Another newbie-question

[cnewbie1]

Anothe newbie question, I'm afraid. (Shold be a forum called Total Newbies o something.)

I might be a bit tired at the present time, but this I don't get:

The following code:

char correctname[7] = "Myname";
int correctlength = strnlen(correctname);
printf("Name: %s\n", correctname);
printf("Length: %d\n", correctlength);

works like I expect it too. It creates a string, then creates an int variable with the string's length as value. Then it prints the string and the value. 100% obvious an basic.

Output:
Name: Myname
Length: 6

However, when I want to add another string:

char correctname[6] = "Frank";
int correctlength = strnlen(correctname);
char newname[10];       <--------------------------------- another string variable
printf("Name: %s\n", correctname);
printf("Length: %d\n", correctlength);

suddenly, the int variable correctlength becoes zero.

Output:
Name: MyName
Length: 0

Wha-what-what?! By declaring another char variable, the unrelated int variable counting the length of the previous cha variable, is changed?

[quzah]

#include<stdio.h>
#include<string.h>
int main( void )
{
    char correctname[6] = "Frank";
    int correctlength = strlen(correctname);
    char newname[10];
    printf("Name: %s\n", correctname);
    printf("Length: %d\n", correctlength);
    
    return 0;
}

It works fine, assuming you actually use the right function strlen not strnlen.


Quzah.

[itCbitC]

Problem may stem from the fact that strnlen() is being invoked incorrectly ie it takes two arguments not one.

[cnewbie1]

Oh my. I WAS wondering a bit about that, but obviously I didn't wonder enough. Somehow I started using strnlen instead of strlen - not sure why. I guess i AM too tired :-)

However - one last question: This line:

int correctlength = strlen(correctname);

gives this warning:

warning: incompatible implicit declaration of built-in function ‘strlen’;

And I don't like warnings. How obvious is the answer to that?

[quzah]

Did you include string.h as shown in my working example?


Quzah.

[cnewbie1]

Of course I didn't - when I was testing this exclusively. Thanks :-)