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Why does "unsigned int" + "unsigned int" return an "unsigned int"?


By : ZacAmsden
Date : October 18 2020, 06:10 AM
will help you This would have truly evil consequences:
Would you really want 1 + 1 to be a long type? And (1 + 1) + (1 + 1) would become a long long type? It would wreak havoc with the type system.
code :
0UL + "unsigned int" + "unsigned int"


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What's an efficient way to convert "unsigned char" array to its "unsigned short" counterpart?


By : Kevin
Date : March 29 2020, 07:55 AM
With these it helps Addressing short on byte boundary may (or may not) cause alignment issues, depending on platform.
Also, multiplying is very ineffective, why not use shifting instead? (some compilers may optimize x * 0x100, but if they don't - it's a huge performance hit when all you want is just x << 8...)

Standard guarantees for aliases of integer types in C/C++? E.g.: Is "unsigned" always equal to "unsigned


By : Henry
Date : March 29 2020, 07:55 AM
help you fix your problem Yes, these are guaranteed. In C++11, see §7.1.6.2[dcl.type.simple]/table 10, which lists all of the simple type specifiers (and combinations thereof) and what they mean. For example, the table includes the following:
code :
unsigned      => unsigned int
unsigned int  => unsigned int

signed        => int
signed int    => int
int           => int

Cast from "int" to "unsigned short" after applying bitwise operator "~"


By : Hannibal
Date : March 29 2020, 07:55 AM
will help you The operands of the arithmetic and bitwise operators always undergo the standard promotions before the value is computed. Anything shorter than an int is promoted to either int or unsigned int, depending on the platform (i.e. depending on whether int can represent all values of the type that's being promoted).
On your platform, uint16_t is standard-promoted to int, since your int can represent all values of a uint16_t. Then the bitwise negation is applied to that int value, which is the cause of the problem.
code :
 uint16_t var2 = (uint16_t) ~((unsigned int) var1);

Issues with converting between "string" "const unsigned char" and "utf8proc_uint8_t"


By : Mellinnal Coder
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , There appear to be a couple problems here. The biggest is the assignment:
code :
char b = (char)a;
char* b = (char*)a;
string d = string(b);
string d = string((char*)a);
char tmp = *pStr;     // a single character (first in the string)
char tmp2 = pStr[1];  // a single character (second in the string)
char* pTmp = pStr;    // a pointer to a null terminated string
char tmp3 = (char)pStr;     // compiles, but makes no sense to treat pointer as a character.
utf8proc_uint8_t* pUTF = utf8proc_NFC( (const unsigned char*)e.c_str() );
string strUTF;
if (pUTF)
{
   strUTF = (char*)pUTF;
   free pUTF;
}

When converting to unsigned, the standard says "the least unsigned integer" is the result. Why does "leas


By : Mr.988
Date : March 29 2020, 07:55 AM
hope this fix your issue There are an infinite number of integers equal to any value k modulo 2n. There is k, k+2n, k+2*2n, k+3*2n, k-2n, k-2*kn, etc.
Of these, one is the least unsigned (positive) value.
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