S

sprintf is a wrapper for a C function that uses %s to represent "insert a string here" and %i to represent "insert an integer here". Arguments after the string get pasted into the string in order.

%s most likely means a string. Sort of like a placeholder or variable.

In some languages you can write something like:

string name = "Joe";
printf("Hi %s", name);

Which would print out: "Hi Joe".

What you are seeing there is the string substitution patterns that are built into console.log() or console.debug().

The pattern goes as I have presented below:

%s for a String

%d or %i for Number

%f for Floating points

%o for an Object

%j for an JSON

So essentially you are replacing the specifier with the values supplied as so:

var name = 'Chris';
console.log('Hi, my name is %s.', name);
// Hi, my name is Chris.

console.debug('Hi, my name is %s.', name);
// Hi, my name is Chris.

Docs:

• Docs for Chrome
• Docs for Firefox
• Docs for IE
• Docs for Node.js
• Docs for Spec

"/s" is suppose to imply sarcasm.

It is a string formatting syntax (which it borrows from C).

Please see "PyFormat":

Python supports formatting values into strings. Although this can include very complicated expressions, the most basic usage is to insert values into a string with the %s placeholder.

Here is a really simple example:

#Python 2
name = raw_input("who are you? ")
print "hello %s" % (name,)

#Python 3+
name = input("who are you? ")
print("hello %s" % (name,))

The %s token allows me to insert (and potentially format) a string. Notice that the %s token is replaced by whatever I pass to the string after the % symbol. Notice also that I am using a tuple here as well (when you only have one string using a tuple is optional) to illustrate that multiple strings can be inserted and formatted in one statement.

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For both *printf and *scanf, %s expects the corresponding argument to be of type char *, and for scanf, it had better point to a writable buffer (i.e., not a string literal).

char *str_constant = "I point to a string literal";
char str_buf[] = "I am an array of char initialized with a string literal";

printf("string literal = %s\n", "I am a string literal");
printf("str_constant = %s\n", str_constant);
printf("str_buf = %s\n", str_buf);

scanf("%55s", str_buf);

Using %s in scanf without an explcit field width opens the same buffer overflow exploit that gets did; namely, if there are more characters in the input stream than the target buffer is sized to hold, scanf will happily write those extra characters to memory outside the buffer, potentially clobbering something important. Unfortunately, unlike in printf, you can't supply the field with as a run time argument:

printf("%*s\n", field_width, string);

One option is to build the format string dynamically:

char fmt[10];
sprintf(fmt, "%%%lus", (unsigned long) (sizeof str_buf) - 1);
...
scanf(fmt, target_buffer); // fmt = "%55s"

EDIT

Using scanf with the %s conversion specifier will stop scanning at the first whitespace character; for example, if your input stream looks like

"This is a test"

then scanf("%55s", str_buf) will read and assign "This" to str_buf. Note that the field with specifier doesn't make a difference in this case.