void print_array not working as exprected

how to call a void method in java
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I have this code written in C:

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

void print_array(char** array, int height, int width);

int main()
{
    int height, width;
    char **array = 0;

    printf("Give height board size:");
    scanf("%d", &height);
    printf("Give width board size:");
    scanf("%d", &width);

    print_array(array, height, width);


    return 0;
}

void print_array(char** array, int height, int width)
{
    int i, j;

    printf("\n |"); for (j = 0; j < width; j++) printf("-");
    printf("|\n");
    for (i = 0; i < height; i++)
    {
        printf(" |");
        for (j = 0; j < width; j++) printf("%c", array[i][j]);
        printf("|\n");
    }
    printf(" |"); for (j = 0; j < width; j++) printf("-");
    printf("|");
}

The expected result was this for 10x10

|----------|
|          |
|          |
|          |
|          |
|          |
|          |
|          |
|          |
|          |
|          |
|----------|

But the actual result for whatever number I give to Height is

E.g. Height 10 and Width 20

|--------------------|
|

If I run with Visual Studio I actually get and error code on line 32 which is the following

Exception thrown: read access violation. array was 0x1110112.

Line 32

for (j = 0; j < width; j++) printf("%c", array[i][j]);

This answer is probably helpful to you. It gives an answer similar to @Govind Parmar's, but with this solution the array won't allocate a contiguous region of memory, which could give problems with functions that assume this.

Instead, you could do:

// ConsoleApplication1.cpp : Defines the entry point for the console application.
//


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

void print_array(char** array, int height, int width);

int main()
{
    int height, width;
    char** array;
    char* temp;

    printf("Give height board size:");
    scanf_s("%d", &height);
    printf("Give width board size:");
    scanf_s("%d", &width);

    array = malloc(height * sizeof(char*));
    temp = malloc(height * width * sizeof(char*));

    for (int i = 0; i < height; i++) {
        array[i] = temp + (i * width);
    }

    print_array(array, height, width);

    free(temp);
    free(array);

    return 0;
}

void print_array(char** array, int height, int width)
{
    int i, j;

    printf("\n |"); for (j = 0; j < width; j++) printf("-");
    printf("|\n");
    for (i = 0; i < height; i++)
    {
        printf(" |");
        for (j = 0; j < width; j++) printf("%c", array[i][j]);
        printf("|\n");
    }
    printf(" |"); for (j = 0; j < width; j++) printf("-");
    printf("|");
}

60 Tips On Object Oriented Programming, This is a defensive programming technique' to avoid accessing null pointer by and using null objects can even be done for avoiding null pointer access problem length arrays instead of passing null when a method expects an array object. explicit null check we need not worry: // Java code public static void printArray  Hello. I am trying to return an array and print the values. Help. I can't figure it out. Thank you again. The problem is with the main method. You're creating a new array of the same length as that of the one in the "getNames()" method. No need to do that. When you call the "getNames()" method, you

Since you are declaring array as a pointer to a pointer to char, but not defining it in read-only memory in the same line as declaration, you have to use the functions malloc to first allocate the array, and then later free it:

int i;
array = malloc(sizeof(char *) * height);
for(i = 0; i < height; i++)
    array[i] = malloc(width);

// Use array....

for(i = 0; i < height; i++)
    free(array[i]);

free(array);

That's if you actually need a 2-dimensional array. For your use case of printing out a square of arbitrary width and height, you don't.

Also note that the choice of allocating height as the first dimension, and then width as the second dimension, is completely arbitrary and it could just as easily be the other way around.

Scripting in Java: Integrating with Groovy and JavaScript, You can use a Nashorn array when a Java array is expected. Suppose that you have a PrintArray class, as shown in Listing 7-6, that contains a PrintArray.java package com.jdojo.script; public class PrintArray { public void print(String[] list)  1st void* arr1 is not array - it just pointer to anything. 2nd - u can start work get address from some multibyte object convert it type as (void*) so u get void* arr1 that containt some value in it. 3nd for arr1[i] or (arr1+i) u must convert pointer arr1 to some type (int ) (float*) etc (struct ustract*)

array is an uninitialized pointer-to-pointer. Attempting to dereference that array via the [] operator invokes undefined behavior.

You don't need an array here at all. Just print a space:

printf(" |");
for (j = 0; j < width; j++) printf(" ");
printf("|\n");

Beginning Java 8 APIs, Extensions and Libraries: Swing, JavaFX, , You can use a JavaScript array when a Java array is expected. Suppose you have a PrintArray class, as shown in Listing 10-22, that contains a print() method package com.jdojo.script; public class PrintArray { public void print(String[] list)​  Arduino digital pins are inputs by default, although it does no harm to explicitly declare inputs in setup. Your loop code could look like this:

Arrays | Think Java, public static void printArray(int[] a) { System.out.print("{" + a[0]); for (int i = 1; Arrays that provides methods for working with arrays. If you actually want to copy the array, not just a reference, you have to create a new array Usually determinism is a good thing, since we expect the same calculation to yield the same result. As with most things it was very simple, and so I over looked it because I was so frustrated with the Sainsmart 2004 lcd. What I was doing wrong was I was not including the lcd in the setup. Void setup should have ended at the double c declaration and void loop should have started at the set cursor statement after it.

Arrays and references | Think Java, public static void printArray(int[] a) { System.out.print("{" + a[0]); for (int i = 1; Usually determinism is a good thing, since we expect the same calculation to yield The enhanced for loop does not work with strings directly, but you can convert  r/arduino: A place for all things Arduino! Press J to jump to the feed. Press question mark to learn the rest of the keyboard shortcuts

array_values - Manual, print_r(array_values($array)); ?> The above example will output: Array ( [0] => XL​  Java Print Array Examples. Dec 25, 2015 Array, Core Java, Examples comments Arrays are usually useful when working with arbitrarily large number of data having the same type. It is usually convenient if we can print the contents of an array. Below are some examples on how to print the contents of an Array in Java.

Comments
  • char **array is not an array, it's a pointer to pointer to char.
  • Why do you even need an array? Can't you just use height and width?
  • I suspect the expectation was the (nonexistent) "array" was loaded with spaces already and the goal of the function was to exhibit that misinformed fact. Otherwise the function would have been more aptly named, print_box. =O