#include<stdio.h>
int main(void)
{
char a = 01212;
printf("%d",a);
return 0;
}
On compiling i get a warning and output -118 how? I know any number starting with 0 in c is considered as octal. The octal equivalent of 01212 is 650 then why the output is -118?
The assignment char a = 01212; on most of the systems is out of range and implementation dependent. A system with 8-bit char that implement 2's complement will print -118.
For detail, please read below explanation.
Unlike integer a char is not signed by default; there are three different char types in C.
char,
signed char
and
unsigned char
A char has a range from CHAR_MIN to CHAR_MAX. For a particular compiler, the char will use either an underlying signed or unsigned representation. You can check this value in limits.h of your system.
Here is the text from C99 standard point number 15
6.2.5 Types
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.35)
And again note 35 says
35) CHAR_MIN, defined in , 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.
Having said this char a = 01212; is larger than 8 bit. The C standard allows char size more than 8 bit but I think almost all computers in today's world implement 8 Bit char.
So if char is implemented as unsigned char and the value is more than CHAR_MAX the value will be converted according to Modulo CHAR_MAX+1.
In 8 bit system, the converted value is 650 modulo 256 which is 650-512 = 138
If char is implemented as signed char the conversion is implementation dependent. If it's an 8-bit char system and it implements 2's complement the value will be -118 as you have seen in your result. Note that in this system the Range for char will be from -128 to +127.
The value of 650 is most likely out-of-range for your char type. In C the behavior is implementation-defined in such cases as out-of-range integer conversions. I.e. it is clear that you will not get 650 in your char, and what exactly you will get depends on your compiler. Consult your compiler documentation to figure out why you got -118.
Char is only occupying one byte, or 8 bits, so the maximum number a unsigned char can hold is 2^8 - 1, which is 255, and a signed char has a maximum of 127. When assigned a number that's greater than that, it would cause undefined behavior, in which a negative number may appear.
Related
int a = 0x11223344;
char b = (char)a;
I am new to programming and learning C. Why do I get value of b here as D?
If I want to store an integer into a char type variable, which byte of the integer will be stored?
This is not fully defined by the C standard.
In the particular situation you tried it, what likely happened is that the low eight bits of 0x11223344 were stored in b, producing 4416 (6810) in b, and printing that prints “D” because your system using ASCII character codes, and 68 is the ASCII code for “D”.
However, you should be wary of something like this working, because it is contingent on several things, and variations are possible.
First, the C standard allows char to be signed or unsigned. It also allows char to be any width that is eight bits or greater. In most C implementations today, it is eight bits.
Second, the conversion from int to char depends on whether char is signed or unsigned and may not be fully defined by the C standard.
If char is unsigned, then the conversion is defined to wrap modulo M+1, where M is the largest value representable in char. Effectively, this is the same as taking the low byte of the value. If the unsigned char has eight bits, its M is 255, so M+1 is 256.
If char is signed and the value is out of range of the char type, the conversion is implementation-defined: It may either trap or produce an implementation-defined value. Your C implementation may wrap conversions to signed integer types similarly to how it wraps conversions to unsigned types, but another reasonable behavior is to “clamp” out-of-range values to the limits of the type, CHAR_MIN and CHAR_MAX. For example, converting −8000 to char could yield the minimum, −128, while converting 0x11223344 to char could yield the maximum, +127.
Third, the C standard does not require implementations to use ASCII. It is very common to use ASCII. (Usually, the character encoding is not just ASCII, because ASCII covers only values from 0 to 127. C implementations often use some extension beyond ASCII for values from 128 to 255.)
I'm referring this question because I can't understand how ASCII characters from 0 to 255 can be represented with a signed char if the range of it is from -128 to 127.
Being char = sizeof(char)= 1 byte, it is also reasonable to think that it can easily represent values up to the maximum of 255;
So why the assignment: char a = 128 has nothing wrong and also why shouldn't I use unsigned char for it.
Thank you in advance!
char c = 128; by itself is correct in C. The standard says that a char contains CHAR_BIT bits, which can be greater than 8. Also, a char can be signed or unsigned, implementation defined, and an unsigned char has to contain at least the range [0, 255].
So an implementation where a char is bigger than 8 bits, or the char is unsigned by default, this line is valid and relevant.
Even in a common 8 bit signed char implementation, the expression is still well-defined in how it will convert the 128 to fit in a char, so there is no problem.
In real cases, the compiler will often issue a warning for these, clang for example :
warning: implicit conversion from 'int' to 'char' changes value from 128 to -128 [-Wconstant-conversion].
signed or unsigned - it takes 8bits. 8bits can contain 256 values. Just question how we use them.
Please see the following C code.
#include <stdio.h>
int main(void)
{
char c1 = 3000;
char c2 = 250;
printf("%d\n",c1);
printf("%d\n",c2);
}
The output of the above code is
-72
-6
Please explain the integer to char conversion rule applied here as both 3000 and 250 are outside of the range of char (-128 to 127).
Please explain the integer to char conversion rule applied here as both 3000 and 250 are outside the range of char(-128 to 127).
Note first that C does not specify whether char is signed or unsigned. That is left to implementations to decide, and they are not consistent on that. On implementations where char is unsigned, 250 is within its range.
Supposing, however, that your chars are signed, which indeed seems consistent with your results, the C rule for the conversions implicit in the assignment statements will not satisfy you:
the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.
(C2011 6.3.1.3/3)
Evidently no signal was raised, so the result is implementation-defined. Among the possibilities is that the least-significant CHAR_BIT bits of each assigned value are stored in the target variable.
There is then an additional conversion when you call printf(). The arguments are promoted from char to int, and since int can represent all values of type char, that one is value-preserving. That allows us to conclude that it is indeed plausible that your implementation converts int to char by keeping only the least-significant bits, and interpreting them as 8-bit two's complement.
Integer uses 4 byte and char uses 1 byte. Numbers in C are represented as signed and that means first bit from the left is for sign (positive, negative) and the rest is number in full complement. So number 3000 is represent like this 00000000000000000000000010111000 in binary and for int it is stored like this. Because char is only 1 byte last 8 bits represent saved number in char variable and that is 10111000. When you convert this into decimal you will have -72.
I have a code like this:
#include <stdio.h>
int main()
{
char a=20,b=30;
char c=a*b;
printf("%c\n",c);
return 0;
}
The output of this program is X .
How is this output possible if a*b=600 which overflows as char values lies between -128 and 127 ?
Whether char is signed or unsigned is implementation defined. Either way, it is an integer type.
Anyway, the multiplication is done as int due to integer promotions and the result is converted to char.
If the value does not fit into the "smaller" type, it is implementation defined for a signed char how this is done. Far by most (if not all) implementations simply cut off the upper bits.
For an unsigned char, the standard actually requires (briefly) cutting of the upper bits.
So:
(int)20 * (int)20 -> (int)600 -> (char)(600 % 256) -> 88 == 'X'
(Assuming 8 bit char).
See the link and its surrounding paragraphs for more details.
Note: If you enable compiler warnings (as always recommended), you should get a truncation warning for the assignment. This can be avoided by an explicit cast (only if you are really sure about all implications). The gcc option is -Wconversion.
First off, the behavior is implementation-defined here. A char may be either unsigned char or signed char, so it may be able to hold 0 to 255 or -128 to 127, assuming CHAR_BIT == 8.
600 in decimal is 0x258. What happens is the least significant eight bits are stored, the value is 0x58 a.k.a. X in ASCII.
This code will cause undefined behavior if char is signed.
I thought overflow of signed integer is undefined behavior, but conversion to smaller type is implementation-defined.
quote from N1256 6.3.1.3 Signed and unsigned integers:
3 Otherwise, the new type is signed and the value cannot be represented in it; either the
result is implementation-defined or an implementation-defined signal is raised.
If the value is simply truncated to 8 bits, (20 * 30) & 0xff == 0x58 and 0x58 is ASCII code for X. So, if your system do this and use ASCII code, the output will be X.
First, looks like you have unsigned char with a range from 0 to 255.
You're right about the overflow.
600 - 256 - 256 = 88
This is just an ASCII code of 'X'.
This question already has answers here:
What is an unsigned char?
(16 answers)
char!=(signed char), char!=(unsigned char)
(4 answers)
Closed 5 years ago.
So I know that the difference between a signed int and unsigned int is that a bit is used to signify if the number if positive or negative, but how does this apply to a char? How can a character be positive or negative?
There's no dedicated "character type" in C language. char is an integer type, same (in that regard) as int, short and other integer types. char just happens to be the smallest integer type. So, just like any other integer type, it can be signed or unsigned.
It is true that (as the name suggests) char is mostly intended to be used to represent characters. But characters in C are represented by their integer "codes", so there's nothing unusual in the fact that an integer type char is used to serve that purpose.
The only general difference between char and other integer types is that plain char is not synonymous with signed char, while with other integer types the signed modifier is optional/implied.
I slightly disagree with the above. The unsigned char simply means: Use the most significant bit instead of treating it as a bit flag for +/- sign when performing arithmetic operations.
It makes significance if you use char as a number for instance:
typedef char BYTE1;
typedef unsigned char BYTE2;
BYTE1 a;
BYTE2 b;
For variable a, only 7 bits are available and its range is (-127 to 127) = (+/-)2^7 -1.
For variable b all 8 bits are available and the range is 0 to 255 (2^8 -1).
If you use char as character, "unsigned" is completely ignored by the compiler just as comments are removed from your program.
There are three char types: (plain) char, signed char and unsigned char. Any char is usually an 8-bit integer* and in that sense, a signed and unsigned char have a useful meaning (generally equivalent to uint8_t and int8_t). When used as a character in the sense of text, use a char (also referred to as a plain char). This is typically a signed char but can be implemented either way by the compiler.
* Technically, a char can be any size as long as sizeof(char) is 1, but it is usually an 8-bit integer.
Representation is the same, the meaning is different. e.g, 0xFF, it both represented as "FF". When it is treated as "char", it is negative number -1; but it is 255 as unsigned. When it comes to bit shifting, it is a big difference since the sign bit is not shifted. e.g, if you shift 255 right 1 bit, it will get 127; shifting "-1" right will be no effect.
A signed char is a signed value which is typically smaller than, and is guaranteed not to be bigger than, a short. An unsigned char is an unsigned value which is typically smaller than, and is guaranteed not to be bigger than, a short. A type char without a signed or unsigned qualifier may behave as either a signed or unsigned char; this is usually implementation-defined, but there are a couple of cases where it is not:
If, in the target platform's character set, any of the characters required by standard C would map to a code higher than the maximum `signed char`, then `char` must be unsigned.
If `char` and `short` are the same size, then `char` must be signed.
Part of the reason there are two dialects of "C" (those where char is signed, and those where it is unsigned) is that there are some implementations where char must be unsigned, and others where it must be signed.
The same way -- e.g. if you have an 8-bit char, 7 bits can be used for magnitude and 1 for sign. So an unsigned char might range from 0 to 255, whilst a signed char might range from -128 to 127 (for example).
This because a char is stored at all effects as a 8-bit number. Speaking about a negative or positive char doesn't make sense if you consider it an ASCII code (which can be just signed*) but makes sense if you use that char to store a number, which could be in range 0-255 or in -128..127 according to the 2-complement representation.
*: it can be also unsigned, it actually depends on the implementation I think, in that case you will have access to extended ASCII charset provided by the encoding used
The same way how an int can be positive or negative. There is no difference. Actually on many platforms unqualified char is signed.