I'm trying to pass an entire array to the function but the last value i get is always a garbage value and not 3. Please point out where I've made a mistake.
#include<stdio.h>
#include<conio.h>
main() {
int array[3] = {0, 1, 2, 3};
display(&array[0], 3);
}
display(int *j, int n) {
int i;
for(i=0; i<=n; i++) {
printf("\n%d", *j);
j++;
}
}
You have an off-by-one error in your for loop. Your array has only three elements. You try to read four elements instead of three.
Change your code to this (note the < instead of <=):
for (i = 0; i < n; i++) {
You also try to initialize the array with four elements when it only has room for three:
int array[3] = {0, 1, 2, 3};
Maybe you can try to change your compiler settings to flag this as a warning/error. Then you might get something like this that would notify you of the problem:
prog.c: In function ‘main’:
prog.c:13: error: excess elements in array initializer
prog.c:13: error: (near initialization for ‘array’)
The array variable represent an address of first element so no need to use & address operator with the function call and another issue is with size of an array.
main() {
int array[4] = {0, 1, 2, 3};
display(array, 3);
}
display(int *j, int n) {
int i;
for(i=0; i<=n; i++) {
printf("\n%d", *j);
j++;
}
}
There a few observations in the code:
1. main() should be int main(void) or int main(int argc, char **argv). Also you need to return an int value from main when to change it to int main(void) or int main(int argc, char **argv)
2. You are initializing an array of 3 elements with 4 values.
Either declare as int array[3] = {0, 1, 2}; or as int array[] = {0, 1, 2, 3}; (this will create an array with 4 elements)
3. display(int *j, int n) should have a return type say void display(int *j, int n) (otherwise it default to int in which case the function is not returning any value). In this case you will have to define the function before using.
4. In current case, to the function display you are passing an array of 3 elements and accessing element at index 3 (when i equals n) in the for loop which is the 4th element (i.e. out of bounds which is undefined behavior). Remember that array in C is zero indexed. So for(i=0; i<=n; i++) should be for(i=0; i<n; i++)
Hope this helps!
You cannot pass "an entire array to a function" -- C does not do this. Instead, you are passing a pointer to the first element of the array.
main() {
int array[3] = {0, 1, 2, 3};
main should be declared int main(int argc, char* argv[]). It's not that horrible to type, and you can even get your text editor to do it for you, if you care enough.
You have declared array to contain only three items, but store four items into the array. This should have thrown a compile warning at the least. Pay attention to those warnings. Don't specify the size of your array unless that size is vital (Say, if you wanted to keep track of US states, then int states[50] = { ... }; might make sense. For now. Maybe this is a bad example.)
for(i=0; i<=n; i++) {
printf("\n%d", *j);
j++;
for(i=0; i<=n; i++) is very often a bug -- C array indexes run from 0 to n-1, not 0 to n. You're reading one-past-the-array, and that value is not surprisingly garbage.
Further this code is awkward; you should just use printf("\n%d", j[i]) and not increment j each time. Incrementing two variables as "loop variables" this way is a recipe for writing bugs in the future. (If you need to update two variables with every loop iteration, place both in the last section of the for(;;i++, j++) loop.)
Your array is declared to have 3 elements yet you initialize it with 4:
int array[3] = {0, 1, 2, 3};
You create an array for only 3 integers, but put 4 in it. But the error is when you print the array, you print four entries, though only three exist.
Related
#include "stdio.h"
int printsomething(int *array, int arrayreturn[5]) {
int i;
for(i = 0; i < 10; ++i) {
printf("%d\n", array[i]);
}
for(i = 0; i < 5; ++i) {
int &arrayreturn[i] = {i};
}
return 0;
}
int main() {
int array[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
// int *arraypointer = &array;
int arrayp[5];
int i;
printsomething(array, arrayp);
for(i = 0; i < 5; ++i) {
printf("%d\n", arrayp[i]);
}
return 0;
}
I am learning C and right now just playing with arrays and pointers trying to get comfortable. This bit of code has the goal of passing an array to a function, which was successful before I added the second part. That second part being assigning values in the called function to an already initialized array. Since we can't directly return an array I understood this was the way to do it. What exactly do you all think is going wrong here? And I just completely off the target?
If you want to assign values to the array elements you need to use [] to access the elements and = to assign them. I cannot really explain your code because it is unclear how you came to the conclusion that you need to write int &arrayreturn[i] = {i};. Your loop can be this:
for(i = 0; i < 5; ++i) {
arrayreturn[i] = i;
}
the first problem is that when you have a parameter of the form int arrayreturn[5] you actually just pass an int pointer not an entire array of 5 elements. int arrayreturn[5] and int *arrayreturn compile to exactly the same cpu instructions. I never use the int arrayreturn[5] syntax because i think it is confusing so i rather just pass a pointer and this is common practice as far as i know.
secondly in the second part of your code you try to declare a new array of size i by calling int &arrayreturn[i] = {i} this is not possible because of multiple reasons mostly because you cant dynamically allocate arrays on the stack. it should be arrayreturn[i] = i
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.
I use to program with FORTRAN, but I decided to learn C and C++. I started with C language, and the one thing that I never used are pointers, because FORTRAN pass values by reference. I built the sample code below to understand how pointers work with multidimensional arrays:
#include <stdio.h>
#include <stdlib.h>
#define DIM1 3
#define DIM2 2
#define DIM3 4
void display3DArray1(int, int , int n, int (*arr)[][n]);
void display3DArray2(int rows, int cols1, int cols2,int arr[][cols1][cols2]);
int main(void)
{
int matrix3D[DIM1][DIM2][DIM3] = {
{{1, 2, 3, 4}, {5, 6, 7, 8}},
{{9, 10, 11, 12}, {13, 14, 15, 16}},
{{17, 18, 19, 20}, {21, 22, 23, 24}}
};
int (*pmatrix3D)[DIM2][DIM3] = matrix3D;
display3DArray1(DIM1, DIM2, DIM3,pmatrix3D);
display3DArray2(DIM1, DIM2, DIM3,pmatrix3D);
return 0;
}
void display3DArray1(int rows, int cols1, int cols2,int (*arr)[][cols2]) {
printf("\n");
for(int i=0; i<rows; i++) {
for(int j=0; j<cols1; j++) {
for(int k=0; k<cols2; k++) {
printf("*arr : %d adress: %p\n",*(*((*arr+i*cols1))+j*cols2+k),*((*arr+i*cols1))+j*cols2+k);
}
}
}
}
void display3DArray2(int rows, int cols1, int cols2,int arr[][cols1][cols2]) {
printf("\n");
for(int i=0; i<rows; i++) {
for(int j=0; j<cols1; j++) {
for(int k=0; k<cols2; k++) {
printf("*arr : %d adress: %p\n", *(*(*(arr+i)+j) + k), *(*(arr+i)+j) + k) ;
}
}
}
}
The code works, but there is something that I wasn't able to understand. When I try to use the second printf of the second function in the first one I get a compilation error:
"invalid use of array with unspecified bounds" -- under gcc.
Why *(arr + i) doesn't work in the first function?
You can use the following two ways to pass/print the matrix:
void display3DArray1(int rows, int cols1, int cols2, int *A) {
int *a, i, j, k;
printf("\n");
for(i=0; i<rows; i++) {
for(j=0; j<cols1; j++) {
for(k=0; k<cols2; k++) {
a= A+(i*cols1*cols2)+(j*cols2)+k;
printf("%d, %p\n", *a, a);
}
}
}
}
void display3DArray2(int A[DIM1][DIM2][DIM3]) {
int i, j, k;
printf("\n");
for(i=0; i<DIM1; i++) {
for(j=0; j<DIM2; j++) {
for(k=0; k<DIM3; k++) {
printf("%d, %p\n", A[i][j][k], &A[i][j][k]);
}
}
}
}
The first method does not rely on the dimensions of the matrix; the second one does. As a result, the first one needs explicit address calculations (row i, col j, cell k).
Use calls respectively:
display3DArray1(DIM1, DIM2, DIM3, (int *)matrix3D);
display3DArray2(matrix3D);
Note the cast of the matrix to an int pointer.
In your code, you used parameter names to specify the dimensions of the matrix. In my C version, that is not legal; they must be constants.
Just a complement to Paul Ogilvie's answer.
The correct usage of Variable Length Arrays would be:
void display3DArray3(int rows, int cols1, int cols2,int arr[][cols1][cols2]) {
printf("\n");
for(int i=0; i<rows; i++) {
for(int j=0; j<cols1; j++) {
for(int k=0; k<cols2; k++) {
printf("*arr : %d adress: %p\n", arr[i][j][k], &arr[i][j][k]);
}
}
}
}
I was a bit puzzled, to be honest. The core issue is that the declaration of a function parameter like in f(T arr[]) declares an incomplete type whose size is not known (neither at compile time nor at run time). Originally I thought empty square brackets in function parameter declarations simply declare a pointer — notation notwithstanding —, but that is not the case. The parameter still has array type, albeit incomplete.1
When you write
void display3DArray1(int rows, int cols1, int cols2,int (*arr)[][cols2])
you declare a pointer to such an incomplete type of unknown size. This pointer cannot be manipulated in all the usual ways; in particular, adding to it in order to jump to the next element is impossible because we don't know where the current element ends (and hence the next element starts). But you try that in
printf("*arr : %d adress: %p\n", *(*(*(arr+i)+j) + k), *(*(arr+i)+j) + k) ;
with the innermost arr+1. Just dereferencing it works, because the variable holds the address of the first element all right. This is what the print in the first function does:
printf("*arr : %d adress: %p\n",*(*((*arr+i*cols1))+j*cols2+k),*((*arr+i*cols1))+j*cols2+k);
with *arr. The element size of the incomplete array to which arr points is known (these elements are arrays of cols2 ints), so that we can add to *arr, even if we can't add to arr proper.
For completeness: Why can you access arr that way in the second function? Well:
void display3DArray2(int rows, int cols1, int cols2,int arr[][cols1][cols2])
declares arr as an incomplete array type, true; but the size of its elements is well known: Those are cols1 x cols2 int matrices. The declaration just doesn't specify how many are there, but we can surely iterate them if we somehow know when to stop.
1 Of course this array, like any other, "decays" to a pointer in most contexts, so that the missing type information doesn't matter. But it matters if we have pointers to them.
I had trouble finding the correct syntax for dereferencing matrices of rank2 and rank3. The intent is to remove a level of indexing for faster execution. Here's some pseudo code that will get you started. This is more about getting the syntax right (MSVC 2019) than an example of working code.
int recurse(int n, unsigned char rank2[13][4], unsigned char rank3[13][4][13])
{
if (n < 13)
{
unsigned char (*deRefRank1)[4] = &rank2[n]; // deref matrix rank 2 to rank 1
unsigned char(*deRefRank2)[4][13] = &rank3[n]; // deref matrix rank 3 to 2
// insert pre-recurse code here ... use (*deRefRank1)[] and (*deRefRank2)[][]
if (recurse(n + 1, rank2[n], rank3[n])
return -1;
// insert post-recurse code here ...
}
return 0;
}
Your array indexing in the 'print' routines is invalid. The declaration of 'matrix3D' implies that the type of that name is 'pointer to int'. NOTE: Only one level of indirection there. The indexing expressions have a bunch of ''-s in front of terms; in C that means 'the item to the right of the '' is a pointer which must be dereferenced to get to the value'. That means you are in effect treating 'matrix3D' as 'pointer to (pointer to (pointer to int))' which is too many indirections.
#include <iostream>
#include <math.h>
#include <stdio.h>
#define column 3
#define row 3
#define share 3
int matrix_multiplication(int left_matrix[][column], int right_matrix[][column], int result_matrix[][column], int rows, int cols, int shared);
int A[][3] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
},
B[][3] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
}, C[3][3]; //initialize "hard coded" three matrices
int main() {
matrix_multiplication(A, B, C, row, column, share); //passes address of each matrix to function
return 0;
}
int matrix_multiplication(int left_matrix[][column], int right_matrix[][column], int result_matrix[][column], int rows, int cols, int shared) {
int i, j, k;
for (i = 0; i < 3; i++) {
for (j = 0; j < 3; j++) {//stays within first column and row through first iteration
for (k = 0; k < 3; k++)//ensures inner dimensions match with variable k i.e. ixk * kxj =ixj or A*B=C
result_matrix[i][j] += right_matrix[i][k] * left_matrix[k][j]; //C programming has ability to perform matrix multiplication
//Adds result of each operation to C matrix i.e. +=
printf("%d\t", result_matrix[i][j]); //matrix C is composed of rows(i) and columns(j)
}//new tab after each column iteration
printf("\n"); //new line for each row iteration
}
return 0;
}
This code is an good example of passing multidimensional arrays to a function using pointers and printing multidimensional arrays after mutliplication. There are multiple ways to indicate pointers to the compiler. I recommend seeing the "Correct way of passing a 2 dimensional array into a function." for example:
/*void display(int (*p)[numcols],int numRows,int numCols)//First method//
void dispaly(int *p,int numRows,int numCols) //Second Method//
void dispaly (int p[][numCols],int numRows,int numCols) //Third Method*/
Remove the column variable, and add this above the matrix_multiplication function declaration:
#define column 3
(You may also want to rename column to COLUMNS.)
In C++ you can also do this:
static const int column = 3;
or, in C++11:
constexpr int column = 3;
The idea behind all this is that all but the very first size of a multidimensional array must be known at compile time.
To fix the expected primary-expression before ']' token" error, change your inner assignment to something like this:
result_matrix[i][j] += right_matrix[i][k] * left_matrix[k][j];
Also you should initialize result_matrix with 0s first.
Also remove the * from int *result_matrix[][column].
Most modern compilers display warnings if you pass an int instead of an int*. Please enable all warnings in your compiler, recompile, fix them, and update your question stating that the example code compiles cleanly, without warnings.
To print an element of the matrix, you have to specify which element you want to print:
printf("%d\t", result_matrix[i][j]);
I can't believe that your compiler didn't display a warning when you omitted the [i][j]. Warnings are for your benefit: they indicate possible bugs in your code.
when I am compiling this small program I am getting different values as output, instead of getting numbers from 0 to 5. And the size of array is always 8. The different values I am getting are:
-981774704
32767
0
0
4195728
0
Any tips would be really valuable. Thank you
#include <stdio.h>
int main() {
int array[10];
int i;
for (i = 0; i < 6; i++) {
printf("%d\n", array[i]);
}
int z = sizeof(&array);
printf("\n Size of array is %d", z);
return 0;
}
You aren't assigning any values to the array, so you're getting the uninitialized values.
You need to do something like array[0] = 5; //or some value etc.
If you want an array of size 8, with the numbers indexing the array stored in it, so {0, 1, 2, 3, 4, 5, 6, 7}, you could do something like:
int array[8];
for(int i = 0; i < 8; ++i)
{
array[i] = i;
}
the elements of the array are uninitialized therefore it's printing garbage value...first initialize the elements with values...you can do this...
#include<stdio.h>
int main()
{
int array[10] ;
int i;
for (i = 0; i < 6; i++)
{
array[i]=i;
printf("%d\n", array[i]);
}
int z = sizeof(&array);
printf("\n Size of array is %d", z);
return 0;
}
if you want the array to be initialized automatically then declare the array globally...
if you want to get the size of the whole array remove &
sizeof(array);
C will not initialize your array to any default value. When you create the array it will be full of garbage values (nothing relevant or defined). Always initialize your data using memset or something equivalnet.
So if you are expecting a value of 0 on a new array then initialize it like this:
memset(array, 0, sizeof(array));
you have to initialize your array first, memory may not be filled with 0 by default