How does printf() works without argument count being explicitly mentioned - c

In function with variable number of arguments, the "first argument" is "total number of arguments" being passed. But in printf() we never mention argument count. So how does it get to know about the total argument list ? How does printf() works ?

Let's look on printf declaration structure:
int printf(const char *format, ...)
format is actually the string that contains the text to be written to stdout.
The contained embedded format tags are later replaced by the values specified in subsequent additional arguments, and format is set accordingly as required.

You don't supply an argument count to printf - You do supply a format string, however - And that specifies how many arguments printfshould expect.
Very roughly speaking, the number of % signs in the format string is the argument count (although reality is a bit more complicated).

Conversion specifiers in the format string tell printf the number and types of arguments it should expect - for example, the format string "there are %d vowels in %s\n" tells printf to expect two arguments in addition to the format string, the first one being type int and the second being char *.
It's up to you to make sure the arguments match up with the format string. If you don't pass enough arguments, or the argument types don't match what the format string expects, then the behavior is undefined (most likely garbled output or a runtime error). If you pass too many arguments, the additional arguments are evaluated, but otherwise the function will operate normally as long as the format string is satisfied.
Edit
Chapter and verse
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.

Related

printf with multiple pointers as arguments in C

What is the behavior of printf() when we supply multiple arguments to it without a format specifier?
Example:
int main()
{
printf("hello", "hi");
return 0;
}
Why does the compiler produce a warning on compilation of the above program? :
warning: too many arguments for format [-Wformat-extra-args]
If we compile the similar program below:
int main()
{
char *s1 = "hello";
char *s2 = "hi";
printf(s1, s2);
}
No warnings are produced. What is the reason for this?
Also, why do both programs output hello only, and don't also print hi?
The C 2018 standard specifies the behavior of printf in clause 7.21.6.3, in which paragraph 2 says “The printf function is equivalent to fprintf with the argument stdout interposed before the arguments to printf.”
The standard specifies the behavior of fprintf in 7.21.6.1, which tells us the second argument (the first argument of printf) is a format string and that it may contain various conversion specifications introduced by the character “%”. Thus, in printf("hello", "hi"), "hello" is a format string with no conversion specifications. In this case, paragraph 2 tells us what happens:
If the format is exhausted [fully processed] while arguments remain, the excess arguments are evaluated (as always) but are otherwise ignored.
Thus, in printf("hello", "hi"), "hi" is ignored, and "hello" is a format string that contains only ordinary characters, which are copied to the output stream per paragraph 3.
The compiler warns about printf("hello", "hi") because it is able to see that this call contains an excess argument because the format string does not contain a conversion specification for it.
Your compiler does not warn about printf(s1,s2); because it does not analyze what s1 will contain during this call. This sort of analysis is not impossible in this situation, but situations like this are rare: When a programmer uses a pointer to a string as the format string for printf, it is usually a string or pointer that is computed, constructed, or selected during program execution, and the manner of this computation is often beyond the ability of a compiler to analyze. Situations where the pointer is clearly a pointer to a fixed string are rare, since they are not frequently useful, so presumably compiler implementors have not found it valuable to implement the code necessary for the compiler to handle these situations.
tl;dr: Extra arguments to printf() are ignored.
The official C language standard (the link is to a draft of the C11 version) says the following:
§ 7.21.6.1 The fprintf function
...
... 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.
... and printf() is simply fprintf() targeted at the standard output file.
About your two code snippets:
The compiler is giving you a hint, for the first snippet, that the number of arguments doesn't match the number of specifiers in the format string. It's just a courtesy - it's not required to notice this. This also explains why the compiler does not notice it for the second snippet. It could, but it's too much effort to chase your pointers and check what they point at.
In both cases, your format string is your first argument to printf(), i.e. "hello". That string has no format specifiers, so the printf() looks at the "hello", and understands it only needs to print that and doesn't need process any other arguments. That's whi it ignores "hi".
The first parameter of printf is the format string, because printf is about printing formatted data. To specify how to format the data, printf uses the first argument. This is different from other languages and libraries where all the parameters (like Python's print) are used in the same way and formatting is done through other means.
The first and second examples you provide are both "incorrect" although technically valid because you are passing a format string that does not need any extra argument, so "hi" is unused.
What you may want to do instead is:
printf("%s %s", "hello", "hi");
many compilers know well very well the printf function family and read compile time the format string analysing the parameters. printf("hello",s2); compiler see that there is no %... in the format string and does not expect any other parameters. Warning is issued
if you call printf(s1,s2); compiler does not know what is the content of the s1 and it cannot go through the format string and no warning issued.
Many compilers have special extension to inform them that your function is printf like and you want compiler to read the format string - gcc:
extern int
my_printf (void *my_object, const char *my_format, ...)
__attribute__ ((format (printf, 2, 3)));

What does scanf("%d/%d/%d) mean in C?

int currD, currM, currY;
scanf("%d/%d/%d", &currD, &currM, &currY);
I saw this code receiving birth date in the format DD/MM/YYYY, but I wonder what's the meaning of putting '/', I know without this, it will lead to bad input because of the character '/'. So what does it actually mean?
When encountering code that you don’t understand, and which is calling a function from a library, your first order of business is to research the documentation for that function. For C standard functions it’s enough to google the function name.
A good reference in this case is cppreference (don’t be misled by the website name, this is the C reference, not the C++ reference). It gives the function’s definition as
int scanf( const char *format, ... );​
Now look for the parameter description of the format parameter:
pointer to a null-terminated character string specifying how to read the input.
The subsequent text explains how to read the format string. In particular:
[…] character [except %] in the format string consumes exactly one identical character from the input stream, or causes the function to fail if the next character on the stream does not compare equal.
conversion specifications [in] the following format
introductory % character
conversion format specifier
d — matches a decimal integer.
In other words:
scanf parses a textual input based on the format string. Inside the format string, / matches a slash in the user input literally. %d matches a decimal integer.
Therefore, scanf("%d/%d/%d", …) will match a string consisting of three integers separated by slashes, and store the number values inside the pointed-to variables.
Is just the separator in the date format. The error must raise when some function searchs for those /.
The first parameter of scanf is a string specifying the format of the string you want to use to store the informations in the further arguments. You can see this format string as a pattern : %d means an integer, and without the '%' it means it just has to match exactly the characters.
Input is expected to provide like 04/07/2019.
If input is provided only 04072019. currD alone hold the value 04072019, currM and currY might garbage value as it is not initialised.
It expects the input to be in the format three integers separated by two slashes ("/"). For example: 10/11/1999.

Why is printf() printing out in this code?

This simple code is puzzling me - I am deliberately printing out more integers than I passed to printf. I expected an error. I got weird numbers - where are they coming from?
#include <stdio.h>
/* learn arrays */
void main(){
int pout;
pout = 6;
printf("%i %i %i\n%i %i %i\n%i %i %i\n", pout);
}
One example of the output:
6 608728840 0
-885621664 -885543392 608728816
0 0 -889304251
The single digits do not change with repeated runs, but the large integers do.
It's one of printf string format vulnerability. You are trying to call more argument than there actually are, so printf takes whatever he can on the stack.
It was (and still is) very used to exploit programs into exploring stacks to access hidden information or bypass authentication for example.
Viewing the stack
printf ("%08x %08x %08x %08x %08x\n");
This
instructs the printf-function to retrieve five parameters from the
stack and display them as 8-digit padded hexadecimal numbers. So a
possible output may look like:
40012980 080628c4 bffff7a4 00000005 08059c04
See this for a more complete explanation.
Because it's undefined behavior. If the number of specifiers is larger than the number of matching parameters or their types are incompatible, the behavior is undefined.
This qoute is from the c11 standard draft
7.21.6.1 The fprintf function
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.
If a conversion specification is invalid, the behavior is undefined.282) If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.
I highlighted the relevant parts making them bold.
The int reserves some RAM but you didn't wrote anything in so it shows you what numbers are random somewhere in your RAM

Use of % in scanf and printf [duplicate]

This question already has an answer here:
Code for printf function in C [duplicate]
(1 answer)
Closed 8 years ago.
What is the exact use of % in scanf and printf? And would scanf and printf work without the % sign? All I could find is that % is the conversion specifier but I want to know how it works actually?
% is simply the symbol used to identify the beginning of a conversion specifier in the format string; why % as opposed to any other symbol is an open question, and probably doesn't have that interesting an answer. The printf and scanf functions search the format string for conversion specifiers to tell them the number and types of additional arguments to expect, and how to format the output (for printf) or interpret the input (for scanf).
To print a literal %, you need to use %%.
printf can work without using a conversion specifier, but you'll be limited to writing literal strings. scanf is pretty useless without it.
The '%' character is used in the format string of the scanf() and print()-like functions from the standard header <stdio.h>, to indicate place holders for variable data of several kinds. For example, the format specifier "%d" is a place holder for a value having type int.
Thus, the variadic function printf() expects additional parameters passed as arguments to the function, the first of them having type int.
The value of this int argument is converted to string, and it will be replace to the place holder "%d".
In the case of scanf(), the situation is similar, but now scanf() is an input function that expects that the user enters in command-line a value fitting on the type indicated by the format specifier. Thus, a format specifier "%d" will expect that the user enters a value of type int.
Since all the arguments in C are passed by value, the input data requires you use the address of the variable, to mimic a by-reference mode of passing arguments.
There are a lot of options and details related to these format specifiers.
Yo need to take a look at the bibliography.
For example, start in Wikipedia:
printf() format string
scanf() format string
The % in a scanf() or printf() is a keyword whose purpose is identify the type of data that will be stored in the named variable. So, in the following example, the compiler would build instructions to accept input data of type integer and store it in the memory location at the address assigned to num1:
int num1;
scanf("%d",&num1);

What's the standard definition of printf in C?

What is the function definition of the printf() function as defined in the standard C library?
I need the definition to solve the following question:
Give the output of the following:
int main()
{
int a = 2;
int b = 5;
int c = 10;
printf("%d ",a,b,c);
return 0;
}
The C language standard declares printf as follows:
int printf(const char *format, ...);
It returns an integer and takes a first parameter of a pointer to a constant character and an arbitrary number of subsequent parameters of arbitrary type.
If you happen to pass in more parameters than are required by the format string you pass in, then the extra parameters are ignored (though they are still evaluated). From the C89 standard §4.9.6.1:
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.
You pass an array of chars (or pointer) as the first argument (which includes format placeholders) and additional arguments to be substituted into the string.
The output for your example would be 2 1 to the standard output. %d is the placeholder for a signed decimal integer. The extra space will be taken literally as it is not a valid placeholder. a is passed as the first placeholder argument, and it has been assigned 2. The extra arguments won't be examined (see below).
printf() is a variadic function and only knows its number of additional arguments by counting the placeholders in the first argument.
1 Markdown does not allow trailing spaces in inline code examples. I had to use an alternate space, but the space you will see will be a normal one (ASCII 0x20).
Its
int printf(const char *format, ...);
format is a pointer to the format string
... is the ellipsis operator , with which you can pass variable number of arguments, which depends on how many place holders we have in the format string.
Return value is the number of characters that were printed
Have a look here about the ellipsis operator: http://bobobobo.wordpress.com/2008/01/28/how-to-use-variable-argument-lists-va_list/
man 3 printf gives...
int printf(const char *restrict format, ...);
Writes to the standard output (stdout) a sequence of data formatted as the format argument specifies. After the format parameter, the function expects at least as many additional arguments as specified in format.
%d = Signed decimal integer
printf("%d ",a,b,c);
For every %(something) you need add one referining variable, therefore
printf("%d ",a+b+c); //would work (a+b+c), best case with (int) before that
printf("%d %d %d",a,b,c); //would print all 3 integers.

Resources