%s is a C format specifier for a string. means “where you see the first %s, replace it with the contents of command as a string, and where you see the second %s, replace it with the contents of context->user as a string.
Edit: I can’t reproduce the problem. Everything works as expected when I use the following code compiled with gcc: The output of gcc –version is as follows: gcc (Debian 4.7.2-5) 4.7.2
Edit: I can’t reproduce the problem. Everything works as expected when I use the following code compiled with gcc: The output of gcc –version is as follows: gcc (Debian 4.7.2-5) 4.7.2
When reading input using scanf, the input is read after the return key is pressed but the newline generated by the return key is not consumed by scanf, which means the next time you read a char from standard input there will be a newline ready to be read. One way to avoid is to … Read more
Short answer – yes, long answer: not how you want it. You can use the %* form of printf, which accepts a variable width. And, if you use ‘0’ as your value to print, combined with the right-aligned text that’s zero padded on the left.. produces: With my tongue firmly planted in my cheek, I … Read more
I have this but once it reaches the supposed EOF it just repeats the loop and scanf again.
Declare your ch as And your printf will work.
The printf() family of functions uses % character as a placeholder. When a % is encountered, printf reads the characters following the % to determine what to do: See this Wikipedia article for a nice picture: printf format string The \n at the end of the string is for a newline/carriage-return character.