Storing an integer into a char* in C++

Question

I'm writing some code that returns an integer, which then needs to be outputted using printw from the ncurses library. However, since printw only takes char*, I can't figure out how to output it.

Essentially, is there a way to store a integer into a char array, or output an integer using printw?

Michael Krelin - hacker · Accepted Answer · 2009-12-20 20:32:19Z

10

printw() accepts const char * as a format specifier. What you want is

printw("%d",yournumber);

answered Dec 20, 2009 at 20:32

Michael Krelin - hacker

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2 Comments

Galileo Over a year ago

out of curiosity, what exactly is "%d"?

Michael Krelin - hacker Over a year ago

You might want to look up the printf manpage - linux.die.net/man/3/printf to learn the full power of formatting. in particular %d means signed integer as the first parameter after format string here. But that's not even the top of the iceberg ;-)

stiank81 · Accepted Answer · 2009-12-20 20:30:33Z

1

The itoa function converts an int to char*.

answered Dec 20, 2009 at 20:30

stiank81

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Comments

Pavel Radzivilovsky · Accepted Answer · 2009-12-20 20:30:14Z

0

Use itoa() or sprintf() to convert integer to ascii string.

Example:

char s[50]; sprintf(s, "%d", someInteger);

now u can pass s as char*

answered Dec 20, 2009 at 20:30

Pavel Radzivilovsky

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2 Comments

Pavel Radzivilovsky Over a year ago

Sorry, my answer is crap. Use Michael's. I just wasn't sure what ncurses printw does, so I wrote a workaround.

Artyom Over a year ago

You can delete answer. Several "bad" points: itoa -- there is no such thing in standard C (only atoi), it is better to use snprintf(s,sizeof(s),"%d",someInteger) -- safer.

Li0liQ · Accepted Answer · 2009-12-20 20:30:18Z

0

itoa will help you.

answered Dec 20, 2009 at 20:30

Li0liQ

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