C array of pointers on string

Question

Well I don't work a lot with a pointers but now I get confused. I'm trying create array of pointers at strings (in this example just 2 pointers).

    char str[120];
    char* st[2];
    int i=1;
    while (fgets(str,120,f)) {

    printf("%s",str);
    st[i]=&str;

    i++;
    if (i==3)
        break;
}
for (i=1;i<=2;i++) {
    puts(st[i]);
}

warning: assignment from incompatible pointer type [enabled by default]

so can you help me understand all this situation with pointers on a strings. Please

st[i] is a char*, or a pointer to a character. &str is a char(*)[120], or a pointer to an array of 120 characters. Notice those are different. — Stack Exchange Broke The Law
– Stack Exchange Broke The Law, Commented May 13, 2016 at 0:18
so I need change &str to str, when I did it I got this: 12 1211 ��!�� !�� — ZeroVash
– ZeroVash, Commented May 13, 2016 at 0:20
int i=0;..st[i]=strdup(str);.. if (i==2).. for (i=0;i<2;i++) { — BLUEPIXY
– BLUEPIXY, Commented May 13, 2016 at 0:20
The second thing you're doing is that you're using st[1] and st[2]. st[1] is the second element in st. st[2] is the third element in st, which is two elements long, and therefore doesn't have a third element. — Stack Exchange Broke The Law
– Stack Exchange Broke The Law, Commented May 13, 2016 at 0:22
@immibis now i got it, it is very interesting that indexing is very important for pointeer. it will be good if you will write an answer with how array of pointer works — ZeroVash
– ZeroVash, Commented May 13, 2016 at 0:26

Mrunmoy · Accepted Answer · 2016-05-14 11:51:25Z

Arrays always use 0-based index.

for example:

int my_arr[] = { 10, 11, 12, 13, 14, 15 };

means that the array my_arr has 6 elements starting from index 0 till 6-1 (5). So, an N element array will have its indices from 0 to N-1. Also, note that my_arr by itself is also a pointer to the array.

define a int pointer:

int *iPtr = my_arr; // note that we do not use &my_arr here. my_arr is already a pointer to the array it defined.

or

int *iPtr;
iPtr= my_arr;

they are the same.

In the figure, the box represent the values and their addresses are written besides the boxes.

When you iterate over the array using variable i as index. You basically do the following indirectly:

i = 0
my_arr[i]

is same as

my_arr[0]

is same as

*(my_arr + 0)

is same as (substitute the value of box named my_arr)

*(20014000+0)

The * operator gets you the value at the address 0x20014000 in this example which is 10.

The above operation is also same as

iPtr[0] or *(iPtr+0)

Look at the box named iPtr, it also contains the same value 0x20014000.

Now lets define an array of pointers:

int *myptr_arr[3];
int varA = 25;
int varB = 34;
int varC = 67;

myptr_arr[0] = &varA; // note that varA is not a pointer, to get its address use &
myptr_arr[1] = &varB; // note that varB is not a pointer, to get its address use &
myptr_arr[2] = &varC; // note that varC is not a pointer, to get its address use &

once again, when you write myptr_arr[0] it translates to

*(myptr_arr + 0)

*(20152000 + 0)

which will get you 0x20162000 i.e. the value of box named [0] = the address of varA. Now to get the value at this address, you need to dereference it with * operator.

*myptr_arr[0]

which is same as * ( *(myptr_arr + 0) ) which is same as *(0x20162000) which is the value of box varA

Please see : Example usage of pointers

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