<argc; i++){
printf("%s\n", *argv); }

Should be:


for (i=0; i<argc; i++){
printf("%s\n", *argv++); }

[Pithikos]

By the way can someone explain me this?:
char *food[] = { "sausage", "bacon", "eggs" };Does this mean we have an array "food" that has pointers and each pointer points to a string(sausage, bacon, eggs)?
so food[0] has the adress of string "sausage"
so food[1] has the adress of string "bacon"
so food[2] has the adress of string "eggs"
Is my assumption right?
And if it is then shouldnt "printf("%s", *food[1])" print "bacon" on the screen? As food[1] has the adress of string "bacon" and adding the asterisk infront of it should get the value where it's pointing?
I am really confused :S

[dwhitney67]

I am really confused :S
Why don't you develop a small little program and test your questions? You learn much more by doing, than merely by reading.

[lloyd_b]

By the way can someone explain me this?:
char *food[] = { "sausage", "bacon", "eggs" };Does this mean we have an array "food" that has pointers and each pointer points to a string(sausage, bacon, eggs)?
so food[0] has the adress of string "sausage"
so food[1] has the adress of string "bacon"
so food[2] has the adress of string "eggs"
Is my assumption right?
And if it is then shouldnt "printf("%s", *food[1])" print "bacon" on the screen? As food[1] has the adress of string "bacon" and adding the asterisk infront of it should get the value where it's pointing?
I am really confused :S

Pointers and arrays (which are basically the same thing, expressed in two different ways) are generally pretty confusing to newcomers. But they are *very* powerful, so it's necessary to learn them.

First tip - you need to realize there's no such thing as a "pointer to a string" in C. Or a "pointer to an array" for that matter. What you are pointing to is to the first element of that array. "food[1]" is exactly the same thing as "&food[1][0]", a pointer to the character at the start of the food[1] array. Calling something a "pointer to an array" is just easier than saying "pointer to the first element of an array" :)

For printf(), the "%s" format expects a character pointer, which is expected to be the a pointer to the first element of a zero-terminated string. 'printf("%s\n", food[1])' fulfills this requirement, while 'printf("%s\n", *food[1])' does not, as *food[1] is not a pointer, but the contents of the memory pointed to by food[1], the letter 'b'.

Hope I didn't confuse you further :)

Lloyd B.

[Pithikos]

No in fact that last post helped me a lot :P
I tested a lot following what you said and it seems to make some sense now. I just had a hard time understanding that a pointer points only to the first character of a word. Noone ever mentions it in any book or tutorial so clearly out as you did.
Secondly from some assumption I understood that
printf("%s", x) requires an address in the place of x, something that is totally different for the other cases that don't deal with arrays.
Then I found written somewhere that according to ANSI standard
ptr=&array[0] is the same thing as
ptr=array So that also explains why in the hell you can print a string with just giving the array name.
Well I guess I will be able to finish that exercise now after fighting with pointers and arrays for hours :rolleyes:

[Pithikos]

btw I found this cool program in Ubuntu repository called "cdecl". You can run it and type any declaration of variables, arrays etc and it outputs the declaration in plain english.

example:

cdecl> explain int (*arr)[3];
declare arr as pointer to array 3 of int
cdecl> explain int **arr[3];
declare arr as array 3 of pointer to pointer to int
Cheers

[Zymus]

int main(int argc, char** argv)
{
printf("Number of arguments: %d\n", argc);
for(int i = 0; i < argc; i++)
{
printf("argv[%d]: %s\n", i, argv[i]);
}
}