1.The operands from << and >> may be any of integer type (including char) The integer promotions are performed on both operands the result has the type of the left operand after promotion.
It means that if we have z = x >> y then sizeof(z) == sizeof(x) ?
2. The ~ operator is unary the integer promotions are performed on its operand.
So if I have short int y; and int x=1; y = ~x what is the meaning here?
what order a CPU would process the following arithmetic problem: 5 - (-9) = 14? Would the CPU recognize that the 'minus a minus' combination simply represents 5 + 9, and proceed with that addition, or would the CPU have to first calculate the 2's complement of -9, and then proceed to take the 2's complement of that first result in order to complete the calculation of the addition of the 'double negative'?
View 2 Replies View RelatedI have a project assignment for school to write a program that does number conversions using bitwise operators. The premise is that the user enters a number with one of three letter prefixes -- Q1232, O6322, H762FA, etc. -- and the program will take that number and convert it to the other two number bases. Q is for quarternary, O is for octal, and H is for hexadecimal. The transformations should be done using bitwise operators and bit shifting. I am guessing I need to scan the number, convert it to binary, then convert it to the other two bases.
However, I am completely new to bitwise operators and bit shifting, so how to convert numbers of different bases to binary and then binary to other bases using these bit and bitwise functions. I don't have much code done yet, since I am still unsure of how to approach it, but I'll post what little I have.
Here it is:
#include <stdio.h>
#include <string.h>
int main() {
char numType;
printf("
The user will enter a number up to 32 digits in quarternary
");
printf("(base 4), octal (base 8), or hexadecimal (base 16). If in
");
[Code] ....
I figure in each case I can write a function that converts the entered number to binary, then maybe two more functions that convert said binary number to the other bases. For default in the switch I will tell the user they entered an invalid number. I don't have the program looping until the user types 'EXIT' yet, but I will once I figure out anything about these bitwise operators.
What is the difference between at performance level, if any, between the following cases, during assignment?
case 1: #define Value_16 16
and
case 2: #define Value_16 (1<<4)e.
I have a 32 bit integer variable with some value (eg: 4545) in it, now I want to read first 8 bits into uint8_t and second 8 bits into another uint8_t and so on till the last 8 bits.
I am thinking of using bitwise operators...
I would like to ask about how we calculates the following bitwise expression.
Code:
unsigned char ch[2] = {0x49,0x49};
ch[0] | ch[1] << 8; I'm thinking ch[1] << 8 as 0x00 ...
So, I think that the above expression converts to 0x49 | 0x00 ... and the complete expression should be 0x49 for me.
But, the compiler gives me the result of 0x4949 as two bytes.How does the compiler calculate this expression as two bytes?show me the steps included in the calculation of this expression?
How to do this program i can easily do it in a simple for loop but i have to do this program with the following directions:
1. Write a function called bitN() that returns the value of bit N in number, where number is the first parameter, and N is the second. Assume N of the least significant bit is zero and that both parameters are unsigned int's. (A simple one-liner will suffice)
2. Write a main() function that uses bitN() to convert a decimal integer into its binary equivalent. Obtain the integer to convert from the first command-line argument.
3. Use the expression
unsigned int numBits = sizeof(unsigned int)*CHAR_BIT;
to get the number of bits in an unsigned int. (Include limits.h to get the definition for CHAR_BIT.)
Trying to write 4 bytes ints in a binary file and extract them after... I'm using the exclusive or (^) to isolate single bytes to write to and extract from the file since the write() function accepts only chars, only the beginning and end results are not the same...
#include <iostream>
#include <ctime>
#include <fstream>
#include <cstdlib>
using namespace std;
[Code] .....
Let's examine the code.
int x = 100;
unsigned long answer1 = ~x;
unsigned long long answer2 = ~x;
cout << (bitset<32>) x << "
[Code] .....
Shouldn't the decimal of answer 1 and 2 the same thing?
I get 4294967195 for answer1 and 18446744073709551515 for answer 2.
I'm doing a bitwise operations on 2 bytes in a buffer, then storing the result in a variable. However, I sometimes get a non-zero value for the variable even though I'm expecting a zero value.
The relevant portion of the code is as follows.
unsigned int result = 0;
long j = 0, length;
unsigned char *data;
data = (unsigned char *)malloc(sizeof(unsigned char)*800000);
[Code] ......
I'm expecting result to be zero when my data[j] and data[j+1] are 0xb6 and 0xab respectively, which is the case for most of the time. However, for certain values of j, my result is strangely not zero.
j = 62910, result = 64
j = 78670, result = 64
j = 100594, result = 64
j = 165658, result = 512
j = 247990, result = 128
j = 268330, result = 512
j = 326754, result = 1
j = 415874, result = 256
j = 456654, result = 1024
j = 477366, result = 512
It appears that these strange result values are all powers of 2, with a 1 bit appearing somewhere in the unsigned int.
I'm not changing the value of result anywhere else in the code, and when I print out (unsigned int)(((data[j]^0xb6)<<8)|(data[j+1]^0xab)), I get 0, but somehow when it gets stored in result, it's no longer zero.
I have a doubt in Left Shift Operator
int i = 1;
i <<= (sizeof (int) *8);
cout << i;
It prints 1.
Doubt:
i has been initialized to 1.
And while moving the bits till the size of the integer, it fills the LSB with 0's and as 1 crosses the limit of integer, i was expecting the output to be 0.
How and Why it is 1?
how to show all the bits of a number using bitwise shift operator....and hence represent the number in 2's complement representation
View 1 Replies View RelatedIn a .h file there is a function that takes in this parameter:
void (^callback)(float * arg)=NULL
as in a function definition:
void func(void (^callback)(float * arg)=NULL);
What I am able to read is that it takes a function pointer and if not defined it overrides with NULL. The part I do not get is the ^ in (^callback). I only know ^ as a bitwise XOR operator. It also generates issues in my VS2012 compiler (something with CLR). So I would really like to rewrite this part to something else, without the bitwise operator...
I'm doing a bitwise operations on 2 bytes in a buffer, then storing the result in a variable. However, I sometimes get a non-zero value for the variable even though I'm expecting a zero value. The relevant portion of the code is as follows.
Code:
unsigned int result = 0;
long j = 0, length;
unsigned char *data;
data = (unsigned char *)malloc(sizeof(unsigned char)*800000);
[Code] ....
I'm expecting result to be zero when my data[j] and data[j+1] are 0xb6 and 0xab respectively, which is the case for most of the time. However, for certain values of j, my result is strangely not zero.
Code:
j = 62910, result = 64
j = 78670, result = 64
j = 100594, result = 64
j = 165658, result = 512
j = 247990, result = 128
j = 268330, result = 512
j = 326754, result = 1
j = 415874, result = 256
j = 456654, result = 1024
j = 477366, result = 512
It appears that these strange result values are all powers of 2, with a 1 bit appearing somewhere in the unsigned int.
I'm not changing the value of result anywhere else in the code, and when I print out
Code: (unsigned int)(((data[j]^0xb6)<<8)|(data[j+1]^0xab))
I get 0, but somehow when it gets stored in result, it's no longer zero. I really don't understand what could be wrong.
The + operator has left to right associativity then for the program f1() + f2(), why f1() is not called first compared to f2()?
View 5 Replies View RelatedI wrote this program using an online compiler i am getting a lot of errors.
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main() {
string birthmonth;
string birthday;
[Code] ....
I have a code that uses multi threads using OMP. However, Inside the code I allocate too many memory using "new".
I know that the memory allocation using new must be done in serial. Therefore I am not getting a good performance out of the multi-threaded program.
I thought about allocating a big memory only once, and then advance the pointer every time I need to allocate a small memory.
For example, if I need to allocate chunks of 7 doubles:
Code:
main_memory = new double [10000];
double* x1 = main_memory;
main_memory+= 7;
double* x2 = main_memory;
main_memory+= 7
However, I do not know how to handle the memory deallocation. Beside, this is not thread safe method because main_memory is shared by all threads.
May be there is a boost library that does this, which I am not aware of.
I have two class GameOfLife and Cell and i want to overload square braket for class GameOfLife."if g is a GameOfLife object, g[10][5] will return the Cell at row 10 and column 5. If there is no such Cells, then it will return a new Cell with position (-1000,- 1000)."
but if g[10][1000] and 1000>cols,then it returns different Cell exp (3,2) How i do control the col ( [row][col] )?
Code: vector<Cell> GameOfLife::operator [](const int row){
vector<Cell> rowCell;
for(int i=0; i<cols; ++i)
{
if( isLive(row,i) )
rowCell.push_back( Cell(row,i) );
else
rowCell.push_back( Cell(-1000,-1000) );
}
return rowCell;
}
Came across the foll code:
#include<stdio.h>
main() {
int i=4,j=7;
j=j||(printf("you can")&&(i=5));
printf("%d %d",i,j);
}
output: 4 1
Although I am specifying the braces for the && operator so that it gets executed first..Then also the value of i remains 4 only..Why does not it gets changed to 5??Also the printf does not execute??
Switch case statements are a substitute for long if statements that compare a variable to several "integral" values ("integral" values are simply values that can be expressed as an integer, such as the value of a char).
So does that mean switch statements can only test if variable == value and nothing more, like > < >= <= != etc... ? I tried to make a program to test this and it seems like switch statements are limited to == but I'm not sure, maybe I'm doing something wrong.
This is the program I tried to make to test this:
Code:
#include <stdio.h>
int main () {
int n;
[Code]....
So is it true that switch statements only work with the built in == operator? if that was the case then I would feel rather meh about switch statements.
I have a small piece of code that used the set::insert function on a set of myClass. For that, I need to overload the < and > operators, which I have, but I still get a huge error saying it can't compare.
set<MyClass> mySet;
MyClass myClass
All the class information gets filled in. Then, I try to insert...
mySet.insert(myClass);
bool operator<(MyClass &lhs, MyClass &rhs) {
return lhs.name < rhs.name; //name is a string
}
The error says
...stl_function.h:230:22: error: no match for 'operator<' in '__x < __y'
MyFile.h:80:6: note: candidate is bool operator<(MyClass&, MyClass&)
I'm experimenting with a custom memory-pool for my application, and I initially planned to override the global new and delete operators to allocate memory in this pool. But since I'm using QT, this will apply to all the QT-related stuff as well. Should I instead just override the new and delete operators per class?
View 2 Replies View RelatedI have two files NumDays.h and ClientProgram.cpp
clientprogram.cpp basically has the main module below
int main(){
// Initialized UDT object Declarations
NumDays hoursWorked_John; // Instantiate with Default Constructor
NumDays hoursWorked_Sue(36.9); // Instantiate with Initializing Constructor
NumDays hoursUsed_Sue(4.5); // Instantiate with Initializing Constructor
[Code] .....
I can't figure out anything to put in for NumDays.cpp so it's not there.
In the following program.
void main() {
int a=1;
cout<<a++<<" "<<++a<<" "<<a++<<endl;
}
If I execute the above program i should get 1 3 3. But I'm getting different values when I executed this program. The values that I get after execution are 3 3 1.
I made a program that allows the user to enter information of credit cards on an array of size 5, however I need to allow the user to compare the five credit cards with each other and I am having problems with this particular part. I made my bool operator functions for the operator< and the operator> but how to make the user be able to select which cards he wants to compare and how to compare them. My code is the following:
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
const int SIZE = 5;
enum OPCIONES {CARGAR=1, ABONAR, NADA};
[Code] ......
Why can't I take the address of operators for primitives?
#include <iostream>
#include <string>
int main()
{
{
std::string (&plus)(std::string const&, std::string const&) = &std::operator+;
std::string a ("Hello, "), b("World!");
std::cout << plus(a, b) << std::endl;
[Code]...
[URL]....
I'm using this functionality in a templated class, do I really have to specialize for primitives or use std::enable_if?
