The below question may sound repetitive to the community but sometimes you need to ask the exact question in your mind to understand things correctly. I am trying to get a thorough understanding of pointers. I was watching a video when the presenter says print the address of each element of array by using &pointername[subscript]. My question is why not just pointer. In the below code after doing j++ wont J have address of the array elements? why should I use &j[k]? what is the difference between &j[k],j(if there is)
#include <stdio.h>
void main()
{
int i[] = {1, 2, 3, 4, 5}, *j, k;
j = i;
for (k = 0; k < 5; k++)
{
printf("%p, %p, %p\n", &j[k], j, i);
j++;
}
}
Thanks,
Sahana
In most cases when you refer to an array in C, what you automatically get is a pointer to the array's first element. In other words, when you wrote
j = i;
it was exactly as if you had written
j = &i[0];
And this further means that when you write
i[k]
it is exactly equivalent to
*(i + k)
This then means that
&i[k]
will always be equal to
i + k /* or &i[0] + k */
And in your program it's also true that j[k] is equal to *(j + k), meaning that &j[k] is equal to j + k.
It looks like your program is trying to demonstrate these equivalences, but it's confusing because you're both incrementing j and indexing j by k. Here's a version of your program that should make it more clear what's going on. (Among other things, I've changed the array's name to a, because "i" is a very confusing name for an array! Similarly, I've changed j to p.)
int main()
{
int a[] = {11, 22, 33, 44, 55}, *p, k;
printf("a = %p\n", a);
p = a;
for(k=0; k<5; k++)
{
printf("%d %p %p %d %d\n", k, &a[k], p, a[k], *p);
p++;
}
}
You are partially right:
With the assignment j=i, j points to the first element in the array i. From now on j[k] will point to the same memory location as i[k].
With j[k] you are retrieving the data of the k-th cell in the array. So j[k] is an int.
If you want to get the address of it you have to place a & in front of it.
In C, &ptr[index] is exactly the same as ptr+index or even &index[ptr]
and arrays are decayed into pointers, in particular when passing them as arguments (e.g. to printf).
Read exactly what decaying an array means in your favorite C programming book. Notice that sizeof, when applied to an array or to a pointer, is not the same. So with int arr[5]; on my Linux/x86-64 desktop: when passing arr as an argument, it is decayed to a pointer so is the same as &arr or &arr[0]; but sizeof(arr) is 20 but sizeof(&arr) is 8 (like sizeof(ptr) after some int*ptr = &arr[3];)and sizeof(arr[0])is 4.
Simply
#include <stdio.h>
int main() {
int i[] = { 1, 2, 3, 4, 5 }, j;
for (j = 0; j < 5; ++j)
printf("%p\n", &i[j]);
return 0;
}
Output
0xbff22d8c
0xbff22d90
0xbff22d94
0xbff22d98
0xbff22d9c
Related
I have written code which allows me to modify the elements of a 1D array within my function by passing the element of the array:
I print the original array
I pass each element of the array to the function
Within the function I add the value 50 to each element of the array
I then call the function, and print out to screen the modified element value (i.e the value of each element +50)
I have been able to do this for a 1D array, with example values in the array being (10,20,30) and the valued printed after modification being (60,70,80).
What I am hoping to do is adapt that code to work for 2D arrays, you will see my attempt at doing this below. This code focuses on the use of int, but once I understand how to achieve this I am hoping to adapt for the use of a 2D string as well.
With the code below:
My objective is
Print to screen the original 2D array
Pass each element of the 2D array to the function
Within the function add the value 50 to each element of the array
Then call the function, and print out the modified element values to the screen(expected result displayed on screen 60,61,etc,.)
So far I have been able to print the original 2D array to the screen. It is the function I think I am messing up and would appreciate any advice. Thank you.
#include <stdio.h>
#include <string.h>
#define M 4
#define N 2
int function(int **arr);
int main() {
int i, a;
int arr[N][M] = {10, 11, 12, 13, 14, 15, 16, 17};
// the int array first
for(i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
// Accessing each variable
printf("value of arr[%d] is %d\n", i, arr[i][j]);
}
}
printf("\n ***values after modification***\n");
a = function(&arr[i][0]);
// int array print results
for(int i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
printf("value of arr %d\n", arr[i][j]);
}
}
return 0;
}
int function(int **arr) {
int i;
int j;
for(int i = 0; i < 3; i++) {
for(size_t j = 0; j < 5; j++) {
arr[i][j] = arr[i][j] + 50;
}
}
}
My apologies in advance for silly mistakes I am very new to C.
Thank you in advance.
The function int function(int **arr) does not return an int so make it void.
When you call it, a = function(&arr[i][0]);, you do not use a after the assignment. I suggest that you remove a from the program completely since it's not used anywhere.
The call to the function, function(&arr[i][0]);, should simply be function(arr);
The function signature needs to include the extent of all but the outermost dimension:
void function(int arr[][M])
Inside the function, you use 3 and 5 instead of N and M. That accesses the array out of bounds.
In function, the i and j you declare at the start of the function are unused. Remove them.
arr[i][j] = arr[i][j] + 50; is better written as arr[i][j] += 50;
When initializing a multidimensional array, use braces to make it simpler to read the code:
int arr[N][M] = {{10, 11, 12, 13}, {14, 15, 16, 17}};
In main you mix int and size_t for the indexing variables. I suggest you settle for one type.
Remove unused header files (string.h)
Example:
#include <stdio.h>
#define N 2
#define M 4
void function(int arr[][M]) {
for(int i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
arr[i][j] += 50;
}
}
}
int main() {
int arr[N][M] = {{10, 11, 12, 13}, {14, 15, 16, 17}};
for(size_t i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
printf("value of arr[%zu][%zu] is %d\n", i, j, arr[i][j]);
}
}
printf("\n ***values after modification***\n");
function(arr);
// int array print results
for(size_t i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
printf("value of arr[%zu][%zu] is %d\n", i, j, arr[i][j]);
}
}
}
Since you print the array more than once, you could also add a function to do so to not have to repeat that code in main:
void print(int arr[][M]) {
for(size_t i = 0; i < N; i++) {
for(size_t j = 0; j < M; j++) {
printf("value of arr[%zu][%zu] is %d\n", i, j, arr[i][j]);
}
}
}
Two-Dimensional arrays in C (and C++) are actually one-dimensional arrays whose elements are one-dimensional arrays. The indexing operator [] has left-to-right semantics, so for a type arr[N][M] the first index (with N elements) is evaluated first. The resulting expression, e.g. arr[0], the first element in arr, is a one-dimensional array with M elements. Of course that array can be indexed again , e.g. arr[0][1], resulting in the second int in the first sub-array.
One of the quirks in the C language is that if you use an array as a function argument, what the function sees is a pointer to the first element. An array used as an argument "decays" or, as the standard says, is "adjusted" that way. This is no different for two-dimensional arrays, except that the elements of a two-dimensional array are themselves arrays. Therefore, what the receiving function gets is a pointer to int arr[M].
Consider: If you want to pass a simple integer array, say intArr[3], to a function, what the function sees is a pointer to the first element. Such a function declaration might look like void f(int *intPtr) and for this example is simply called with f(intArr). An alternative way to write this is void f(int intPtr[]). It means exactly the same: The parameter is a pointer to an int, not an array. It is pointing to the first — maybe even only — element in a succession of ints.
The logic with 2-dimensional arrays is exactly the same — except that the elements, as discussed, have the type "array of M ints", e.g. int subArr[M]. A pointer argument to such a type can be written in two ways, like with the simple int array: As a pointer like void f(int (*subArrPtr)[M]) or in array notation with the number of top-level elements unknown, like void f(int arr[][M]). Like with the simple int array the two parameter notations are entirely equivalent and interchangeable. Both actually declare a pointer, so (*subArrPtr)[M] is, so to speak, more to the point(er) but perhaps more obscure.
The reason for the funny parentheses in (*subArrPtr)is that we must dereference the pointer first in order to obtain the actual array, and only then index that. Without the parentheses the indexing operator [] would have precedence. You can look up precedences in this table. [] is in group 1 with the highest priority while the dereferencing operator * (not the multiplication!) is in group 2. Without the parentheses we would index first and only then dereference the array element (which must therefore be a pointer), that is, we would declare an array of pointers instead of a pointer to an array.
The two possible, interchangeable signatures for your function therefore are
void function( int (*arrArg)[M] ); // pointer notation
void function( int arrArg[][M] ); // "array" notation (but actually a pointer)
The entire program, also correcting the problems Ted mentioned, and without printing the original values (we know them, after all), is below. I have also adapted the initialization of the two-dimensional array so that the sub-arrays become visible. C is very lenient with initializing structures and arrays; it simply lets you write consecutive values and fills the elements of nested subobjects as the come. But I think showing the structure helps understanding the code and also reveals mistakes, like having the wrong number of elements in the subarrays. I have declared the function one way and defined it the other way to show that the function signatures are equivalent. I also changed the names of the defines and of the function to give them more meaning.
#include<stdio.h>
#define NUM_ELEMS_SUBARRAY 4
#define NUM_ELEMS_ARRAY 2
/// #arrArg Is a pointer to the first in a row of one-dimensional
/// arrays with NUM_ELEMS_SUBARRAY ints each.
void add50ToElems(int arrArg[][NUM_ELEMS_SUBARRAY]);
int main()
{
// Show the nested structure of the 2-dimensional array.
int arr[NUM_ELEMS_ARRAY][NUM_ELEMS_SUBARRAY] =
{
{10, 11, 12, 13},
{14, 15, 16, 17}
};
// Modify the array
add50ToElems(arr);
// print results
for (int i = 0; i < NUM_ELEMS_ARRAY; i++) {
for (int j = 0; j < NUM_ELEMS_SUBARRAY; j++)
{
printf("value of arr[%d][%d]: %d\n", i, j, arr[i][j]);
}
}
return 0;
}
// Equivalent to declaration above
void add50ToElems(int (*arrArg)[NUM_ELEMS_SUBARRAY])
{
for (int i = 0; i < NUM_ELEMS_ARRAY; i++)
{
for (size_t j = 0; j < NUM_ELEMS_SUBARRAY; j++)
{
//arrArg[i][j] = arrArg[i][j] + 50;
arrArg[i][j] += 50; // more idiomatic
}
}
}
Why is it wrong to pass a two-dimensional array to a function expecting a pointer-to-pointer? Let's consider what void f(int *p) means. It receives a pointer to an int which often is the beginning of an array, that is, of a succession of ints lying one after the other in memory. For example
void f(int *p) { for(int i=0; i<3; ++i) { printf("%d ", p[i]); }
may be called with a pointer to the first element of an array:
static int arr[3];
void g() { f(arr); }
Of course this minimal example is unsafe (how does f know there are three ints?) but it serves the purpose.
So what would void f(int **p); mean? Analogously it is a pointer, pointing to the first in a succession of pointers which are lying one after the other in memory. We see already why this will spell disaster if we pass the address of a 2-dimensional array: The objects there are not pointers, but all ints! Consider:
int arr1[2] = { 1,2 };
int arr2[2] = { 2,3 };
int arr3[2] = { 3,4 };
// This array contains addresses which point
// to the first element in each of the above arrays.
int *arrOfPtrToStartOfArrays[3] // The array of pointers
= { arr1, arr2, arr3 }; // arrays decay to pointers
int **ptrToArrOfPtrs = arrOfPtrToStartOfArrays;
void f(int **pp)
{
for(int pi=0; pi<3; pi++) // iterate the pointers in the array
{
int *p = pp[pi]; // pp element is a pointer
// iterate through the ints starting at each address
// pointed to by pp[pi]
for(int i=0; i<2; i++) // two ints in each arr
{
printf("%d ", pp[pi][i]); // show double indexing of array of pointers
// Since pp[pi] is now p, we can also say:
printf("%d\n", p[i]); // index int pointer
}
}
}
int main()
{
f(ptrToArrOfPtrs);
}
f iterates through an array of pointers. It thinks that the value at that address, and at the subsequent addresses, are pointers! That is what the declaration int **pp means.
Now if we pass the address of an array full of ints instead, f will still think that the memory there is full of pointers. An expression like int *p = pp[i]; above will read an integer number (e.g., 1) and think it is an address. p[i] in the printf call will then attempt to access the memory at address 1.
Let's end with a discussion of why the idea that one should pass a 2-dimensional array as a pointer to a pointer is so common. One reason is that while declaring a 2-dimensional array argument as void f(int **arr); is dead wrong, you can access the first (but only the first) element of it with e.g. int i = **arr. The reason this works is that the first dereferencing gives you the first sub-array, to which you can in turn apply the dereferencing operator, yielding its first element. But if you pass the array as an argument to a function it does not decay to a pointer to a pointer, but instead, as discussed, to a pointer to its first element.
The second source of confusion is that accessing elements the array-of-pointers uses the same double-indexing as accessing elements in a true two-dimensional array: pp[pi][i] vs. arr[i][j]. But the code produced by these expressions is entirely different and spells disaster if the wrong type is passed. Your compiler warns about that, by the way.
Trying to work on leetcode #497 in C on my vscode. When writing main(), I am not sure how to deal with int** that leetcode provides. Is it possible to pass a 2D array using int**?
#include <stdio.h>
#include <stdlib.h>
typedef struct {
int rectsSize;
int * rectsColSize;
int** rects;
} Solution;
int points[100];
Solution* solutionCreate(int** rects, int rectsSize, int* rectsColSize) {
Solution* sol = malloc(sizeof(Solution));
sol->rects = rects;
sol->rectsSize = rectsSize;
sol->rectsColSize = rectsColSize;
//some codes
}
return sol;
}
int* solutionPick(Solution* obj, int* retSize) {
//some codes
return ret;
}
void solutionFree(Solution* obj) {
free(obj);
}
int main(void)
{
int rects[2][4] = {{1, 1, 5, 5}, {6, 6, 9, 9}};
int rectsSize = 2;
int rectsColSize = 4;
int retSize;
Solution* obj = solutionCreate(rects, rectsSize, &rectsColSize);
int* param_1 = malloc(sizeof(int));
param_1 = solutionPick(obj, &retSize);
solutionFree(obj);
return 0;
}
While in general there are many different ways to handle 2D array, the simple answer is no. There is a lot of info about 2d arrays in C: 1, 2, 3, etc. In principle, when dealing with 2d arrays, every dimension except first to the left needs to be specified exactly. In your case, every rectangle is defined by 4 integers, so instead int** rects consider int*[4] rects. This makes rectsColSize useless, because now each column has constant size of 4 ints.
Just for completness: what you are trying to do is second approach to arrays, where each column has independent size, and (usually) additional malloc call. While this approach is also valid and requires int** type, it is not needed for your task. Nice description of the difference here.
Edit
Here is how to loop through 2d arrays:
#define col 4
void print_2d(int (*a)[col], int aSize){
for(size_t i = 0; i < aSize; i++){
for(size_t j = 0; j < col; j++){
printf("%d ", a[i][j]);
}
printf("\n");
}
}
and here for int**:
void print_pp(int** a, int aSize, int* aiSize){
for(size_t i = 0; i < aSize; i++){
for(size_t j = 0; j < aiSize[i]; j++){
printf("%d ", a[i][j]);
}
printf("\n");
}
}
It seems that you want to convert int*[4] to int**, or more precisely, int*[4] arr2d with it's size int arr2dSize to structure Solution. In that case, here is wrapper to solutionCreate.
Solution* solutionCreateWrap(int (*arr2d)[4], int arr2dSize) {
int* rectsColSize = malloc(arr2dSize * sizeof(int));
int** rects = malloc(arr2dSize * sizeof(int*));
size_t arr2dMem = arr2dSize * 4 * sizeof(int);
rects[0] = malloc(arr2dMem);
memcpy(rects[0], arr2d, arr2dMem);
rectsColSize[0] = 4;
for(size_t i = 1; i < arr2dSize; i++){
rects[i] = rects[0] + i*4;
rectsColSize[i] = 4;
}
sol->rects = rects;
sol->rectsSize = rectsSize;
sol->rectsColSize = rectsColSize;
//some codes
}
return solutionCreate(rects, arr2dSize, rectsColSize);
}
Now for int rects[2][4] = {{1, 1, 5, 5}, {6, 6, 9, 9}}; call solutionCreateWrap(rects, 2) will return initialised structure Solution. It looks gruesome, and it's details are even worse, so if it just works, you may skip the explanation. Understanding low level C details isn't neccesarily to write in it, and this (or any other) explanation cannot possibly cover this matter, so don't be discouraged, if you won't get it all.
arr2d is contiguous block of memory of arr2dSize*4 integers. When multiplied by sizeof(int) we get size in bytes - arr2dMem in my code. Declaration int (*arr2d)[4] means, that arr2d is of type int*[4]. Knowing this we can cast it to int* like so: int* arr = (int*)arr2d and expression arr2d[i][j] is translated as arr[i*4+j].
The translation to rects is as follows; int** is array of pointers, so every rect[i] has to be pointer to i-th row of arr2d. Knowing this, everything else is pointer arithmetic. rects[0] = malloc(arr2dMem); and memcpy(rects[0], arr2d, arr2dMem); copies whole arr2d to rect[0], then every next rects[i] = rects[0] + i*4; is shifted 4 integers forward. Because rect is of type int**, the expression rects[i][j] translates to *(rects[i]+j), and replacing rects[i] by rects[0] + i*4, we get *((rects[0] + 4*i)+j), that is rects[0][4*i+j]. Note striking similarity between last expression, and arr[i*4+j]. rectsColSize is somewhat superfluous in this case, but it is essential in general int** array, when every subarray could have different sizes. After wrap function is done, rects is exact copy of arr2d, but with type appropriate for your Solution structure, so we can call solutionCreate().
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.
Learning C and this confusing me:
#include <stdio.h>
int main(){
int m[5][5];
int count = 0;
for(int i = 0; i < 5; i++){
for(int j = 0; j < 5; j++){
m[i][j] = count++;
}
}
int **p = m;
int (*k)[5] = m;
printf("%p\t%p\t%p", *p, *k, *m);
return 0;
}
that's what has been printed:
0x100000000 0x7ffff1fc9d70 0x7ffff1fc9d70
I'm really confused why dereference *p is 0x100000000, shouldn't it be 0x7ffff1fc9d70?
You’re thinking that m is of type int **, right?
It’s not. If you had done this, it would be:
int *m[5];
That make an array of five pointers to int, so m would be pointer to pointer to int.
However, if you do this:
int m[5][5];
You get enough space for 25 ints, which the compiler will access as a 5x5 two-dimensional array. p points to the beginning of that array. Dereference it, and the compiler reads the beginning of that array and interprets it as a pointer to int. Change the numbers you’re filling it with, and you’ll get different pseudo-pointers.
I'm having some difficulties understanding two dimensional arrays in C.
Let's look at this example:
#include <stdio.h>
void foo(int arr[2][3]) {
printf("%d", *arr);
}
int main() {
int arr[2][3] = { {10, 20, 30},
{40, 50, 60}
};
foo(arr);
return 0;
}
I have a few questions:
What is the value of arr? Is it the address of arr[0][0]?
If arr is the address of arr[0][0], then why the line:
printf("%d", *arr);doesn't print the value 10?
Each time I run it, I get a strange number. what is the meaning of this number?
Thanks :)
In answer to your questions:
Used in an expression, the value of arr is a pointer to its first element. Since it's an array of arrays, the pointer to its first element is &arr[0]. This value has an unusual type, "pointer to array of 3 ints".
Because arr is not the address of arr[0][0].
This is a crazy situation, hard to understand and hard to explain. In brief: since arr is a pointer to an array, *arr is that array. But when you try to pass it to printf, the compiler turns around and generates a pointer to the array's first element again. I suspect that pointer value differs because your compiler and OS are putting main (and therefore arr) in a different place on the stack each time. (And then there's the additional problem that since we're talking about pointers, it doesn't necessarily work to print them %d, especially if your machine has 32-bit ints and 64-bit pointers.)
My advice to you is not to worry about why the incorrect code printed changing values. Rather, please just print the array correctly, with code like this:
int i, j;
for(i = 0; i < 2; i++) {
for(j = 0; j < 3; j++)
printf("%d ", arr[i][j]);
printf("\n");
}
It is possible by doing like this:
#include <stdio.h>
int main()
{
int arr[2][2] = {{2,3},{5,6}};
for (int i = 0; i < 2; i++) {
for (int j = 0;j < 2; j++) {
printf("%d\n" , *(&arr[i][j]));
}
}
}