[expr.unary.op] The operand of ~ shall have integral or unscoped enumeration type; the result is the one’s complement of its operand. Integral promotions are performed. [expr.shift] The shift operators << and >> group left-to-right. […] The operands shall be of integral or unscoped enumeration type and integral promotions are performed. What’s the integral promotion of uint8_t … Read more
C has the “union” to handle this type of view of data: typedef union { int i; float f; } u; u u1; u1.f = 45.6789; /* now u1.i refers to the int version of the float */ printf(“%d”,u1.i);
I’ve linked to this answer for the full details from several other “rotate” questions, including this community wiki question, which should be kept up to date with best-practices. I found a blog post about this issue, and it looks like it’s finally a solved problem (with new-enough compiler versions). John Regehr at the University of … Read more
5 is 101. 4 is 100. 5 & 4 is not 0: 101 100 & ↓↓↓ 100 Problem solved ✓ Clarification: In C, every non-zero value satisfies the if condition. Meaning, if you write: if (-5) { if (100) { // reachable code } } Whereas: if (0) { destroyTheWorld(); // we are safe }
They’re compound assignment operators, translating (very loosely) x |= y; into x = x | y; and the same for &. There’s a bit more detail in a few cases regarding an implicit cast, and the target variable is only evaluated once, but that’s basically the gist of it. In terms of the non-compound operators, … Read more
Let’s begin by looking the multiplication code. The idea is actually pretty clever. Suppose that you have n1 and n2 written in binary. Then you can think of n1 as a sum of powers of two: n2 = c30 230 + c29 229 + … + c1 21 + c0 20, where each ci is … Read more
& is the bitwise logical and operator – It performs the operation on 2 integer values. WHERE (sc.Attributes & 1) = 0 The above code checks to see if sc.Attributes is an even number. Which is the same as saying that the first bit is not set. Because of the name of the column though: … Read more
#include <stdio.h> int main(void) { int a, b, result; printf(“Enter the numbers to be multiplied:”); scanf(“%d%d”, &a, &b); // a > b result = 0; while (b != 0) // Iterate the loop till b == 0 { if (b & 1) // Bitwise & of the value of b with 1 { result = … Read more
You can convert the characters to integers and xor those instead: l = [ord(a) ^ ord(b) for a,b in zip(s1,s2)] Here’s an updated function in case you need a string as a result of the XOR: def sxor(s1,s2): # convert strings to a list of character pair tuples # go through each tuple, converting them … Read more
|| and && are boolean operators and the built-in ones are guaranteed to return either true or false. Nothing else. |, & and ^ are bitwise operators. When the domain of numbers you operate on is just 1 and 0, then they are exactly the same, but in cases where your booleans are not strictly … Read more