the length of my array is not what I expected

Hi All

I just tried the following program

Code:

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#include <stdio.h>
#include <string.h>

main() {
        char c[5] ;
        int x ;
        for( x= 0; x < 5; x++ ) {
                        c[x] = 'c' ;
        }
        printf("\n%s l=%d\n", c, strlen(c)) ;

}

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If I run this code it tells me the length of the array is 6!! Also when I let the loop go from 0 to 4 the length is still 6. What happened with the length of 5 ?

The result of the following example did also surprise me:

Code:

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        char d[5] ;
        strcpy(d, "1234567890123" ) ;
        printf("\n%s l=%d\n", d, strlen(d)) ;

---

The length was 13, how is that possible. Interesting to say is that when I make that string 1 character longer the progrem executes, telling me the length is 14, but it ends with an segmentation fault

How is all this possible ?

Thnx a lot
LuCa

strlen goes from the array pointer to the first NULL character. It has no way of knowing that d is defined to fit 5 characters. The fact that your first function gave you a length of six says that their happened to be a NULL after the array's memory. Both calls to strlen go out of the array bounds and could cause a segmentation fault.

So I should declare this array 1 element longer than I actually need, for this NULL character ?

LuCa

If you're planning on using string functions, yes.

Quote:

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Originally Posted by luca

So I should declare this array 1 element longer than I actually need, for this NULL character ?

LuCa

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Yes
And you should put this null character there

Quote:

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Originally Posted by vart

Yes
And you should put this null character there

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Though keep in mind that string functions such as strcpy and strcat put it in for you. Only if you were to assign to the string character by character would you have to put the null in yourself.

However, functions like strncpy do not guarantee a null byte.


Quzah.

thnx a lot!!