Traversing structs using pointer++

Question

This might be a dumb question but I'm just learning C.

I have declared:

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

typedef struct STRUCTURE{
    char name[30];
}STRUCT;

int main(void){
    STRUCT struct1[30] // declaring 30 structs.
    STRUCT *pointer;
    pointer = &struct1;
    return 0;
}

Now, if I wanted to reference to specific struct (of the 30 structs) I could do it like:

&struct[0] // first struct's memory location
&struct[1] // second struct's memory location 
...
&struct[i] // where i = 0, 1, 2, 3, 4, 5 ... 29

But how would I do it using the pointer?

I thought by incrementing the pointer I could traverse the structs:

pointer++

But I can't seem to be getting from struct1[i] to struct1[i+1] with pointer++

pointer = &struct1; is invalid C. Didn't you get a message from the compiler? Which compiler are you using and what warning levels are used? Also, no semicolon here: STRUCT struct1[30] — Lundin
– Lundin, Commented Feb 13, 2023 at 8:32
"But I can't seem to be getting from struct1[i] to struct1[i+1] with pointer++" Well, that's how it works.... But tells how you reached that conclusion. You must have some code!!! Post it — 4386427
– 4386427, Commented Feb 13, 2023 at 8:34
name of an array is already a pointer to its first element no need for & and taking its address with &struct1 would give you a pointer to a pointer, which is not what you want. — mantissa
– mantissa, Commented Feb 13, 2023 at 8:39
OT: naming your personal datatype "STRUCT" is legal, but still not a good idea. Use your imagination!! — user17592432
– user17592432, Commented Feb 13, 2023 at 8:51
"... but I'm just learning C." Okay, here is the very first and very important thing to learn about C: Make sure to set your compiler to a high warning level and fix all warnings before running the program. For gcc I will at least recommend: gcc -Wall -Wextra -Werror — 4386427
– 4386427, Commented Feb 13, 2023 at 9:21

Vlad from Moscow · Accepted Answer · 2023-02-13 10:06:44Z

For starters this assignment

pointer = &struct1;

is incorrect. The expression in the right side of the assignment &struct1 has the pointer type STRUCT ( * )[30] while the operand in the left side has the pointer type STRUCT * and there is no implicit conversion from one type to another.

Instead you need to write

STRUCT struct1[30] // declaring 30 structs.
STRUCT *pointer;
pointer = struct1;

In this case the array designator struct1 is implicitly converted to a pointer to its first element and the pointer pointer will point to that first element of the array struct1 and using the pointer arithmetic you can access any element of the array.

For example, to output strings stored in the data member name of elements of the array (provided that they are initialized) using for loop you can write

for ( STRUCT *p = struct1; p != struct1 + 30; ++p )
{
    printf( "%s ", p->name );
}
putchar( '\n' );

Or instead of using the magic number 30 you could introduce a named constant as for example

enum { N = 30 };
STRUCT struct1[N];

and the n to write the loop like

for ( STRUCT *p = struct1; p != struct1 + N; ++p )
{
    printf( "%s ", p->name );
}
putchar( '\n' );

In this case if you will want to change the size of the array it will be enough to change the program only in one place in the enumeration declaration.

Pay attention to that the expression

pointer[i] is equivalent to the expression *( pointer + i] ). And the expression &pointer[i] is correspondingly equivalent to pointer + i.

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