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Signedness of char and Unicode in C++0x


By : user3852207
Date : October 15 2020, 01:19 AM
To fix the issue you can do char will be the type used for UTF-8 because it's redefined to be sure it can be used with it:
code :


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Is the signedness of char an interface issue?


By : user2396224
Date : March 29 2020, 07:55 AM
To fix the issue you can do According to ISO/IEC 9899:TC3,
calling a function through an expression of incompatible type is undefined behaviour (6.5.2.2 §9) compatible function types must have compatible parameter types (6.7.5.3 §15) compatible pointer types must point to compatible types (6.7.5.1 §2) char, signed char and unsigned char are different basic types (6.2.5 §14) and thus incompatible (6.2.7 §1), which is also explicitly mentioned in footnote 35 on page 35

Remove un-/signedness of char type


By : Adrian Ziemkowski
Date : March 29 2020, 07:55 AM
help you fix your problem The "sign-ness" of char is implementation defined. It doesn't have to be able to store negative numbers. In strict standardese, char even is never a signed integer type, even if it can store negative numbers on an implementation. Still, the class template is_signed will report true for char if it can store negative numbers, because that's a useful thing for it to do.
Anyway, the boost docs say the following about make_unsigned, making it look like you can use it for your purpose.

C cast and char signedness


By : Eytyixs
Date : March 29 2020, 07:55 AM
will help you So lately, I read on an issue regarding the three distinct types in C, char/unsigned char/signed char. The problem that I now encounter is not something I have experienced up till now (my program works correctly on all tested computers and only targets little-endian (basically all modern desktops and servers using Windows/Linux right?). I frequently reuse a char array I defined for holding a "string" (not a real string of course) as temporary variables. E.g. instead of adding another char to the stack I just reuse one of the members like array[0]. However, I based this tactic on the fact that a char would always be signed, until I read today that it actually depends on the implementation. What will happen if I now have a char and I assign a negative value to it? , The following code prints No:
code :
#include <stdio.h>

int
main()
{
    unsigned char a;

    a = -1;

    if(a == -1)
        printf("Yes\n");
    else
        printf("No\n");

    return 0;
}
`(int)a == -1`

The signedness of char in g++/gcc and its history


By : Sushant
Date : March 29 2020, 07:55 AM
To fix this issue I see no undefined behavior in your program. Negative array indices are not necessarily invalid, as long as the result of adding the index to the prefix refers to a valid memory location. (A negative array index is invalid (i.e., has undefined behavior) if the prefix is the name of an array object or a pointer to the 0th element of an array object, but that's not the case here.)
In this case, aptr points to element 256 of a 512-element array, so the valid indices go from -256 to +255 (+256 yields a valid address just past the end of the array, but it can't be dereferenced). Assuming CHAR_BIT==8, any of signed char, unsigned char, or plain char has a range that's a subset of the array's valid index range.
code :
aptr[0] = 0xFF;

Issues about the signedness of char


By : Angel
Date : March 29 2020, 07:55 AM
will help you 1) testing MSB: (x | 0x7F) != 0x7F (or reinterpret_cast(x) & 0x80)
2) reinterpret_cast(x) = 0xFF;
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