Oct 19, 2016 · I want to read a number from stdin. I don't understand why scanf requires the use of & before the name of my variable: int i; scanf("%d", &i); Why does scanf need the …

The & in C is an operator that returns the address of the operand. Think of it this way, if you would simply give scanf the variable a without the &, it will be passed to it by-value, which means …

10 From scanf: On success, the function returns the number of items successfully read. This count can match the expected number of readings or fewer, even zero, if a matching failure …

Nov 24, 2012 · scanf(" %c", &c); The blank in the format string tells scanf to skip leading whitespace, and the first non-whitespace character will be read with the %c conversion specifier.

When the scanf() function completes, the input is pointing at the next newline character. The body of the loop reads and prints that character, so that when the loop restarts, the input is looking …

You are reading input using scanf, and it is leaving the user's newline on the input stream, and that stray newline is wrongly getting read by a later call to scanf("%c").

So snprintf writes some data to a string buffer, and sscanf reads from that created string. Where exactly does this replace scanf in that it reads from stdin?

Jan 30, 2016 · char str[25]; And you cannot use scanf to read sentences--it stops reading at the first whitespace, so use fgets to read the sentence instead. And in your last printf, you need …

Apr 15, 2017 · This scanf format string consists of two parts: a space character, which skips space characters (' ', '\t', '\n', etcetera) in the input, and the %[^\n] conversion specification, …

The first scanf() doesn't return when a user has hit space, even though that's where the number ends (because at that point the line is still in the terminal's line buffer); and the second scanf() …