void print_array not working as exprected

-1

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]);

asked Nov 19 '18 at 20:57

user9874845

char **array is not an array, it's a pointer to pointer to char.
– Fiddling Bits
Nov 19 '18 at 20:58

3

Why do you even need an array? Can't you just use height and width?
– Fiddling Bits
Nov 19 '18 at 20:59

add a comment |

-1

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]);

asked Nov 19 '18 at 20:57

user9874845

char **array is not an array, it's a pointer to pointer to char.
– Fiddling Bits
Nov 19 '18 at 20:58

3

Why do you even need an array? Can't you just use height and width?
– Fiddling Bits
Nov 19 '18 at 20:59

add a comment |

-1

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]);

asked Nov 19 '18 at 20:57

user9874845

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]);

c arrays

asked Nov 19 '18 at 20:57

user9874845

asked Nov 19 '18 at 20:57

user9874845

asked Nov 19 '18 at 20:57

user9874845

asked Nov 19 '18 at 20:57

user9874845

asked Nov 19 '18 at 20:57

user9874845

char **array is not an array, it's a pointer to pointer to char.
– Fiddling Bits
Nov 19 '18 at 20:58

3

Why do you even need an array? Can't you just use height and width?
– Fiddling Bits
Nov 19 '18 at 20:59

add a comment |

char **array is not an array, it's a pointer to pointer to char.
– Fiddling Bits
Nov 19 '18 at 20:58

3

Why do you even need an array? Can't you just use height and width?
– Fiddling Bits
Nov 19 '18 at 20:59

char **array is not an array, it's a pointer to pointer to char.
– Fiddling Bits
Nov 19 '18 at 20:58

Why do you even need an array? Can't you just use height and width?
– Fiddling Bits
Nov 19 '18 at 20:59

add a comment |

3 Answers
3

active

oldest

votes

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("|");

}

answered Nov 19 '18 at 22:55

pooh17

444

add a comment |

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.

edited Nov 19 '18 at 21:25

answered Nov 19 '18 at 21:04

Govind Parmar

7,17553055

add a comment |

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");

answered Nov 19 '18 at 21:01

dbush

93.7k12101134

2

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
– WhozCraig
Nov 19 '18 at 21:05

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53382523%2fvoid-print-array-not-working-as-exprected%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

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("|");

}

answered Nov 19 '18 at 22:55

pooh17

444

add a comment |

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("|");

}

answered Nov 19 '18 at 22:55

pooh17

444

add a comment |

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("|");

}

answered Nov 19 '18 at 22:55

pooh17

444

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("|");

}

answered Nov 19 '18 at 22:55

pooh17

444

answered Nov 19 '18 at 22:55

pooh17

444

answered Nov 19 '18 at 22:55

pooh17

444

answered Nov 19 '18 at 22:55

pooh17

444

add a comment |

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.

edited Nov 19 '18 at 21:25

answered Nov 19 '18 at 21:04

Govind Parmar

7,17553055

add a comment |

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.

edited Nov 19 '18 at 21:25

answered Nov 19 '18 at 21:04

Govind Parmar

7,17553055

add a comment |

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.

edited Nov 19 '18 at 21:25

answered Nov 19 '18 at 21:04

Govind Parmar

7,17553055

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.

edited Nov 19 '18 at 21:25

answered Nov 19 '18 at 21:04

Govind Parmar

7,17553055

edited Nov 19 '18 at 21:25

answered Nov 19 '18 at 21:04

Govind Parmar

7,17553055

answered Nov 19 '18 at 21:04

Govind Parmar

7,17553055

answered Nov 19 '18 at 21:04

Govind Parmar

7,17553055

add a comment |

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");

answered Nov 19 '18 at 21:01

dbush

93.7k12101134

2

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
– WhozCraig
Nov 19 '18 at 21:05

add a comment |

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");

answered Nov 19 '18 at 21:01

dbush

93.7k12101134

2

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
– WhozCraig
Nov 19 '18 at 21:05

add a comment |

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");

answered Nov 19 '18 at 21:01

dbush

93.7k12101134

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");

answered Nov 19 '18 at 21:01

dbush

93.7k12101134

answered Nov 19 '18 at 21:01

dbush

93.7k12101134

answered Nov 19 '18 at 21:01

dbush

93.7k12101134

answered Nov 19 '18 at 21:01

dbush

93.7k12101134

2

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
– WhozCraig
Nov 19 '18 at 21:05

add a comment |

2

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
– WhozCraig
Nov 19 '18 at 21:05

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
– WhozCraig
Nov 19 '18 at 21:05

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

Search This Blog

Ufyukyu