Assigning character in array of strings segmentation fault in C

Very confused on why I'm getting a "Segmentation Fault (core dumped)".

const int bufSize = 32;

char* splitStrings[bufSize];

splitStrings[0][0] = 'a';

splitStrings[0][1] = '';

printf("testchar: %c",splitStrings[0][0]);

printf("teststr: %s",splitStrings[0]);

Thanks for any help

asked Nov 21 '18 at 12:26

Coder77

65221323

2

You declared a list of 32 char *s. And what they point to is undefined.

– usr2564301
Nov 21 '18 at 12:32

malloc is your friend.

– Gerhardh
Nov 21 '18 at 12:41

amen to malloc, thanks guys

– Coder77
Nov 21 '18 at 13:07

add a comment |

Very confused on why I'm getting a "Segmentation Fault (core dumped)".

const int bufSize = 32;

char* splitStrings[bufSize];

splitStrings[0][0] = 'a';

splitStrings[0][1] = '';

printf("testchar: %c",splitStrings[0][0]);

printf("teststr: %s",splitStrings[0]);

Thanks for any help

asked Nov 21 '18 at 12:26

Coder77

65221323

2

You declared a list of 32 char *s. And what they point to is undefined.

– usr2564301
Nov 21 '18 at 12:32

malloc is your friend.

– Gerhardh
Nov 21 '18 at 12:41

amen to malloc, thanks guys

– Coder77
Nov 21 '18 at 13:07

add a comment |

Very confused on why I'm getting a "Segmentation Fault (core dumped)".

const int bufSize = 32;

char* splitStrings[bufSize];

splitStrings[0][0] = 'a';

splitStrings[0][1] = '';

printf("testchar: %c",splitStrings[0][0]);

printf("teststr: %s",splitStrings[0]);

Thanks for any help

asked Nov 21 '18 at 12:26

Coder77

65221323

Very confused on why I'm getting a "Segmentation Fault (core dumped)".

const int bufSize = 32;

char* splitStrings[bufSize];

splitStrings[0][0] = 'a';

splitStrings[0][1] = '';

printf("testchar: %c",splitStrings[0][0]);

printf("teststr: %s",splitStrings[0]);

Thanks for any help

c string segmentation-fault

asked Nov 21 '18 at 12:26

Coder77

65221323

asked Nov 21 '18 at 12:26

Coder77

65221323

asked Nov 21 '18 at 12:26

Coder77

65221323

asked Nov 21 '18 at 12:26

Coder77

65221323

asked Nov 21 '18 at 12:26

Coder77

65221323

2

You declared a list of 32 char *s. And what they point to is undefined.

– usr2564301
Nov 21 '18 at 12:32

malloc is your friend.

– Gerhardh
Nov 21 '18 at 12:41

amen to malloc, thanks guys

– Coder77
Nov 21 '18 at 13:07

add a comment |

2

You declared a list of 32 char *s. And what they point to is undefined.

– usr2564301
Nov 21 '18 at 12:32

malloc is your friend.

– Gerhardh
Nov 21 '18 at 12:41

amen to malloc, thanks guys

– Coder77
Nov 21 '18 at 13:07

You declared a list of 32 char *s. And what they point to is undefined.

– usr2564301
Nov 21 '18 at 12:32

malloc is your friend.

– Gerhardh
Nov 21 '18 at 12:41

amen to malloc, thanks guys

– Coder77
Nov 21 '18 at 13:07

add a comment |

4 Answers
4

active

oldest

votes

I suggest you read more about how to allocate dynamic memory in C with malloc() function.

I think you want to define a 2-D char array. To do this make these steps:

First

Define a char** pointer.
char **splitStrings[bufSize];

Second

Allocate memory for row: splitStrings = malloc(bufSize*sizeof(char));

Finally

For each row allocate memory for its columns.

for (int i=0;i<bufSize;i++){ splitStrings[i]=malloc(bufSize*sizeof(char)); }

Code

#include <stdio.h>///for printf()

#include <stdlib.h>///for malloc()



int main(){



    const int bufSize = 32;

    char **splitStrings;



    splitStrings = malloc(bufSize*sizeof(char*));



    for (int i=0;i<bufSize;i++){

        splitStrings[i]=malloc(bufSize*sizeof(char*));

    }



    splitStrings[0][0] = 'a';

    splitStrings[0][1] = '';



    printf("testchar: %cn", splitStrings[0][0]);

    printf("teststr: %sn", splitStrings[0]);



    free(splitStrings);



}

Further Notes

more information about 2-D dynamic array see

By using malloc() function in C never cast the result of it see

After allocating dynamic memory in Heap never forget to release it by using free().

Last Edition on code

Change malloc(bufSize*sizeof(char)); to malloc(bufSize*sizeof(char*)); this is true way of allocating memory

Because in first case calling free() on splitStrings pointer cause memory error. Why?

edited Dec 26 '18 at 16:13

answered Nov 21 '18 at 12:45

EsmaeelE

1,13321121

2

I believe OP is still a begginer in terms of C, and given the question's content, I would assume he will have a lot of trouble understanding this answer without any explanation or comment.

– Ian A.B.K.
Nov 21 '18 at 12:49

add a comment |

You missed a layer of declaration. You have a table of pointer on char splitStrings. Then you would need to book some memory space and point to it using your pointer. Here is the example for the first pointer:

splitStrings[0] = malloc(2*sizeof(char));

splitStrings[0][0] = 'a';

splitStrings[0][1] = '';

printf("testchar: %c",splitStrings[0][0]);

printf("teststr: %s",splitStrings[0]);

Read how malloc is working and especially how to de-allocate memory with free once your done to avoid memory leak.

answered Nov 21 '18 at 12:48

Puck

1,4131224

Because by definition sizeof(char) will always == 1, your statement can be simplified to ...malloc(2);

– ryyker
Nov 21 '18 at 13:29

add a comment |

char* splitStrings[bufSize]; //is expecting address, if it is derefencing directely will through segmentation fault.

we have two options one is dynamic memory allocation otherwise declare like below.

#include<stdio.h>

void main()

{

    const int bufSize = 32;

    char splitStrings[1][bufSize];

    splitStrings[0][0] = 'a';

    splitStrings[0][1] = '';

    printf("testchar: %c",splitStrings[0][0]);

    printf("teststr: %s",splitStrings[0]);

}

answered Nov 21 '18 at 13:08

Ramya Muralidharan

1015

add a comment |

Others have answered how to fix the error, so I won't go over that. Instead, I will answer the question of why you are getting a segmentation fault.

So, let's step through your code and see what's happening!

const int buffsize = 32;

char* splitStrings[buffsize];

Alright! So far, you have declared a const int of size 32 (good use of const btw!), and you have created a pointer to an array of characters of size 32!

So let's look at the next line!

splitStrings[0][0] = 'a';

So now, you are trying to look at the first item in the array that splitStrings points to, and then looking at the first item in that array. And this is where you get your segmentation fault!

In C, a segmentation fault occurs when something tries to access memory that it is not allowed to access. In this case, it's alright for you to access splitStrings[0], but not splitStrings[0][0]. This is because you currently don't have an array at splitStrings[0] - instead, you just have an unassigned pointer to a character.

So, when the compiler tries to work through that, it says "Okay, let's look at the first item in splitStrings! Alright, that's a pointer to a character! Now, let's look at the first item under that character! It's - wait, hold on. I haven't assigned that yet, it isn't allowed to have its own array yet! Segmentation fault!"

To fix this, you will need to create a 2D array (that's an array of arrays) and allocate that memory. EsmaeelE has given good instructions on how to do that in their answer here. I also highly recommend the TutorialsPoint segments on multi-dimensional arrays and arrays of pointers. The site is an excellent resource for anyone trying to learn the C programming language!

I hope that helped you understand the segmentation fault better! Feel free to ask for any clarifications!

answered Dec 26 '18 at 16:59

MrSpudtastic

1468

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

I suggest you read more about how to allocate dynamic memory in C with malloc() function.

I think you want to define a 2-D char array. To do this make these steps:

First

Define a char** pointer.
char **splitStrings[bufSize];

Second

Allocate memory for row: splitStrings = malloc(bufSize*sizeof(char));

Finally

For each row allocate memory for its columns.

for (int i=0;i<bufSize;i++){ splitStrings[i]=malloc(bufSize*sizeof(char)); }

Code

#include <stdio.h>///for printf()

#include <stdlib.h>///for malloc()



int main(){



    const int bufSize = 32;

    char **splitStrings;



    splitStrings = malloc(bufSize*sizeof(char*));



    for (int i=0;i<bufSize;i++){

        splitStrings[i]=malloc(bufSize*sizeof(char*));

    }



    splitStrings[0][0] = 'a';

    splitStrings[0][1] = '';



    printf("testchar: %cn", splitStrings[0][0]);

    printf("teststr: %sn", splitStrings[0]);



    free(splitStrings);



}

Further Notes

more information about 2-D dynamic array see

By using malloc() function in C never cast the result of it see

After allocating dynamic memory in Heap never forget to release it by using free().

Last Edition on code

Change malloc(bufSize*sizeof(char)); to malloc(bufSize*sizeof(char*)); this is true way of allocating memory

Because in first case calling free() on splitStrings pointer cause memory error. Why?

edited Dec 26 '18 at 16:13

answered Nov 21 '18 at 12:45

EsmaeelE

1,13321121

2

I believe OP is still a begginer in terms of C, and given the question's content, I would assume he will have a lot of trouble understanding this answer without any explanation or comment.

– Ian A.B.K.
Nov 21 '18 at 12:49

add a comment |

I suggest you read more about how to allocate dynamic memory in C with malloc() function.

I think you want to define a 2-D char array. To do this make these steps:

First

Define a char** pointer.
char **splitStrings[bufSize];

Second

Allocate memory for row: splitStrings = malloc(bufSize*sizeof(char));

Finally

For each row allocate memory for its columns.

for (int i=0;i<bufSize;i++){ splitStrings[i]=malloc(bufSize*sizeof(char)); }

Code

#include <stdio.h>///for printf()

#include <stdlib.h>///for malloc()



int main(){



    const int bufSize = 32;

    char **splitStrings;



    splitStrings = malloc(bufSize*sizeof(char*));



    for (int i=0;i<bufSize;i++){

        splitStrings[i]=malloc(bufSize*sizeof(char*));

    }



    splitStrings[0][0] = 'a';

    splitStrings[0][1] = '';



    printf("testchar: %cn", splitStrings[0][0]);

    printf("teststr: %sn", splitStrings[0]);



    free(splitStrings);



}

Further Notes

more information about 2-D dynamic array see

By using malloc() function in C never cast the result of it see

After allocating dynamic memory in Heap never forget to release it by using free().

Last Edition on code

Change malloc(bufSize*sizeof(char)); to malloc(bufSize*sizeof(char*)); this is true way of allocating memory

Because in first case calling free() on splitStrings pointer cause memory error. Why?

edited Dec 26 '18 at 16:13

answered Nov 21 '18 at 12:45

EsmaeelE

1,13321121

2

I believe OP is still a begginer in terms of C, and given the question's content, I would assume he will have a lot of trouble understanding this answer without any explanation or comment.

– Ian A.B.K.
Nov 21 '18 at 12:49

add a comment |

I suggest you read more about how to allocate dynamic memory in C with malloc() function.

I think you want to define a 2-D char array. To do this make these steps:

First

Define a char** pointer.
char **splitStrings[bufSize];

Second

Allocate memory for row: splitStrings = malloc(bufSize*sizeof(char));

Finally

For each row allocate memory for its columns.

for (int i=0;i<bufSize;i++){ splitStrings[i]=malloc(bufSize*sizeof(char)); }

Code

#include <stdio.h>///for printf()

#include <stdlib.h>///for malloc()



int main(){



    const int bufSize = 32;

    char **splitStrings;



    splitStrings = malloc(bufSize*sizeof(char*));



    for (int i=0;i<bufSize;i++){

        splitStrings[i]=malloc(bufSize*sizeof(char*));

    }



    splitStrings[0][0] = 'a';

    splitStrings[0][1] = '';



    printf("testchar: %cn", splitStrings[0][0]);

    printf("teststr: %sn", splitStrings[0]);



    free(splitStrings);



}

Further Notes

more information about 2-D dynamic array see

By using malloc() function in C never cast the result of it see

After allocating dynamic memory in Heap never forget to release it by using free().

Last Edition on code

Change malloc(bufSize*sizeof(char)); to malloc(bufSize*sizeof(char*)); this is true way of allocating memory

Because in first case calling free() on splitStrings pointer cause memory error. Why?

edited Dec 26 '18 at 16:13

answered Nov 21 '18 at 12:45

EsmaeelE

1,13321121

I suggest you read more about how to allocate dynamic memory in C with malloc() function.

I think you want to define a 2-D char array. To do this make these steps:

First

Define a char** pointer.
char **splitStrings[bufSize];

Second

Allocate memory for row: splitStrings = malloc(bufSize*sizeof(char));

Finally

For each row allocate memory for its columns.

for (int i=0;i<bufSize;i++){ splitStrings[i]=malloc(bufSize*sizeof(char)); }

Code

#include <stdio.h>///for printf()

#include <stdlib.h>///for malloc()



int main(){



    const int bufSize = 32;

    char **splitStrings;



    splitStrings = malloc(bufSize*sizeof(char*));



    for (int i=0;i<bufSize;i++){

        splitStrings[i]=malloc(bufSize*sizeof(char*));

    }



    splitStrings[0][0] = 'a';

    splitStrings[0][1] = '';



    printf("testchar: %cn", splitStrings[0][0]);

    printf("teststr: %sn", splitStrings[0]);



    free(splitStrings);



}

Further Notes

more information about 2-D dynamic array see

By using malloc() function in C never cast the result of it see

After allocating dynamic memory in Heap never forget to release it by using free().

Last Edition on code

Change malloc(bufSize*sizeof(char)); to malloc(bufSize*sizeof(char*)); this is true way of allocating memory

Because in first case calling free() on splitStrings pointer cause memory error. Why?

edited Dec 26 '18 at 16:13

answered Nov 21 '18 at 12:45

EsmaeelE

1,13321121

edited Dec 26 '18 at 16:13

answered Nov 21 '18 at 12:45

EsmaeelE

1,13321121

answered Nov 21 '18 at 12:45

EsmaeelE

1,13321121

answered Nov 21 '18 at 12:45

EsmaeelE

1,13321121

2

I believe OP is still a begginer in terms of C, and given the question's content, I would assume he will have a lot of trouble understanding this answer without any explanation or comment.

– Ian A.B.K.
Nov 21 '18 at 12:49

add a comment |

2

I believe OP is still a begginer in terms of C, and given the question's content, I would assume he will have a lot of trouble understanding this answer without any explanation or comment.

– Ian A.B.K.
Nov 21 '18 at 12:49

I believe OP is still a begginer in terms of C, and given the question's content, I would assume he will have a lot of trouble understanding this answer without any explanation or comment.

– Ian A.B.K.
Nov 21 '18 at 12:49

add a comment |

splitStrings[0] = malloc(2*sizeof(char));

splitStrings[0][0] = 'a';

splitStrings[0][1] = '';

printf("testchar: %c",splitStrings[0][0]);

printf("teststr: %s",splitStrings[0]);

Read how malloc is working and especially how to de-allocate memory with free once your done to avoid memory leak.

answered Nov 21 '18 at 12:48

Puck

1,4131224

Because by definition sizeof(char) will always == 1, your statement can be simplified to ...malloc(2);

– ryyker
Nov 21 '18 at 13:29

add a comment |

splitStrings[0] = malloc(2*sizeof(char));

splitStrings[0][0] = 'a';

splitStrings[0][1] = '';

printf("testchar: %c",splitStrings[0][0]);

printf("teststr: %s",splitStrings[0]);

Read how malloc is working and especially how to de-allocate memory with free once your done to avoid memory leak.

answered Nov 21 '18 at 12:48

Puck

1,4131224

Because by definition sizeof(char) will always == 1, your statement can be simplified to ...malloc(2);

– ryyker
Nov 21 '18 at 13:29

add a comment |

splitStrings[0] = malloc(2*sizeof(char));

splitStrings[0][0] = 'a';

splitStrings[0][1] = '';

printf("testchar: %c",splitStrings[0][0]);

printf("teststr: %s",splitStrings[0]);

Read how malloc is working and especially how to de-allocate memory with free once your done to avoid memory leak.

answered Nov 21 '18 at 12:48

Puck

1,4131224

splitStrings[0] = malloc(2*sizeof(char));

splitStrings[0][0] = 'a';

splitStrings[0][1] = '';

printf("testchar: %c",splitStrings[0][0]);

printf("teststr: %s",splitStrings[0]);

Read how malloc is working and especially how to de-allocate memory with free once your done to avoid memory leak.

answered Nov 21 '18 at 12:48

Puck

1,4131224

answered Nov 21 '18 at 12:48

Puck

1,4131224

answered Nov 21 '18 at 12:48

Puck

1,4131224

answered Nov 21 '18 at 12:48

Puck

1,4131224

Because by definition sizeof(char) will always == 1, your statement can be simplified to ...malloc(2);

– ryyker
Nov 21 '18 at 13:29

add a comment |

Because by definition sizeof(char) will always == 1, your statement can be simplified to ...malloc(2);

– ryyker
Nov 21 '18 at 13:29

Because by definition sizeof(char) will always == 1, your statement can be simplified to ...malloc(2);

– ryyker
Nov 21 '18 at 13:29

add a comment |

char* splitStrings[bufSize]; //is expecting address, if it is derefencing directely will through segmentation fault.

we have two options one is dynamic memory allocation otherwise declare like below.

#include<stdio.h>

void main()

{

    const int bufSize = 32;

    char splitStrings[1][bufSize];

    splitStrings[0][0] = 'a';

    splitStrings[0][1] = '';

    printf("testchar: %c",splitStrings[0][0]);

    printf("teststr: %s",splitStrings[0]);

}

answered Nov 21 '18 at 13:08

Ramya Muralidharan

1015

add a comment |

char* splitStrings[bufSize]; //is expecting address, if it is derefencing directely will through segmentation fault.

we have two options one is dynamic memory allocation otherwise declare like below.

#include<stdio.h>

void main()

{

    const int bufSize = 32;

    char splitStrings[1][bufSize];

    splitStrings[0][0] = 'a';

    splitStrings[0][1] = '';

    printf("testchar: %c",splitStrings[0][0]);

    printf("teststr: %s",splitStrings[0]);

}

answered Nov 21 '18 at 13:08

Ramya Muralidharan

1015

add a comment |

char* splitStrings[bufSize]; //is expecting address, if it is derefencing directely will through segmentation fault.

we have two options one is dynamic memory allocation otherwise declare like below.

#include<stdio.h>

void main()

{

    const int bufSize = 32;

    char splitStrings[1][bufSize];

    splitStrings[0][0] = 'a';

    splitStrings[0][1] = '';

    printf("testchar: %c",splitStrings[0][0]);

    printf("teststr: %s",splitStrings[0]);

}

answered Nov 21 '18 at 13:08

Ramya Muralidharan

1015

char* splitStrings[bufSize]; //is expecting address, if it is derefencing directely will through segmentation fault.

we have two options one is dynamic memory allocation otherwise declare like below.

#include<stdio.h>

void main()

{

    const int bufSize = 32;

    char splitStrings[1][bufSize];

    splitStrings[0][0] = 'a';

    splitStrings[0][1] = '';

    printf("testchar: %c",splitStrings[0][0]);

    printf("teststr: %s",splitStrings[0]);

}

answered Nov 21 '18 at 13:08

Ramya Muralidharan

1015

answered Nov 21 '18 at 13:08

Ramya Muralidharan

1015

answered Nov 21 '18 at 13:08

Ramya Muralidharan

1015

answered Nov 21 '18 at 13:08

Ramya Muralidharan

1015

add a comment |

Others have answered how to fix the error, so I won't go over that. Instead, I will answer the question of why you are getting a segmentation fault.

So, let's step through your code and see what's happening!

const int buffsize = 32;

char* splitStrings[buffsize];

Alright! So far, you have declared a const int of size 32 (good use of const btw!), and you have created a pointer to an array of characters of size 32!

So let's look at the next line!

splitStrings[0][0] = 'a';

So now, you are trying to look at the first item in the array that splitStrings points to, and then looking at the first item in that array. And this is where you get your segmentation fault!

I hope that helped you understand the segmentation fault better! Feel free to ask for any clarifications!

answered Dec 26 '18 at 16:59

MrSpudtastic

1468

add a comment |

Others have answered how to fix the error, so I won't go over that. Instead, I will answer the question of why you are getting a segmentation fault.

So, let's step through your code and see what's happening!

const int buffsize = 32;

char* splitStrings[buffsize];

Alright! So far, you have declared a const int of size 32 (good use of const btw!), and you have created a pointer to an array of characters of size 32!

So let's look at the next line!

splitStrings[0][0] = 'a';

So now, you are trying to look at the first item in the array that splitStrings points to, and then looking at the first item in that array. And this is where you get your segmentation fault!

I hope that helped you understand the segmentation fault better! Feel free to ask for any clarifications!

answered Dec 26 '18 at 16:59

MrSpudtastic

1468

add a comment |

Others have answered how to fix the error, so I won't go over that. Instead, I will answer the question of why you are getting a segmentation fault.

So, let's step through your code and see what's happening!

const int buffsize = 32;

char* splitStrings[buffsize];

Alright! So far, you have declared a const int of size 32 (good use of const btw!), and you have created a pointer to an array of characters of size 32!

So let's look at the next line!

splitStrings[0][0] = 'a';

So now, you are trying to look at the first item in the array that splitStrings points to, and then looking at the first item in that array. And this is where you get your segmentation fault!

I hope that helped you understand the segmentation fault better! Feel free to ask for any clarifications!

answered Dec 26 '18 at 16:59

MrSpudtastic

1468

Others have answered how to fix the error, so I won't go over that. Instead, I will answer the question of why you are getting a segmentation fault.

So, let's step through your code and see what's happening!

const int buffsize = 32;

char* splitStrings[buffsize];

Alright! So far, you have declared a const int of size 32 (good use of const btw!), and you have created a pointer to an array of characters of size 32!

So let's look at the next line!

splitStrings[0][0] = 'a';

So now, you are trying to look at the first item in the array that splitStrings points to, and then looking at the first item in that array. And this is where you get your segmentation fault!

I hope that helped you understand the segmentation fault better! Feel free to ask for any clarifications!

answered Dec 26 '18 at 16:59

MrSpudtastic

1468

answered Dec 26 '18 at 16:59

MrSpudtastic

1468

answered Dec 26 '18 at 16:59

MrSpudtastic

1468

answered Dec 26 '18 at 16:59

MrSpudtastic

1468

add a comment |

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