Related
I call the recv() that receive data from socket and print the end of buffer content by hex
char nbuff[BUFSZ];
while ((r_n=recv(sfd,rbuff,B_BUF,MSG_EOF))>-1)
{
printf("r_n:%d eob_p:%x\n",r_n,rbuff[r_n-1]);
if (r_n==0)
{
break;
}
memset(rbuff,0,B_BUF);
}
the result is
r_n:1674 eob_p:3c
r_n:1228 eob_p:76
r_n:2456 eob_p:ffffff81
r_n:1228 eob_p:4b
r_n:1228 eob_p:49
r_n:2456 eob_p:57
r_n:1417 eob_p:ffffff82
I am confused about why the result is 4 bytes.
I create another code to print the file that saved from buff
int main ()
{
char buff[11686];
memset(buff,0,11686);
FILE *in =fopen("web/www.sse.com.cn.html","r");
fread(buff,11686,1,in);
for (int i = 0; i < 11686 ; i++)
{
printf("%x\n",buff[i]);
}
}
the result is
....
buff[11684]:60
buff[11685]:ffffff82
why the char buff 's contents size is 4 bytes buff[11685]:ffffff82
Diagnosis
In the second example, buff is a char buffer and plain char is a signed type on your machine, and you're storing values which are negative in buff, so when they're converted to int in the call to printf(), they are negative integers (of small magnitude), printed in hex.
ISO/IEC 9899:2018
Actually, the links are to an online draft of C11, not C18, in HTML which allows links to the relevant paragraphs in the standard. AFAIK, these details have not changed between C90, C99, C11 and C18 anyway.
The standard says that the plain char type is equivalent to either signed char or unsigned char.
§6.2.5 Types ¶15:
The three types char, signed char, and unsigned char are collectively called the character types. The implementation shall define char to have the same range, representation, and behavior as either signed char or unsigned char.45)
45) CHAR_MIN, defined in <limits.h>, will have one of the values 0 or SCHAR_MIN, and this can be used to distinguish the two options. Irrespective of the choice made, char is a separate type from the other two and is not compatible with either.
§6.3.1.1 Boolean, characters and integers ¶2,3:
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.58) All other types are unchanged by the integer promotions.
3 The integer promotions preserve value including sign. As discussed earlier, whether a "plain" char is treated as signed is implementation-defined.
58) The integer promotions are applied only: as part of the usual arithmetic conversions, to certain argument expressions, to the operands of the unary +, -, and ~ operators, and to both operands of the shift operators, as specified by their respective subclasses.
§6.5.2.6 Function calls ¶6,7:
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. If the number of arguments does not equal the number of parameters, the behavior is undefined. If the function is defined with a type that includes a prototype, and either the prototype ends with an ellipsis (, ...) or the types of the arguments after promotion are not compatible with the types of the parameters, the behavior is undefined. If the function is defined with a type that does not include a prototype, and 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.
7 If the expression that denotes the called function has a type that does include a prototype, the arguments are implicitly converted, as if by assignment, to the types of the corresponding parameters, taking the type of each parameter to be the unqualified version of its declared type. 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.
Exegesis
Note the last two sentences of §6.5.2.6 ¶7 — when the char values are promoted by the 'integer promotions', they are promoted to a (signed) int, and the negative values remain negative. Since an int has 4 bytes, and all the machines you're likely to have available use two's-complement arithmetic, the most significant 3 bytes of the value will be 0xFF each.
Prescription
To always print 2-digit hex for the characters, use %.2X (or %.2x if you prefer; you can also use either %02X or %02x) and pass either (unsigned char)rbuff[r_n-1] or rbuff[r_n-1] & 0xFF as the argument (using the variables from the first example). Or, using the variables from the second example:
printf("%.2X\n", (unsigned char)buff[i]);
printf("%.2X\n", buff[i] & 0xFF);
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!
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.
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.)
unsigned short int i = 0;
printf("%u\n",~i);
Why does this code return a 32 bit number in the console? It should be 16 bit, since short is 2 bytes.
The output is 4,294,967,295 and it should be 65,535.
%u expects an unsigned int; if you want to print an unsigned short int, use %hu.
EDIT
Lundin is correct that ~i will be converted to type int before being passed to printf. i is also converted to int by virtue of being passed to a variadic function. However, printf will convert the argument back to unsigned short before printing if the %hu conversion specifier is used:
7.21.6.1 The fprintf function
...
3 The format shall be a multibyte character sequence, beginning and ending in its initial
shift state. The format is composed of zero or more directives: ordinary multibyte
characters (not %), which are copied unchanged to the output stream; and conversion
specifications, each of which results in fetching zero or more subsequent arguments,
converting them, if applicable, according to the corresponding conversion specifier, and
then writing the result to the output stream.
...
7 The length modifiers and their meanings are:
...
h Specifies that a following d, i, o, u, x, or X conversion specifier applies to a
short int or unsigned short int argument (the argument will
have been promoted according to the integer promotions, but its value shall
be converted to short int or unsigned short int before printing);
or that a following n conversion specifier applies to a pointer to a short
int argument.
Emphasis mine.
So, the behavior is not undefined; it would only be undefined if either i or ~i were not integral types.
When you pass an argument to printf and that argument is of integer type shorter than int, it is implicitly promoted to int as per K&R argument promotion rules. Thus your printf-call actually behaves like:
printf("%u\n", (int)~i);
Notice that this is undefined behavior since you told printf that the argument has an unsigned type whereas int is actually a signed type. Convert i to unsigned short and then to unsignedto resolve the undefined behavior and your problem:
printf("%u\n", (unsigned)(unsigned short)~i);
N1570 6.5.3.3 Unary arithmetic operators p4:
The result of the ~ operator is the bitwise complement of its (promoted) operand (that is,
each bit in the result is set if and only if the corresponding bit in the converted operand is
not set). The integer promotions are performed on the operand, and the result has the
promoted type. ...
Integer type smaller than int are promoted to int. If sizeof(unsigned short) == 2 and sizeof(int) == 4, then resulting type is int.
And what's more, printf conversion specifier %u expects unsigned int, so representation of int is interpreted as unsigned int. You are basically lying to compiler, and this is undefined behaviour.
It's because the arguments to printf() are put into the stack in words, as there is no way inside printf to know that the argument is short. Also by using %u format you are merely stating that you are passing an unsigned number.