how to use memset to fill a char array with space key?

[focusdoit]

Hi, I want to fill out a char array with space. like:

memset(msgArray, '*', 100);

but '*' should be a Space key.

[leon de boer]

memset(msgArray, ' ', 100); memset(msgArray, #32, 100); or a myriad of others

In vino veritas

[focusdoit]

Thanks, ' ' and 32 work. in fact, it's snprintf() reason. snprintf seems fill out the gap in a string with '\0'.

[Lost User]

I find this question a bit mysterious, if you had tried the obvious thing it just should have worked. The "obvious thing" is:

focusdoit wrote:

'*' should be a Space key.

.. literally just replace '*' by ' '.

[focusdoit]

in fact, it's snprintf() reason. snprintf seems fill out the gap in a string with '\0'. the code like:

memset (msgArray, ' ', 100);
snprintf (msgArray, 2, "%d", 1);
snprintf (messageArray+10, 33, "%s", "main");

I assume the string should output like: 1 main but debugged it. it is : 1,\0, ' ', ' ', .. main, \0, ..

[Lost User]

Well snprintf terminates its output with a \0, which is typically useful (it would be a bit dangerous if it didn't), you can just overwrite that \0 with a space.

[Rick York]

Here's the prototype for the function :

int _snprintf( char *buffer, size_t count, const char *format, ... );

I usually use it like this :

const size_t bufferSize = 127
char buffer[bufferSize+1] = {0};

_snprintf( buffer, bufferSize, "%3d %s", index, yourString );

You can combine the _snprintf calls into one since they are going into the same memory buffer.

[leon de boer]

That is because you are doing it all wrong you either join them all up in one sequence or move the first pointer along They are designed to be used a completely different way :-) Single line:

snprintf (msgArray, sizeof(msgArray),"%d%s", 1,"main");

Concat sequence (with buffer safety):

char *cur = &msgArray[0];
char *end = &msgArray[sizeof(msgArray)-1];
cur += snprintf(cur, end-cur, "%d", 1);
snprintf(cur, end-cur, "%s", "main");

In vino veritas