problem with array of strings

This is a discussion on problem with array of strings within the C++ Programming forums, part of the General Programming Boards category; hello i have string array of variable size. could anyone let me know how to make sure that at the ...

  1. #1
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    problem with array of strings

    hello


    i have string array of variable size. could anyone let me know how to make sure that at the null terminater '\0' is at the end of the string

    Code:
    
    here is code :#include <stdio.h>
    #include <math.h>
    
    #define  SIZE  9
    
    int main()
    	{
    
    	char Buff  [ SIZE ] ;
    	int  result;
    
    	result =SIZE + 1 ;
    	Buff[result] = '\0';
    	Buff[result/2] = '|';
    
    	printf("%c\n\n\n",Buff[result]);
    	
    	return 0;
    
    	}
    when i remove the line Buff[result] = '\0' i donot get any error during the execution of porgram. but with that line include i get the following error message

    runtime error #2 the stack around the variable Buff was corrupted

  2. #2
    Registered User C_ntua's Avatar
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    It is Buff[SIZE-1] = '\0', since you have SIZE elements from 0 to SIZE-1

  3. #3
    Registered User whiteflags's Avatar
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    'result' isn't a valid index. I'm not sure what you meant to do but you could possibly do this:

    #include <cstring>
    #define SIZE 9

    char buff[SIZE + 1];
    int i = 0;
    buff[i++] = '|';
    buff[i] = 0;

    Of course since this is c++ another option is to use the string class, which will manage some of the more tedious things for you.

    #include <string>

    std::string buff;
    buff.push_back('|');

    Good luck.

  4. #4
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    i used "for" to fill the blanks in string array with " " (space) , I still get the error message.

  5. #5
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    Quite simply put, you're writing beyond the end of the array.
    If you do...
    char buff[SIZE]
    ...then the indexes range from 0 to SIZE - 1.
    SIZE is out of bounds!

    And if that doesn't help, you need to show the code you used.
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  6. #6
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    I think this is what you're after.

    Code:
    #include <stdio.h>
    #include <math.h>
    
    #define  SIZE  9
    
    int main()
    	{
    
    	char Buff[SIZE] ; // Declaring an array size of 9 elements 0-8.
    	int result;
    
    	result = SIZE - 1; // result = 8
    	Buff[result] = '\0'; // Buff[8] = null terminator.
    	Buff[result/2] = '|'; // Buff[4] will now hold the character '|'.
    
    	printf("&#37;c\n\n\n",Buff[result]); // Are you sure you want to print the null terminator?
    	
    	return 0;
    
    }
    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.

  7. #7
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    thank you guys for replying.

    the way i understnd is that if the char array is declared as follow

    #define SIZE 10

    char buff [SIZE]

    then i can only write in elelments 0 to 8
    and element 9 is automatically filled with '\0'
    and that u can't write to element 9 even if it is '\0';

    if i am correct just let me know.

    although '\0' is automatically placed , now another question is how i ensure that '\0' is at the end of array without writing '\0' at the end of array.

  8. #8
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by cole View Post
    thank you guys for replying.

    the way i understnd is that if the char array is declared as follow

    #define SIZE 10

    char buff [SIZE]

    then i can only write in elelments 0 to 8
    Everything up to here is fine.
    and element 9 is automatically filled with '\0'
    and that u can't write to element 9 even if it is '\0';

    if i am correct just let me know.

    although '\0' is automatically placed , now another question is how i ensure that '\0' is at the end of array without writing '\0' at the end of array.
    And all of this is just wrong. Nothing is automatically written into a char array for you, and therefore you must do buff[9] = '\0' yourself.

  9. #9
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    just to clarify , i declare a character array as follow

    char buff[] = "string";


    would not the character array will have s, t , r , i ,n ,g , \0 ;

    that means total length of 7 elements and

    \0 would be placed automatically because you declared a array of type char,
    if the above statement is wrong, then how do i place \0 myself. As i said i tried to write \0 at element 7 and i get error message (the one i posted at the very begining of thread)

    thanking you

    c

  10. #10
    C++ Witch laserlight's Avatar
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    If that is really what you did, then you are correct, '\0' is indeed at buff[6].
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  11. #11
    Registered User C_ntua's Avatar
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    Quote Originally Posted by cole View Post
    just to clarify , i declare a character array as follow

    char buff[] = "string";


    would not the character array will have s, t , r , i ,n ,g , \0 ;

    that means total length of 7 elements and

    \0 would be placed automatically because you declared a array of type char,
    if the above statement is wrong, then how do i place \0 myself. As i said i tried to write \0 at element 7 and i get error message (the one i posted at the very begining of thread)

    thanking you

    c
    It is not correct actually. It is not because you declared an array of char that the last element is \0. Is because you initialized it with a string literal. A string literal ("something") includes the \0, even though you don't "see" it. If you write this:"a" , anywhere you write this, the compiler will see an array of 2 characters, one 'a' and one '\0'. In whatever function that lets you use strings or char* if you put a string literal then that literal includes the \0.
    But, a char of array does NOT include \0 unless you put it somehow.

  12. #12
    Algorithm Dissector iMalc's Avatar
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    When you use a string literal (something enclosed in double-quotes) then you are implicitly saying that you want a null char at the end. I.e. You asked for the string "string" only, and not the string "string plus some other random crap that happens to follow".
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