Characters that don't appear litterally in printf(" ")

[mcotter222]

When using printf or getchar, there are characters started with a \ that don't apear like how they are written (i.e.

printf("\n");

) I need a list of them, but don't know what they are called properly so I can't do online searches. If anyone knows, the answer or a link to a list would be great!
Thanks,
-Matt

[matsp]

If you want a backslash in your output, you need to give printf() [or any other C function] a string with two backslashes, e.g.

printf("Using '\\n' gives you a newline\n");

will output

Using '\n' gives you a newline

--
Mats

[mc61]

They are called "escape sequences".

http://www.acm.uiuc.edu/webmonkeys/b...guide/1.1.html

mc61

[hk_mp5kpdw]

They are often called "escape/control sequences" or "control characters". This page has a table of some(all?) escape sequences available about a quarter of the way down the page.

[edit]Ooh, beaten! Darn.[/edit]

[Codeplug]

isprint() will tell you if a particular character is printable.

gg

[mcotter222]

Thanks, those were great, but two more questions:
1What is the syntax for isprint()?
2Are there escape sequences for numbers/letters (characters that would register from getchar())?


Matt

[mcotter222]

And one more: is there an easy way to compare a char to a letter?
(Like this:

#include <stdio.h>

int main()
{
  int test;
  scanf("%d", &test);
  if(test == 1)
          printf("Text");
  getchar();
  getchar();
}

, but with char instead of int)

[tabstop]

And one more: is there an easy way to compare a char to a letter?
(Like this:

#include <stdio.h>

int main()
{
  int test;
  scanf("%d", &test);
  if(test == 1)
          printf("Text");
  getchar();
  getchar();
}

, but with char instead of int)

Did you try something like

int main()
{
    char test;
    scanf("%c", &test);
    if(test == '1')
        printf("Text");
    getchar();
    getchar();
}

?

[dwks]

1 is an int. '1' is a char. The former is what you do arithmetic with, and which can consist of multiple digits. The latter can only be one digit, much like 'a' or '#'. You can convert between the two with:

char c = '1';
int i = c - '0';
/* and back again */
char c2 = i + '0';

[mcotter222]

Ahh! Great, I tryed it with " instead of ', that answers all! Thank you so much.
Matt

[dwks]

Double quotes denote strings, which are arrays of chars. They're a bit more complicated.