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.

This is really more of a general PHP programming question and might get closed for that reason. But before that happens, let's see if I can't clear some of this up for you.

Those strings that start with % signs are in "printf format". For example, %2$s translates to "replace this token with the second parameter, and treat it like a string".

When you see a variable in curly braces inside a string, like "Today is {$day}", that's the same as putting the variable there without curly braces, like "Today is $day". But formatting the variable like this makes it stand out, so it's more clear there's a variable there. It also prevents other characters next to the variable name from getting interpreted as part of the name. Without curly braces, would PHP know what to do with "Your robot name would be {$firstName}Number5"? It would see "$firstNameNumber5" and look for a variable named $firstNameNumber5 instead of simply $firstName.

I'm not sure what you're asking about in the last example. Is it the __FILE__ you're confused by? That's a PHP Magic Constant that gets replaced with the full path to the file that bit of code is in.

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.

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.

They are used for formatting strings. %s acts a placeholder for a string while %d acts as a placeholder for a number. Their associated values are passed in via a tuple using the % operator.

name = 'marcog'
number = 42
print '%s %d' % (name, number)

will print marcog 42. Note that name is a string (%s) and number is an integer (%d for decimal).

See https://docs.python.org/3/library/stdtypes.html#printf-style-string-formatting for details.

In Python 3 the example would be:

print('%s %d' % (name, number))

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

This is undefined behavior, anything can happen included something that looks like correct. But it is incorrect. Your compiler can probably tell you the problem if you use correct options.

Standard says (emphasized is mine):

7.21.6.1 The fprintf function

2. The fprintf function writes output to the stream pointed to by stream, under control of the string pointed to by format that specifies how subsequent arguments are converted for output. If there are insufficient arguments for the format, the behavior is undefined. If the format is exhausted while arguments remain, the excess arguments are evaluated (as always) but are otherwise ignored. The fprintf function returns when the end of the format string is encountered.