I am trying to access items in an array of structs and print the structs fields as follows
printList(Album *a, int numOfStructs)
{
int i;
int j;
for(i = 0; i < numOfStructs; i++)
{
printf("number%d\n:", i+1);
printf("%s", a[i]->field2);
printf("%s", a[i]->field2);
printf("%d", a[i]->field3);
for(j = 0; j < a[i]->numOfStrings; j++)
{
printf("%s", a[i]->strings[j]);
}
printf("\n");
}
}
but I get loads of errors as such
invalid type argument of '->'
What am I doing wrong with this pointer?
a is of type Album* which means that a[i] is of type Album (it is the ith element in the array of Album object pointed to by a).
The left operand of -> must be a pointer; the . operator is used if it is not a pointer.
You need to use the . operator. You see, when you apply a * to a pointer, you are dereferencing it. The same goes with the []. The difference between * and [] is that the brackets require an offset from the pointer, which is added to the address in the pointer, before it is dereferenced. Basically, these expressions are identical:
*ptr == ptr[0]
*(ptr + 1) == ptr[1]
*(ptr + 2) == ptr[2]
To connect to your question: Change a[i]->field2 and a[i]->field3 to a[i].field2 and a[i].field3.
Related
So my book is explaining me pointers to an array using ts example
#include <stdio.h>
int main()
{
int s[4][2] = {
{1234,56},{1212,33},{1434,80},{1312,78}
};
int(*p)[2];
int i, j, * pint;
for (i = 0; i <= 3; i++)
{
p = &s[i];
pint = (int*)p;
printf("\n");
for (j = 0; j<= 1; j++)
{
printf("%d ", *(pint + j));
}
}
return 0;
}
The output is Given as
1234 56
1212 33
1434 80
1312 78
No issue I am getting the same output.
My question is what was the need of using another pointer pint ?
Why can't we directly use P?
So When I tried to do it using P directly it didn't work
printf("%d ", *(p + j));
I got garbage values in output, Why is this happening?
I also tried printing p and pint they are the same.
Although p and pint have the same value, p + 1 and pint + 1 do not. p + 1 is the same as (char *)p + sizeof *p, and pint + 1 is the same as (char *)pint + sizeof *pint. Since the size of the object pointed to is different, the arithmetic gives different results.
The pointer p is declared like
int(*p)[2];
So dereferencing the pointer expression with the pointer in this call of printf
printf("%d ", *(p + j));
you will get the j-th "row" of the type int[2] of the two dimensional array that in turn will be implicitly converted to a pointer of the type int * that will point to the first element of the j-th "row".
So instead of outputting elements of each row you will output first elements of each row that moreover results in undefined behavior when i will be greater than 2.
I have been learning C for couple of months and I came across a question which is given below.
#include <stdio.h>
int main() {
int a[2][2] = {{1,2}, {3,4}};
int (**p)[2] = &a;
for (int i=0; i<2; ++i) {
for (int j=0; j<2; ++j) {
printf("%d %u\n", a[i][j], (*(a+i) + j));
}
}
for (int i=0; i<4; ++i) {
printf("%d %u\n", *(*p + i), (*p + i));
}
printf("%u\n", p);
printf("%u\n", p+1);
printf("%u\n", p+2);
printf("%u\n", p+3);
printf("%u\n", *p);
printf("%u\n", *p+1);
printf("%u\n", *p+2);
printf("%u\n", *p+3);
puts("");
}
The output that I am getting on my machine is as follows:
1 3751802992
2 3751802996
3 3751803000
4 3751803004
1 1
9 9
17 17
25 25
3751802992
3751803000
3751803008
3751803016
1
9
17
25
I understand the first four lines of the output where the elements of the 2D array and their respective addresses is getting printed but I have absolutely no clue how the other outputs are happening.
I checked in an online IDE and there also I am getting the same output except the addresses which obviously will differ.
I know that int (**p)[2] is incomparable pointer type to a[2][2] which is a (int *)[2] data type.
But still I want to understand how the p pointer is working.
Can someone please help me understand how this is happening?
I have been eagerly waiting to get the logic behind the code outputs.
I am extremely sorry for the long code snippet and the long output sequence.
Thanks in advance.
N.B - I know that the code is producing a lot of warnings but I want to get the core idea about p.
The problem with this code starts right here:
int main() {
int a[2][2] = {{1,2}, {3,4}};
int (**p)[2] = &a; // <-- Invalid conversion, undefined behaviour
// warning: incompatible pointer types initializing 'int (**)[2]' with an expression of type 'int (*)[2][2]' [-Wincompatible-pointer-types]
// ... Everything past here is undefined behaviour
}
There's a huge difference between int** and what you're attempting to cast, one big enough that this conversion isn't possible.
int** means, specifically, a structure where it's an array of int*, or pointers. Treating int[2] as a pointer is going to be a mess. That any of this code even semi-works is hard to explain. It's the compiler trying to make the best of a bad situation.
I introduced a macro LEN to calculate your array sizes instead of hard-coping the magic numbers, fixed the declaration of p, changed the unsigned format %u to signed %d as you were printed signed values, the last loop, I am sure what the 2nd thing you were trying to print so left it out, and the last section of print statements were pointers so used %p for those in a loop instead of duplicating the code:
#include <stdio.h>
#define LEN(a) sizeof(a) / sizeof(*a)
int main() {
int a[2][2] = {{1,2}, {3,4}};
int (*p)[2] = a;
for (int i=0; i < LEN(a); i++) {
for (int j = 0; j < LEN(*a); j++) {
printf("%d %d\n", a[i][j], *(*(a + i) + j));
}
}
for (int i=0; i < LEN(a) * LEN(*a); i++) {
printf("%d\n", *(*p + i));
}
for(int i = 0; i < LEN(a) * LEN(*a); i++) {
printf("%p\n", p + i);
}
for(int i = 0; i < LEN(a) * LEN(*a); i++) {
printf("%p\n", (void *) *(p + i));
}
puts("");
}
This is a problem:
int (**p)[2] = &a; // int (**)[2] = int (*)[2][2]
The type of &a is int (*)[2][2], not int (**)[2]. Your pointer declaration should be
int (*p)[2][2] = &a;
Unless it is the operand of the sizeof or unary & operators or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T will be converted, or "decay", to an expression of type "pointer to T" and the value of the expression will be the address of the first element of the array.
The expression a "decays" from type "2-element array of 2-element array of int" (int [2][2]) to "pointer to 2-element array of int" (int (*)[2]). However, since a is the operand of the unary & operator that conversion doesn’t take place, so &a has type "pointer to 2-element array of 2-element array of int" (int (*)[2][2]). Thus,
p == &a
(*p) == a
(*p) + i == a + i == &(*p)[i] == &a[i]
*((*p) + i) == *(a + i) == (*p)[i] == a[i]
*((*p) + i) + j == *(a + i) + j == &(*p)[i][j] == &a[i][j]
*(*((*p) + i) + j) == *(*(a + i) + j) == (*p)[i][j] == a[i][j]
A pointer is used to store the address of variables. So, when we define a pointer to pointer, the first pointer is used to store the address of the second pointer. Thus it is known as double pointers.
EXAMPLE:
int main() {
int integerValue = 84;
int *pointer1;
int **pointer2;
pointer1 = &integerValue;
pointer2= &pointer1;
printf("Value of integer = %d\n", integerValue);
printf("Value of integer using single pointer = %d\n", *pointer1);
printf("Value of integer using double pointer = %d\n", **pointer2);
return 0;
}
OUTPUT:
Value of integer = 84
Value of integer using single pointer = 84
Value of integer using double pointer = 84
How do I edit a value in an array with pointer in C?
int *pointer;
int array[3][1];
I tried this:
int *Pointer
int array[2][2];
Pointer[1][1]= 6;
but when compiling, I get a segmentation fault error. What to do?
Given some array int Array[Rows][Columns], to make a pointer to a specific element Array[r][c] in it, define int *Pointer = &Array[r][c];.
Then you may access that element using *Pointer in an expression, including assigning to *Pointer to assign values to that element. You may also refer to the element as Pointer[0], and you may refer to other elements in the same row as Pointer[y], where y is such that 0 ≤ y+c < Columns, i.e., Pointer[y] remains in the same row of the array.
You may also use Pointer[y] to refer to elements of the array in other rows as long as none of the language lawyers see you doing it. (In other words, this behavior is technically not defined by the C standard, but many compilers allow it.) E.g., after Pointer = &Array[r][c];, Pointer[2*Columns+3] will refer to the element Array[r+2][c+3].
To make a pointer you can use to access elements of the array using two dimensions, define int (*Pointer)[Columns] = &Array[r];.
Then Pointer[x][y] will refer to element Array[r+x][y]. In particularly, after int (*Pointer)[Columns] = &Array[0]; or int (*Pointer)[Columns] = Array;, Pointer[x][y] and Array[x][y] will refer to the same element.
You can access any given element with this syntax: array[x][y].
By the same token, you can assign your pointer to any element with this syntax: p = &array[x][y].
In C, you can often treat arrays and pointers as "equivalent". Here is a good explanation:
https://eli.thegreenplace.net/2009/10/21/are-pointers-and-arrays-equivalent-in-c
However, you cannot treat a simple pointer as a 2-d array. Here's a code example:
/*
* Sample output:
*
* array=0x7ffc463d0860
* 1 2 3
* 4 5 6
* 7 8 9
* p=0x7ffc463d0860
* 0x7ffc463d0864:1 0x7ffc463d0868:2 0x7ffc463d086c:3
* 0x7ffc463d0870:4 0x7ffc463d0874:5 0x7ffc463d0878:6
* 0x7ffc463d087c:7 0x7ffc463d0880:8 0x7ffc463d0884:9
*/
#include <stdio.h>
int main()
{
int i, j, *p;
int array[3][3] = {
{1,2,3},
{4,5,6},
{7,8,9}
};
// Dereference 2-D array using indexes
printf("array=%p\n", array);
for (i=0; i < 3; i++) {
for (j=0; j < 3; j++)
printf ("%d ", array[i][j]);
printf ("\n");
}
// Dereference 2-D array using pointer
p = &array[0][0];
printf("p=%p\n", p);
for (i=0; i < 3; i++) {
for (j=0; j < 3; j++)
printf ("%p:%d ", p, *p++);
printf ("\n");
}
/* Compile error: subscripted value p[0][0] is neither array nor pointer nor vector
p = &array[0][0];
printf("p=%p, p[0]=%p, p[0][0]=%p\n", p, &p[0], &p[0][0]);
*/
return 0;
}
Cast the 2D-array into 1D-array to pass it to a pointer,
And then, You are ready to access array with pointer. You can use this method to pass 2D-array to a function too.
#include <stdio.h>
int main()
{
int arr[2][2];
for (int i = 0; i < 2; i++)
{
for (int j = 0; j < 2; j++)
{
arr[i][j] = (2 * i) + j;
}
}
int *Pointer = (int *)arr; // Type conversion
/*
&arr[0][0] = Pointer + 0
&arr[0][1] = Pointer + 1
&arr[1][2] = Pointer + 2
&arr[2][2] = Pointer + 3
Dereference Pointer to access variable behind the address
*(Pointer + 0) = arr[0][0]
*(Pointer + 1) = arr[0][1]
*(Pointer + 2) = arr[1][2]
*(Pointer + 3) = arr[2][2]
*/
for (int i = 0; i < 2; i++)
{
for (int j = 0; j < 2; j++)
{
printf("%d ", *(Pointer + (2 * i) + j)); // Accessing array with pointer
}
printf("\n");
}
return 0;
}
Using the function wv_matalloc from https://www.ratrabbit.nl/ratrabbit/content/sw/matalloc/introduction , you can write the following code:
#include <stdio.h>
#include "wv_matalloc.h"
int main()
{
double **matrix;
int m = 3;
int n = 4;
// allocate m*n matrix:
matrix = wv_matalloc(sizeof(double),0,2,m,n);
// example of usage:
int i,j;
for (i=0; i<m; i++)
for (j=0; j<n; j++)
matrix[i][j] = i*j;
printf("2 3: %f\n",matrix[2][3]);
}
Compile with:
cc -o main main.c wv_matalloc.c
1.
You never assigned a value to Pointer in your example. Thus, attempting to access array by Pointer invokes undefined behavior.
You need to assign Pointer by the address of the first element of array if the pointer shall be a reference:
Pointer = *array;
2.
You can't use 2D notation (p[1][1]) for a pointer to int. This is a C syntax violation.
3.
Since rows of static 2D arrays are allocated subsequent in memory, you also can count the number of array elements until the specific element of desire. You need to subtract the count by 1 since indexing start at 0, not 1.
How does it work?
Each row of array contains 2 elements. a[1][1] (the first element of the second row) is directly stored after the first two.
Note: This is not the best approach. But worth a note beside all other answers as possible solution.
#include <stdio.h>
int main (void)
{
int *Pointer;
static int array[2][2];
Pointer = *array;
Pointer[2] = 6;
printf("array[1][1] (by pointer) = %d\n", Pointer[3]);
printf("array[1][1] (by array istelf) = %d\n", array[1][1]);
}
Output:
array[2][2] (by pointer) = 6
array[2][2] (by array istelf) = 6
Side Notes:
To address the first element of the second row by array[1][2] invokes undefined behavior. You should not use this way.
"but when compiling, I get a segmentation fault error."
Segmentation fault error do not occur at compile time. They occur at run time. It just gives you the impression because high probably your implementation immediately executes the program after compilation.
void get_elemnts(int *array, int max_index){
for(int i = 0; i < max_index; i++){
printf("enter element 0%d: ", i);
scanf("%d", array + i);
}
}
**scanf("%d", array + i);**Can someone explain? I have a code that gets the elements of an array from the user input. And at the moment I have difficulties with understanding what exactly this part of the code does
There are three concepts involved:
scanf("%d", pointer) method will read the integer value (%d) from stdin and write it to the memory referenced by pointer.
Nature of arrays: arrays in C are stored linearly in memory: int array of size n is just n * sizeof(int) bytes in memory and the variable of the array is the same as the pointer to its first element.
Pointer arithmetic: array + i moves the pointer by i memory cells of type int which is i * sizeof(int) bytes. This is exactly where the i'th element of array is. So array + i is a pointer to array[i].
This function definition
void get_elemnts(int *array, int max_index){
for(int i = 0; i < max_index; i++){
printf("enter element 0%d: ", i);
scanf("%d", array + i);
}
}
is equivalent to
void get_elemnts(int *array, int max_index){
for(int i = 0; i < max_index; i++){
printf("enter element 0%d: ", i);
scanf("%d", &array[i] );
}
}
This expression array + i gives pointer to the i-th element of the array.
The format %d used in the function scanf expects a pointer to an object of the type int and this expression array + i yields the pointer using the pointer arithmetic.
The expression array[i] is equivalent to the expression *( array + i ).
So the expression &array[i] is the same as &*( array + i ) where the applied operators &* can be omitted and you will get just ( array + i ).
I want to create an array of strings, so I first use:
char** p = malloc(sizeof(char*) * count); // count is the number of strings
But then trouble comes when I want to initialize those strings:
for (int i = 0; i < count; i++)
{
char* s = malloc(size_of_each_string);
*p + i = s; // THIS STEP INDUCES ERROR
while (*s++ = *input++); // initialize by copy
}
So I'm really confused. *p+i seems to be pointer arithmetic which shifts to another pointer. Pointer assignment in general is allowed (point to the same object). So why does such assignment not allowed? And how could I get around this to accomplish this task?
The error message is: Expression is not assignable.
Use *(p+i)=s instead of *p+i=s.
The things you're trying to do is to keep s in (p+i) address. For this you need to use *(p+i) instead of *p+i.
About Error Message:
Here the arithmetic expression *p+i will derive a value. But you can save a value only into a memory address. For this you're getting that error message.
*p + i = s;
is equivalent to
*(p) + i = s;
due to operator precedence. It yields an rvalue, which is not assignable.
Use parantheses instead:
*(p + i) = s;
or, even better, subscripting:
p[i] = s;
It's a problem with operator precedence.
The expression *p + i is parsed as (*p) + i not as *(p + i), so you need to do it explicitly. Or use normal array indexing syntax p[i].
Why don't you simplify your code by :
char** p = malloc(sizeof(char*) * count); // count is the number of strings
for (int i = 0; i < count; i++)
{
p[i] = malloc(sizeof(char) * size_of_each_string));
strcpy(p[i], input);
}