scanf needs %lf for doubles and printf is okay with just %f
So, why is printf and scanf okay with %d?
This is what I think the reason is:
A floating point (%f) uses exactly 64 bits whereas a double floating-point number (%lf) uses at least 32. The compiler doesn't know how many bits to assign to a variable that is being scanned in by scanf, so we use %lf to tell the compiler that it needs to be at least 32 bits.
Okay... but then why do we use %d for both scanf and printf? Why not %ld and %d? %ld doesn't exist in C for starters. %d is a signed decimal number that needs at least 16 bits. You're already telling the compiler what the lower bound is in how many bits to allocate to it, so it is okay for scanf. This is also why we don't have a %ld.
Please do correct me if I am wrong or inform me what I can do to make this answer better. I'm pretty sure it is not a perfect answer.
You can think scanf as converting input stream into variables that defined in your code. Thus, scanf needs to know the exactly size for each variable. In general, the 32-bit and 64-bit IEEE 754 binary floating-point formats are used in C. So, %f means 32-bit and %lf means 64-bit.
Besides %ld exists and it means 32-bit integer. %lld also exists which means 64-bit integer. The above wiki site explains all C data types very well.
See §6.5.2.2/6-7 in the C99 standard.
§6.5.2.2/6 defines the default argument promotions: (emphasis added)
the integer promotions are performed on each argument, and arguments that have type float are promoted to double.
and specifies that these promotions are performed on arguments to a function declared with no prototype (that is, with an empty parameter list () instead of (void), where the latter is used to indicate no arguments).
Paragraph 7 additionally specifies that if the prototype of the function has a trailing ellipsis (...):
The ellipsis notation in a function prototype declarator causes argument type conversion to stop after the last declared parameter. The default argument promotions are performed on trailing arguments.
The integer promotions are defined in §6.3.1.1/2; they apply to
objects or expressions of an integer type whose "integer conversion rank is less than or equal to the rank of int and unsigned int": roughly speaking, any smaller integer type, such as boolean or character types;
bit-fields of type _Bool, int, signed int or unsigned int.
If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned
int. These are called the integer promotions.
All other types are unchanged by the integer promotions.
In short, if you have varargs function, such as printf or scanf:
Integer arguments which are no larger than int are converted to (possibly unsigned) int. (This does not include long.)
Floating point arguments which are no larger than double are converted to double. (This includes float.)
Pointers are unaltered.
Other non-pointer types are unaltered.
So printf doesn't need to distinguish between float and double, because it will never be passed a float. It does need to distinguish between int and long.
But scanf does need to know whether an argument is a pointer to a float or a pointer to a double, because pointers are unchanged by the default argument promotions.
Related
I'm trying to understand implicit datatype conversions in C. I thought that I had understood this topic, but yet the following code example is still confusing me.
Specifically, I have read about Usual Arithmetic Conversions and Integer Promotion previously from drafts of the C Standard.
unsigned short int a = 0;
printf("\n%lld", (signed int)a - 1);
I am compiling using GCC.
unsigned short int is 2 bytes.
int is 4 bytes.
When I run this code, I get the following result: 4294967295
I expected the result -1.
This is what I expected to happen:
Typecast takes precedence, and LHS of - becomes signed int.
- operation is carried out. No integer promotion or implicit conversions occur here, as LHS and RHS are already both signed int. The result of the operation is -1 with datatype signed int.
Within printf statement, value -1 is retained within the conversion to long long int, and -1 is displayed as the result.
Can someone please explain where the flaw in my logic is?
It's undefined behaviour due to %lld being the inappropriate format specifier for an int type.
Yes indeed (signed int)a - 1 is an int type with value -1, but the printf call is the undefined part. There's nothing in the C standard to suggest that a conversion to long long occurs.
Within printf statement, value -1 is retained within the conversion to long long int
There's no such conversion taking place. printf (family of functions) is dumb and needs a format string that corresponds to the types of the argument list.
printf does not work like an ordinary function void f (long long int x), which would have forced an implicit conversion to the type of the parameter ("as per assignment"/"lvalue conversion"). This would have given you the expected "sign extension".
Notably, there's a another kind of specialized implicit conversion going on here called the default argument promotions, that only applies to variable argument functions and functions with no prototype.
C17 6.5.2.2/6
If the expression that denotes the called function has a type that does not include a
prototype, the integer promotions are performed on each argument, and arguments that
have type float are promoted to double. These are called the default argument
promotions.
C17 6.5.2.2/7 regarding variable argument functions:
The ellipsis notation in a function prototype declarator causes argument type conversion to stop after the last declared parameter. The default argument
promotions are performed on trailing arguments.
In practice this means:
float passed to printf gets implicitly converted to double during function call.
Small integer types passed to printf get implicitly converted during function call as per integer promotions, most likely ending up as int.
Other types passed to printf do not get implicitly promoted during the function call.
And then the passed and potentially converted argument gets treated internally as if it was the type specified by the conversion specifier. If that one doesn't match the actual type, the code has undefined behavior.
In your case you pass an int, it doesn't get implicitly promoted, but as printf treats it as a long long, you get undefined behavior.
Here you can consider yourself lucky. a is a short int that undergoes usual arithmetic conversions to a `signed int', even despite the cast, so
unsigned short int a = 0;
printf("\n%d", (signed int)a - 1);
and
unsigned short int a = 0;
printf("\n%d", a - 1);
would have the same behaviour, if all values of unsigned short are representable in int (as they are in your case). The result of the conversion is an int. Now, for the variable arguments, the default argument promotions are applied and any integers smaller than an int is converted to int if representable, otherwise unsigned int. But lld expects a signed long long int which is 8 bytes wide. Default argument promotions do not promote int implicitly to long long int.
Now comes the luck part - you did get a wrong value. See, since the behaviour is undefined you could have gotten the value that you're expecting, this time - after all it is completely feasible on a 64-bit processor!
int8_t is an 8-bit signed integer. Therefore, its value is anywhere in the range [-128...127].
int8_t num = -1;
printf("%u",num);
Output:
4294967295
Could someone give me a hint?
Your program behaviour is not defined.
%u cannot be used as a format specifier for int8_t since it's a signed type and %u is for unsigned types.
Use %d instead, and rely on the C standard guaranteed automatic promotion of num to an int type.
As others have mentioned, using the incorrect format specifier for printf is undefined behavior. The behavior you experienced cannot be depended on to be consistent between different compilers or even different builds of the same compiler.
That being said, here's is what most likely happened.
Any argument to printf after the first is of an unspecified type. So when num is passed to it, it can't do an exact type check. What ends up happening is that the value of num is promoted to type int.
From section 6.3.1.1 of the C standard:
2 The following may be used in an expression wherever an int or unsigned int may be used:
— An object or expression with an integer type
(other than int or unsigned int) whose integer conversion rank is
less than or equal to the rank of int and unsigned int.
— A bit-field of type _Bool, int, signed int, or unsigned int.
If an int
can represent all values of the original type (as restricted
by the width, for a bit-field), the value is converted to an
int; otherwise, it is converted to an unsigned int. These are
called the integer promotions. All other types are unchanged by
the integer promotions.
Because num was being used in a context where an int could be used, its value in the function call was promoted to int.
Assuming a 32-bit int and 2's compliment representation of negative numbers, the orginal binary representation 11111111 is converted to 11111111 11111111 11111111 11111111. If printed with the %u format specifier, it assumes this representation is unsigned so it prints 4294967295.
Had you used the %d format specifier, which expects a signed value, it would have printed -1.
To reiterate however, what you are seeing is undefined behavior. Other machines / compilers / optimization settings might yield different results.
To print an int8_t, the C standard provids the format specifier macro PRIi8 from <inttypes.h>.
printf("%" PRIi8, num);
Although, %d is fine to use to print an int8_t due to C's default argument promotions, you can simply use it for all signed types. You can see the format specifiers for other fixed width types in POSIX documentation as well.
How exactly do variadic functions treat numeric constants? e.g. consider the following code:
myfunc(5, 0, 1, 2, 3, 4);
The function looks like this:
void myfunc(int count, ...)
{
}
Now, in order to iterate over the single arguments with va_arg, I need to know their sizes, e.g. int, short, char, float, etc. But what size should I assume for numeric constants like I use in the code above?
Tests have shown that just assuming int for them seems to work fine so the compiler seems to push them as int even though these constants could also be represented in a single char or short each.
Nevertheless, I'm looking for an explanation for the behaviour I see. What is the standard type in C for passing numeric constants to variadic functions? Is this clearly defined or is it compiler-dependent? Is there a difference between 32-bit and 64-bit architecture?
Thanks!
I like Jonathan Leffler's answer, but I thought I'd pipe up with some technical details, for those who intend to write a portable library or something providing an API with variadic functions, and thus need to delve in to the details.
Variadic parameters are subject to default argument promotions (C11 draft N1570 as PDF; section 6.5.2.2 Function calls, paragraph 6):
.. the integer promotions are performed on each argument, and arguments that
have type float are promoted to double. These are called the default argument promotions.
[If] .. the types of the arguments after promotion are not compatible with those of the parameters after promotion, the behavior is undefined, except for the following cases:
one promoted type is a signed integer type, the other promoted type is the corresponding unsigned integer type, and the value is representable in both types;
both types are pointers to qualified or unqualified versions of a character type or void
Floating-point constants are of type double, unless they are suffixed with f or F (as in 1.0f), in which case they are of type float.
In C99 and C11, integer constants are of type int if they fit in one; long (AKA long int) if they fit in one otherwise; of long long (AKA long long int) otherwise. Since many compilers assume an integer constant without a size suffix is a human error or typo, it is a good practice to always include the suffix if the integer constant is not of type int.
Integer constants can also have a letter suffix to denote their type:
u or U for unsigned int
l or L for long int
lu or ul or LU or UL or lU or Lu or uL or Ul for unsigned long int
ll or LL or Ll or lL for long long int
llu or LLU (or ULL or any of their uppercase or lowercase variants) for unsigned long long int
The integer promotion rules are in section 6.3.1.1.
To summarize the default argument promotion rules for C11 (there are some additions compared to C89 and C99, but no significant changes):
float are promoted to double
All integer types whose values can be represented by an int are promoted to int. (This includes both unsigned and signed char and short, and bit-fields of types _Bool, int, and smaller unsigned int bit-fields.)
All integer types whose values can be represented by an unsigned int (but not an int) are promoted to unsigned int. (This includes unsigned int bit fields that cannot be represented by an int (of CHAR_BIT * sizeof (unsigned int) bits, in other words), and typedef'd aliases of unsigned int, but that's it, I think.)
Integer types at least as large as int are unchanged. This includes types long/long int, long long/long long int, and size_t, for example.
There is one 'gotcha' in the rules that I'd like to point out: "signed to unsigned is okay, unsigned to signed is iffy":
If the argument is promoted to a signed integer type, but the function obtains the value using the corresponding unsigned integer type, the function obtains the correct value using modulo arithmetic.
That is, negative values will be as if they were incremented by (1 + maximum representable value in the unsigned integer type), making them positive.
If the argument is promoted to an unsigned integer type, but the function obtains the value using the corresponding signed integer type, and the value is representable in both, the function obtains the correct value. If the value is not representable in both, the behaviour is implementation-defined.
In practice, almost all architectures do the opposite of above, i.e. the signed integer value obtained matches the unsigned value substracted by (1 + the largest representable value of the unsigned integer type). I've heard that some strange ones may signal integer overflow or something similarly weird, but I have never gotten my mitts on such machines.
The man 3 printf man page (courtesy of the Linux man pages project) is quite informative, if you compare the above rules to printf specifiers. The make_message() example function at the end (C99, C11, or POSIX required for vsnprintf()) should also be interesting.
When you write 1, that is an int constant. There is no other type that the compiler is allowed to use. If there is a non-variadic prototype for the function that demands a different type, the compiler will convert the integer 1 to the appropriate type, but on its own, 1 is an int constant. So, in your example, all 6 arguments are int.
You have to know the types of the arguments somehow before the called variadic function processes them. With the printf() family of functions, the format string tells it what to expect; similarly with the scanf() family of functions.
Note that the default conversions apply to the arguments corresponding to the ellipsis of a variadic function. For example, given:
char c = '\007';
short s = 0xB0hD;
float f = 3.1415927;
a call to:
int variadic_function(const char *, ...);
using:
int rc = variadic_function("c s f", c, s, f);
actually converts both c and s to int, and f to double.
Here's what libc has to say about variadic functions:
Since the prototype doesn’t specify types for optional arguments, in a call to a variadic function the default argument promotions are performed on the optional argument values. This means the objects of type char or short int (whether signed or not) are promoted to either int or unsigned int, as appropriate; and that objects of type float are promoted to type double. So, if the caller passes a char as an optional argument, it is promoted to an int
Then, why would anyone use "%c", or "%hd" in printf ? they should just use "%d".
I also see that there's no format specifier for float. float has to live with %f which is for double since due to promotions, it's not possible to receive a float as a variadic argument.
I know for scanf, the arguments are pointers and no promotion happens.
Is there any reason I am missing why and when "%c" must exist for printfs?
Then, why would anyone use "%c", or "%hd" in printf ? they should just use "%d".
One would use %c to interpret the integer as its character code (i.e. print 'A' instead of 65). One would use %hd to instruct printf to drop the upper portion of the short that may have been added as part of sign-extending the short value passed in. Both formats offer an alternative interpretation of an int.
I also see that there's no format specifier for float.
That's correct: since the value has been promoted to double, there is no need for a separate flag.
Assuming the following:
sizeof(char) = 1
sizeof(short) = 2
sizeof(int) = 4
sizeof(long) = 8
The printf format for a 2 byte signed number is %hd, for a 4 byte signed number is %d, for an 8 byte signed number is %ld, but what is the correct format for a 1 byte signed number?
what is the correct format for a 1 byte signed number?
%hh and the integer conversion specifier of your choice (for example, %02hhX. See the C11 standard, §7.21.6.1p5:
hh
Specifies that a following d, i, o, u, x, or X conversion specifier applies to a signed char or unsigned char argument (the argument will have been promoted according to the integer promotions, but its value shall be converted to signed char or unsigned char before printing);…
The parenthesized comment is important. Because of integer promotions on the arguments to variadic functions (such as printf), the function never sees a char argument. Many programmers think that that means that it is unnecessary to use h and hh qualifiers. Certainly, you are not creating undefined behaviour by leaving them out, and most of the time it will work.
However, char may well be signed, and the integer promotion will preserve its value, which will make it into a signed integer. Printing the signed integer out with an unsigned format (such as %02X) will present you with the sign-extended Fs. So if you want to display signed char using an unsigned format, you need to tell printf what the original unpromoted width of the integer type was, using hh.
In case that wasn't clear, a simple example (but controversial) example:
/* Read the comments thread to this post; I'll remove
this note when I edit the outcome of the discussion into
the answer
*/
#include <stdio.h>
int main(void) {
char* s = "\u00d1"; /* Ñ */
for (char* p = s; *p; ++p) printf("%02X (%02hhX)\n", *p, *p);
return 0;
}
Output:
$ ./a.out
FFFFFFC3 (C3)
FFFFFF91 (91)
In the comment thread, there is (or possibly was) considerable discussion about whether the above snippet is undefined behaviour because the X format specification requires an unsigned argument, whereas the char argument is (at least on the implementation which produced the presented output) signed. I think this argument relies on §7.12.6.1/p9: "If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined."
However, in the case of char (and short) integer types, the expression in the argument list is promoted to int or unsigned int before the function is called. (It's worth noting that on most architectures, all three character types will be promoted to a signed int; promotion of an unsigned char (or an unsigned char) to an unsigned int will only happen on an implementation where sizeof(int) == 1.)
So on most architectures, the argument to an %hx or an %hhx format conversion will be signed, and that cannot be undefined behaviour without rendering the use of these format codes meaningless.
Furthermore, the standard does not say that fprintf (and friends) will somehow recover the original expression. What it says is that the value "shall be converted to signed char or unsigned char before printing" (§7.21.6.1/p5, quoted above, emphasis added).
Converting a signed value to an unsigned value is not undefined. It is not even unspecified or implementation-dependent. It simply consists of (conceptually) "repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type." (§6.3.1.3/p2)
So there is a well-defined procedure to convert the argument expression to a (possibly signed) int argument, and a well-defined procedure for converting that value to an unsigned char. I therefore argue that a program such as the one presented above is entirely well-defined.
For corroboration, the behaviour of fprintf given a format specifier %c is defined as follows (§7.21.6.8/p8), emphasis added:
the int argument is converted to an unsigned char, and the resulting character is written.
If one were to apply the proposed restrictive interpretation which renders the above program undefined, then I believe that one would be forced to also argue that:
void f(char c) {
printf("This is a '%c'.\n", c);
}
was also UB. Yet, I think almost every C programmer has written something similar to that without thinking twice about it.
The key part of the question is what is meant by "argument" in §7.12.6.1/p9 (and other parts of §7.12.6.1). The C++ standard is slightly more precise; it specifies that if an argument is subject to the default argument promotions, "the value of the argument is converted to the promoted type before the call" which I interpret to mean that when considering the call (for example, the call of fprintf), the arguments are now the promoted values.
I don't think C is actually different, at least in intent. It uses wording like "the arguments&hellips; are promoted", and in at least one place "the argument after promotion". Furthermore, in the description of variadic functions (the va_arg macro, §7.16.1.1), the constraint on the argument type is annotated parenthetically "the type of the actual next argument (as promoted according to the default argument promotions)".
I'll freely agree that all of this is (a) subtle reading of insufficiently precise language, and (b) counting dancing angels. But I don't see any value in declaring that standard usages like the use of %c with char arguments are "technically" UB; that denatures the concept of UB and it is hard to believe that such a prohibition would be intentional, which leads me to believe that the interpretation was not intended. (And, perhaps, should be corrected editorially.)