Why array of struct lets me index to out of boundary array [duplicate]

-1

This question already has an answer here:

How dangerous is it to access an array out of bounds?

10 answers

Sorry for the bad title and i am new to the C but here is the part i don't understand.

I have a simple struct;

struct st{



        int a;

        int b;

};

and i am creating array of struct and indexing values to variable 'a' in main;

int main(){

         struct st st_arr[2];

         st_arr[0].a = 5;

         st_arr[1].a = 10;

         st_arr[4].a = 20;



         printf("%d %d %dn", st_arr[0].a, st_arr[1].a, st_arr[4].a);

}

i have assigned 2 array of structs but it lets me index 4th of the struct why is that? Isn't it suppose to give me an error?

the output is:

5 10 20

asked Nov 20 '18 at 10:20

M. O. Karaköz

marked as duplicate by Lundin arrays
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Nov 20 '18 at 10:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4

Because there is nobody checking that you go out of bounds,,,,

– Paul Ogilvie
Nov 20 '18 at 10:20

2

It is called undefined behavior meaning it can go wrong, it can go right, your program may abort or anything else can happen, either immediately or at a later stage in your program.

– Paul Ogilvie
Nov 20 '18 at 10:22

1

I know you wanted to be sure but it's important to understand that st_arr[2] is already out of bounds, despite its resemblance to the array definition.

– Peter A. Schneider
Nov 20 '18 at 10:31

thank you so much for your answers

– M. O. Karaköz
Nov 20 '18 at 10:44

add a comment |

-1

This question already has an answer here:

How dangerous is it to access an array out of bounds?

10 answers

Sorry for the bad title and i am new to the C but here is the part i don't understand.

I have a simple struct;

struct st{



        int a;

        int b;

};

and i am creating array of struct and indexing values to variable 'a' in main;

int main(){

         struct st st_arr[2];

         st_arr[0].a = 5;

         st_arr[1].a = 10;

         st_arr[4].a = 20;



         printf("%d %d %dn", st_arr[0].a, st_arr[1].a, st_arr[4].a);

}

i have assigned 2 array of structs but it lets me index 4th of the struct why is that? Isn't it suppose to give me an error?

the output is:

5 10 20

asked Nov 20 '18 at 10:20

M. O. Karaköz

marked as duplicate by Lundin arrays
Users with the arrays badge can single-handedly close arrays questions as duplicates and reopen them as needed.

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Nov 20 '18 at 10:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4

Because there is nobody checking that you go out of bounds,,,,

– Paul Ogilvie
Nov 20 '18 at 10:20

2

It is called undefined behavior meaning it can go wrong, it can go right, your program may abort or anything else can happen, either immediately or at a later stage in your program.

– Paul Ogilvie
Nov 20 '18 at 10:22

1

I know you wanted to be sure but it's important to understand that st_arr[2] is already out of bounds, despite its resemblance to the array definition.

– Peter A. Schneider
Nov 20 '18 at 10:31

thank you so much for your answers

– M. O. Karaköz
Nov 20 '18 at 10:44

add a comment |

-1

This question already has an answer here:

How dangerous is it to access an array out of bounds?

10 answers

Sorry for the bad title and i am new to the C but here is the part i don't understand.

I have a simple struct;

struct st{



        int a;

        int b;

};

and i am creating array of struct and indexing values to variable 'a' in main;

int main(){

         struct st st_arr[2];

         st_arr[0].a = 5;

         st_arr[1].a = 10;

         st_arr[4].a = 20;



         printf("%d %d %dn", st_arr[0].a, st_arr[1].a, st_arr[4].a);

}

i have assigned 2 array of structs but it lets me index 4th of the struct why is that? Isn't it suppose to give me an error?

the output is:

5 10 20

asked Nov 20 '18 at 10:20

M. O. Karaköz

This question already has an answer here:

How dangerous is it to access an array out of bounds?

10 answers

Sorry for the bad title and i am new to the C but here is the part i don't understand.

I have a simple struct;

struct st{



        int a;

        int b;

};

and i am creating array of struct and indexing values to variable 'a' in main;

int main(){

         struct st st_arr[2];

         st_arr[0].a = 5;

         st_arr[1].a = 10;

         st_arr[4].a = 20;



         printf("%d %d %dn", st_arr[0].a, st_arr[1].a, st_arr[4].a);

}

i have assigned 2 array of structs but it lets me index 4th of the struct why is that? Isn't it suppose to give me an error?

the output is:

5 10 20

This question already has an answer here:

How dangerous is it to access an array out of bounds?

10 answers

c arrays

asked Nov 20 '18 at 10:20

M. O. Karaköz

asked Nov 20 '18 at 10:20

M. O. Karaköz

asked Nov 20 '18 at 10:20

M. O. Karaköz

asked Nov 20 '18 at 10:20

M. O. Karaköz

asked Nov 20 '18 at 10:20

M. O. Karaköz

marked as duplicate by Lundin arrays
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Nov 20 '18 at 10:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Lundin arrays
Users with the arrays badge can single-handedly close arrays questions as duplicates and reopen them as needed.

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Nov 20 '18 at 10:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4

Because there is nobody checking that you go out of bounds,,,,

– Paul Ogilvie
Nov 20 '18 at 10:20

2

It is called undefined behavior meaning it can go wrong, it can go right, your program may abort or anything else can happen, either immediately or at a later stage in your program.

– Paul Ogilvie
Nov 20 '18 at 10:22

1

I know you wanted to be sure but it's important to understand that st_arr[2] is already out of bounds, despite its resemblance to the array definition.

– Peter A. Schneider
Nov 20 '18 at 10:31

thank you so much for your answers

– M. O. Karaköz
Nov 20 '18 at 10:44

add a comment |

4

Because there is nobody checking that you go out of bounds,,,,

– Paul Ogilvie
Nov 20 '18 at 10:20

2

It is called undefined behavior meaning it can go wrong, it can go right, your program may abort or anything else can happen, either immediately or at a later stage in your program.

– Paul Ogilvie
Nov 20 '18 at 10:22

1

I know you wanted to be sure but it's important to understand that st_arr[2] is already out of bounds, despite its resemblance to the array definition.

– Peter A. Schneider
Nov 20 '18 at 10:31

thank you so much for your answers

– M. O. Karaköz
Nov 20 '18 at 10:44

Because there is nobody checking that you go out of bounds,,,,

– Paul Ogilvie
Nov 20 '18 at 10:20

It is called undefined behavior meaning it can go wrong, it can go right, your program may abort or anything else can happen, either immediately or at a later stage in your program.

– Paul Ogilvie
Nov 20 '18 at 10:22

I know you wanted to be sure but it's important to understand that st_arr[2] is already out of bounds, despite its resemblance to the array definition.

– Peter A. Schneider
Nov 20 '18 at 10:31

thank you so much for your answers

– M. O. Karaköz
Nov 20 '18 at 10:44

add a comment |

1 Answer
1

active

oldest

votes

C does not provide a(ny) bound checking by default.

Accessing out of bounds (i.e., invalid memory) is defined to have undefined behaviour.

answered Nov 20 '18 at 10:23

Sourav Ghosh

109k14129188

Maybe you should change "defined" to "specified" so that I can get back out of the rabbit hole ;-)

– Peter A. Schneider
Nov 20 '18 at 10:28

@PeterA.Schneider That was intentional. :)

– Sourav Ghosh
Nov 20 '18 at 10:30

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

C does not provide a(ny) bound checking by default.

Accessing out of bounds (i.e., invalid memory) is defined to have undefined behaviour.

answered Nov 20 '18 at 10:23

Sourav Ghosh

109k14129188

Maybe you should change "defined" to "specified" so that I can get back out of the rabbit hole ;-)

– Peter A. Schneider
Nov 20 '18 at 10:28

@PeterA.Schneider That was intentional. :)

– Sourav Ghosh
Nov 20 '18 at 10:30

add a comment |

C does not provide a(ny) bound checking by default.

Accessing out of bounds (i.e., invalid memory) is defined to have undefined behaviour.

answered Nov 20 '18 at 10:23

Sourav Ghosh

109k14129188

Maybe you should change "defined" to "specified" so that I can get back out of the rabbit hole ;-)

– Peter A. Schneider
Nov 20 '18 at 10:28

@PeterA.Schneider That was intentional. :)

– Sourav Ghosh
Nov 20 '18 at 10:30

add a comment |

C does not provide a(ny) bound checking by default.

Accessing out of bounds (i.e., invalid memory) is defined to have undefined behaviour.

answered Nov 20 '18 at 10:23

Sourav Ghosh

109k14129188

C does not provide a(ny) bound checking by default.

Accessing out of bounds (i.e., invalid memory) is defined to have undefined behaviour.

answered Nov 20 '18 at 10:23

Sourav Ghosh

109k14129188

answered Nov 20 '18 at 10:23

Sourav Ghosh

109k14129188

answered Nov 20 '18 at 10:23

Sourav Ghosh

109k14129188

answered Nov 20 '18 at 10:23

Sourav Ghosh

109k14129188

Maybe you should change "defined" to "specified" so that I can get back out of the rabbit hole ;-)

– Peter A. Schneider
Nov 20 '18 at 10:28

@PeterA.Schneider That was intentional. :)

– Sourav Ghosh
Nov 20 '18 at 10:30

add a comment |

Maybe you should change "defined" to "specified" so that I can get back out of the rabbit hole ;-)

– Peter A. Schneider
Nov 20 '18 at 10:28

@PeterA.Schneider That was intentional. :)

– Sourav Ghosh
Nov 20 '18 at 10:30

Maybe you should change "defined" to "specified" so that I can get back out of the rabbit hole ;-)

– Peter A. Schneider
Nov 20 '18 at 10:28

@PeterA.Schneider That was intentional. :)

– Sourav Ghosh
Nov 20 '18 at 10:30

add a comment |

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