memset question

just a quick lil question on memset()
if i do something like this

char buf[128];

memset(buf, 'A', sizeof(buf));

will this overwrite the terminating character \0? since im filling the whole buffer
with A or does memset automatically take into account the \0?
because i kno that \0, \r, and \n only take "1 slot" in the buffer

Quote:

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Originally Posted by Alch3mist

will this overwrite the terminating character \0?

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There never was a terminating 0 in your buffer.

Quote:

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Originally Posted by Alch3mist

since im filling the whole buffer with A or does memset automatically take into account the \0?

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Yes, you are filling the whole thing with 'A'; No, memset doesn't take \0 into account

ight so would this work better
char buf[128] = 0;

memset(buf, 'A', sizeof(buf)-1);

or do i have to specify each time after lets say memset, memcpy, strcpy,strcat to place a \0 at the end of my buffer

Quote:

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Originally Posted by Alch3mist

ight so would this work better
char buf[128] = 0;

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Unless buf is declared as having 129 elements, that is a memory access violation.

Regards,

Paul McKenzie

I guess, he meant
char buf[128] = {'\0'};
which will be ok.

Quote:

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Originally Posted by Alch3mist

will this overwrite the terminating character \0? since im filling the whole buffer with A or does memset automatically take into account the \0?
because i kno that \0, \r, and \n only take "1 slot" in the buffer

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Only subset of standard C function (str* functions, formating funcitons as printf/scanf, fgets/fputs etc.) knows or cares about null-terminated 'strings' - for every other function (including memset) it is just array of chars, with \0 having no special meaning. So, when you want to set null-terminated string with memset, you have to add the terminating null yourself. Also keep in mind that you need space for the terminating null, so for 128-char string, you need array of size for 129 chars. It's mess, i know - if you have chance to use C++ and std::string, go for it :D

Quote:

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right so would this work better

char buf[128] = 0;
memset(buf, 'A', sizeof(buf));

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no that won't work cause of coarse your putting your null terminator in the first character of your array... Then you're overwriting that character with the memset... Try this..

Code:

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char buf[128];

memset( buff, '\0', sizeof(buff));
memset( buff, 'A', sizeof( buff) -1 );

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or you could do this..

Code:

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char buff[128];

memset( buff, 'A', sizeof( buff ) -1 );
buff[ sizeof(buff)-1 ] = '\0';

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[/code]

Quote:

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Originally Posted by Paul McKenzie

Unless buf is declared as having 129 elements, that is a memory access violation.

Regards,

Paul McKenzie

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Paul check that... when you initialize an array like that it always begins from the beginning....

you must not have realized it was initializing from the variable declaration..


thus

Code:

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char buff[128] = 0;                          // is fine... possible casting issue
char buff[128]  = "yada yad yada";  // is fine
char buff[128] = '\0';                        // is also fine

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Code:

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char buff[128] = 0;
char buff[128] = '\0';

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no, these are not valid initializers for an array

Quote:

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Originally Posted by spoon!

Code:

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char buff[128] = 0;
char buff[128] = '\0';

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no, these are not valid initializers for an array

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That's correct. They only set the value in pos. 128 to 0, not the whole array. As it was stated before, you need brackets to set the array to 0 when declaring it.

I think the solution from JMS is probably the best

Code:

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char buff[128];

memset( buff, 'A', sizeof( buff ) -1 );
buff[ sizeof(buff)-1 ] = '\0';

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Also, call me crazy, but I recall when initializing arrays, you need to use the initializer list

Code:

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char buff[128] = {'\0'};

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So I don't think either of these are correct for initializing an array

Code:

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char buff[128] = 0;
OR
char buff[128] = '\0'

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Quote:

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Originally Posted by Paul McKenzie

Unless buf is declared as having 129 elements, that is a memory access violation.

Regards,

Paul McKenzie

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compiler won't compile this or have an extension which allows defining that way, in this case it is treated exactly as

Code:

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char buf[128] = { 0 };

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it will never be an access violation but it is for sure not standard. I believe only very few, obscure compilers implement this.

I'm pretty sure that what Paul meant was

Code:

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buff[128] = 0;

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would cause an access violation.
Otherwise you are right, JMS.


-Andy