printf ignores single backslash '\'

Question

I have this code :

int main(int argc, char * argv[])
{
int i;
printf("%d%s",argc,argv[1]);
return 0;
}

If I run this code as a.out a\=b=.I am using C-shell

Its output is "a=b=" is there any way that its output can be changed to "a\=b=".

Answer 1

Update for edited question:

Enclose your command line argument in quotes:

  $ a.out "a\=b="

The quotes prevent the shell from interpreting the command line argument in any way, so just this string is passed to your program. I use the csh/bash ..works with both.

Alternatively, you can "escape" the \ with another one and skip the double quotes:

  $ a.out a\\=b=

---

Previous answer to original question:

Yes, use two \:

char a[]="a\\=b=";

outputs:

a\=b=

Explanation:

\ is an escape character used to indicate a special character sequence, so for instance \t indicates a tab. If you want to actually print \t, you need to "escape" this \ with another \. See this example and output:

printf("\t-->Hi\n");    /* print regular tab via \t                        */
printf("\\t-->Hi\n");   /* want to print "\t", not tab  ..so we use two \\ */

which results in:

    -->Hi
\t-->Hi

This is not unique to the printf() function, nor really to C, may languages use the backslash to indicate "escape sequences" in strings.

Answer 2

printf() is not ignoring your single backslash, it's the way C strings are parsed. A backslash is an escape character to indicate some character that is not easily entered in a string, for example a newline (\n) or embedded quote (\"). Consequently, to include a backslash you must include two backslashes (\\). This is for all strings, and not related to printf().