Using Python, I get
$ python3 -c 'print("\\\n")'
\
$
That is, one backslash and one newline, followed by an extra newline inserted by the interpreter.
Using C, compiling the code
#include <stdio.h>
int main(void)
{
printf("\\\n");
return 0;
}
into a file backslash.out yields
$ ./backslash.out
\
$
That is, one backslash and one newline.
In bash, I get
$ STRING="\\\n"
$ printf "${STRING}"
\n$
What exactly is the bash printf command doing here? What is it doing differently from the python print or C printf commands with respect to the escape character \? And what will I need to put in the variable STRING to obtain the following output on my terminal:
$ printf "${STRING}"
\
$
In your snippet below, you use "double quotes" around the backslash escapes:
$ STRING="\\\n"
$ printf "${STRING}"
\n$
However, Bash still evaluates some backslash-escapes inside double quotes, so the content of your variable after that is really \\n, as "\\" evaluates to \.
Put the string in 'single quotes' to prevent the shell from touching any of the backslashes:
$ STRING='\\\n'
$ printf "$STRING"
\
$
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