I've never seen this syntax before.
#define SNS(s) (s),(sizeof(s)-1)
The way i'm reading this is that SNS(s) = sizeof(s)-1. What is the comma doing? Is it necessary?
int ft_display_fatal(const char *err, unsigned len, int fd, int rcode)
{
UNUSED(write(fd, err, len));
return (rcode);
}
Main
return (ft_display_fatal(SNS("File name missing.\n"), 2, 1));
Macros are just text replacement, so they can expand to just about anything you want. In this case, the macro is being used to expand into two arguments to a function. The function expects a string and the number of characters in the string as arguments, and the SNS() macro generates them. So
ft_display_fatal(SNS("File name missing.\n"), 2, 1)
expands into
ft_display_fatal(("File name missing.\n"),(sizeof("File name missing.\n")-1), 2, 1)
This is basically only useful when the parameter is a string literal: sizeof("string") is the size of the char array including the trailing null byte, and -1 subtracts that byte to get the number of significant characters in the string. This is the len argument to the ft_display_fatal function (I'm not sure why it can't just use strlen() to get this by itself -- I guess it's a performance optimization).
The way i'm reading this is that SNS(s) = sizeof(s)-1.
You are reading it wrong.
What is the comma doing?
Macro expansion results in textual substitution. You can use SNS(a) to pass two arguments to a function.
ft_display_fatal(SNS("File name missing.\n"), 2, 1)
You can see that ft_display_fatal takes 4 arguments, but only 3 are provided. This works because SNS expands to 2 arguments. If it didn't, you'd get a compiler error.
Related
#define Page 5
void printSystemInfo() {
printf ("%i", Page);
}
Thats my code can anyone explain me how to print Page 5 in the console?
For now my console looks like this "5" But I want to have "Page 5"
Thanks for helping !
You can use a little preprocessor trick. We have the # operator, which will convert a symbol into a string.
#define _(a) #a
When you call _(foo), it translates it as "foo". So, in your case, you could do something like:
#include <stdio.h>
#define _(a) # a
#define PAGE 5
int main(int argc, char *argv[])
{
printf("%s: %i\n", _(PAGE), PAGE);
return 0;
}
What this will do is:
We define a macro named _ that takes one parameter a. This macro uses the operator # from the preprocessor (called stringification). This will case a named passed to the macro to be converted into a string. Example: _(foo) gets translated to "foo".
In main, the printf() call is then translated as printf("%s: %i\n", "PAGE", 5);. In a stepwise way, when the preprocessor sees the _(PAGE) symbol, it translates it as "PAGE".
The inner workings of this things is explained in the above link, which I quote (my markings):
Sometimes you may want to convert a macro argument into a string constant. Parameters are not replaced inside string constants, but you can use the ‘#’ preprocessing operator instead. When a macro parameter is used with a leading ‘#’, the preprocessor replaces it with the literal text of the actual argument, converted to a string constant. Unlike normal parameter replacement, the argument is not macro-expanded first. This is called stringification.
Here you go. This is very trivial stuff, but please ask if something is unclear.
#define Page 5
void printSystemInfo()
{
printf((char const[])??<0120,0141,0147,0145,0040,0045,0151,!"bad"??>,Page);
}
I need a function to read a file name, with a max length of MAX_FILE_NAME_SIZE, which is a symbolic constant, I did this the following way:
char * readFileName()
{
char format[6];
char * fileName = malloc(MAX_FILE_NAME_SIZE * sizeof(fileName[0]));
if(fileName== NULL)
return NULL;
sprintf(format, "%%%ds", MAX_FILE_NAME_SIZE-1);
scanf(format, fileName);
fileName= realloc(fileName, strlen(fileName)*sizeof(fileName[0]));
return fileName;
}
I'd really like to get read of the sprintf part (and also the format vector), what's the cleanest and most efficient way to do this?
Solution
You can make a little Preprocessor hack:
#define MAX_BUFFER 30
#define FORMAT(s) "%" #s "s"
#define FMT(s) FORMAT(s)
int main(void)
{
char buffer[MAX_BUFFER + 1];
scanf(FMT(MAX_BUFFER), buffer);
printf("string: %s\n", buffer);
printf("length: %d\n", strlen(buffer));
return 0;
}
The FORMAT and FMT macros are necessary for the preprocessor to translate them correctly. If you call FORMAT directly with FORMAT(MAX_BUFFER), it will translate into "%" "MAX_BUFFER" "s" which is no good.
You can verify that using gcc -E scanf.c. However, if you call it through another macro, which will effectively resolve the macro names for you and translate to "%" "30" "s", which is a fine format string for scanf.
Edit
As correctly pointed out by #Jonathan Leffler in the comments, you can't do any math on that macro, so you need to declare buffer with plus 1 character for the NULL terminating byte, since the macro expands to %30s, which will read 30 characters plus the null byte.
So the correct buffer declaration should be char buffer[MAX_BUFFER + 1];.
Requested Explanation
As asked in the comments, the one macro version won't work because the preprocessor operator # turns an argument into a string (stringification, see bellow). So, when you call it with FORMAT(MAX_BUFFER), it just stringifies MAX_BUFFER instead of macro-expanding it, giving you the result: "%" "MAX_BUFFER" "s".
Section 3.4 Stringification of the C Preprocessor Manual says this:
Sometimes you may want to convert a macro argument into a string constant. Parameters are not replaced inside string constants, but you can use the ‘#’ preprocessing operator instead. When a macro parameter is used with a leading ‘#’, the preprocessor replaces it with the literal text of the actual argument, converted to a string constant. Unlike normal parameter replacement, the argument is not macro-expanded first. This is called stringification.
This is the output of the gcc -E scanf.c command on a file with the one macro version (the last part of it):
int main(void)
{
char buffer[30 + 1];
scanf("%" "MAX_BUFFER" "s", buffer);
printf("string: %s\n", buffer);
printf("length: %d\n", strlen(buffer));
return 0;
}
As expected. Now, for the two levels, I couldn't explain better than the documentation itself, and in the last part of it there's an actual example of this specific case (two macros):
If you want to stringify the result of expansion of a macro argument, you have to use two levels of macros.
#define xstr(s) str(s)
#define str(s) #s
#define foo 4
str (foo)
==> "foo"
xstr (foo)
==> xstr (4)
==> str (4)
==> "4"
s is stringified when it is used in str, so it is not macro-expanded first. But s is an ordinary argument to xstr, so it is completely macro-expanded before xstr itself is expanded (see Argument Prescan). Therefore, by the time str gets to its argument, it has already been macro-expanded.
Resource
The C Preprocessor
I was doing basics of macros. I define a macro as follows:
#define INTTOSTR(int) #int
to convert integer to string.
Does this macro perfectly converts the integer to string? I mean are there some situations where this macro can fail?
Can I use this macro to replace standard library functions like itoa()?
for example:
int main()
{
int a=56;
char ch[]=INTTOSTR(56);
char ch1[10];
itoa(56,ch1,10);
printf("%s %s",ch,ch1);
return 0;
}
The above program works as expected.
Interestingly this macro can even convert float value to string.
for example:
INTTOSTR(53.5);
works nicely.
Till now I was using itoa function for converting int to string in all my projects. Can I replace itoa confidently in all projects. Because I know there is less overhead in using macro than function call.
Macros execute during (before to be exact) compile time, so you can convert a literal number in your sourcecode to a string but not a number stored in a variable
In your example, INTTOSTR(56) uses the stringification operator of the preprocessor which eventually results in "56". If you called it on a variable, you'd get the variable name but not its content.
In C, you can use itoa or if you are desperate and would like to avoid it, use snprintf for instance:
snprintf(my_str, sizeof(int), "%i", my_int);
The problem with your macro is that you are thinking about constants, but of course, your macro will be broken when you need to use a variable holding an integer. Your macro would try to stringify the macro name as opposed to the value it would be holding.
If you are fine with constants, your macro is "good", otherwise it is b0rked.
Your macro does not convert integers to strings, it converts a literal into a string literal, which is something very different.
Literals are any plain numbers or definitions of values in your code. when you do int x = 10; the numeral 10 in an integer literal, while x is a variable and int is the type. const char* ten = "10"; also defines a literal, in this case a string literal, with value "10" and a variable called ten which points to the address where this literal is defined. What your macro actually does is change the way the literal is represented before any actual compilation goes on, from an integer literal into a string literal.
So, the actual change is being done before any compilation, just at source code level. Macros are not functions and cannot inspect memory, and your convertion would not work with variables. If you try:
int x = 10;
const char* ten = INTTOSTR(x);
You would be very puzzled to find that your variable ten would actually hold the value "x". That's because x is treated as a literal, and not as a variable.
If you want to see what's going on, I recommend asking your compiler to stop at preprocessing, and see the output before your code is acutally compiled. You can do this in GCC if you pass the -E flag.
PS. Regarding the apparent "success" with conversion of float values, it just comes to show the danger of macros: they are not type-safe. It does not look at 53.5 as a float, but as a token represented by characters 5, 3, . and 5 in the source code.
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.
I was reading a book and came across a program to read entries from a /proc file.
The program which they mentioned has following line
printf("%.*s", (int) n, line);
I am not clear with meaning of above line
what type of print if above "%.*s used instead of %s
The code can be read here
Abstract from here:
.* - The precision is not specified in
the format string, but as an
additional integer value argument
preceding the argument that has to be
formatted.
So this prints up to n characters from line string.
The cast expression (int) n converts the value of n to type int. This is because the formatting specifier requires a plain int, and I assume (since you didn't include it) the variable n has a different type.
Since a different type, like size_t might have another size, it would create problems with the argument passing to printf() if it wasn't explicitly converted to int.