C++ :: How To Get Half Of The Last Binary Number
Nov 11, 2013For example if we have 101010 First half is 101 which you get by multiplying 101010 * 10 ^-3
now how do you get the second half (010) ??
For example if we have 101010 First half is 101 which you get by multiplying 101010 * 10 ^-3
now how do you get the second half (010) ??
Code: Complete the program below which converts a binary number into a decimal number. Sample outputs are shown belowComplete the program below which converts a binary number into a decimal number. Sample outputs are shown below.
Sample Output 1:
8-bit Binary Number => 11111111
Decimal Number = 255
Sample Output 2:
8-bit Binary Number => 10101010
Decimal Number = 170
Sample Output 3:
8-bit Binary Number => 101010102
Number entered is not a binary number
#include <iostream>
using namespace std;
int main()
{
int num;
[code]....
Here's the part of the codes where I tried to use boolean expression:
Code:
#include <iostream>
using namespace std;
int main() {
int num;
cout << "8-bit Binary Number=";
cin >> num;
[Code] .....
How can I get started with the body?
Here's the part of the codes where I tried to use boolean expression:
#include <iostream>
using namespace std;
int main()
[Code].....
May I know that how can I get started with the body?
whats wrong with this code, I'm trying to parse a .js file and replace all the ";" with "; " i.e add a new line after each ";". So I load the file into a char[] buffer then assign a string to this contents of this buffer. Then loop char by char through using an iterator and check for a ";", if found use replace. So int i gets to about 85898 then crashes with unknown error, 'i' should reach about 175653. It does work up till it crashes. And, is this not a simpler way to load a file into a buffer, there is in C.
Code:
#include <iostream>
#include <fstream>
#include <algorithm>
#include <vector>
#include <stdlib.h>
int main()
[code]....
how to display the other half of the box's border? where should i put the printf? or is there any other method that you can show me using arrays?
Code:
#include <stdio.h>
#define ROW 15
#define COLUMN 15
void disp_box (char b[ROW][COLUMN])
[Code].....
How do I round to the nearest half integer?
Example;
0 to 0.2 = 0
0.3 to 0.5 = 0.5
0.6-0.9 = 1
What's the formula for C++?
I found a way to do it in math but I want to know what I'd have to put in C++. It's (x*2+.5)remove decimal numbers and divide by 2. What do I put in place of "remove decimal"?
For ex; x = 8.4 using (x*2+.5)remove decimal then divide by 2.
8.4 * 2 = 16.8
16.8 + .5 = 17.3
17.3 - decimal = 17
17/2 = 8.5
I want to do something like;
cout << " Half number: << (x*2+.5)remove decimal/2 << endl;
What would I have to put in place of "remove decimal"?
was making a somewhat of a Binary to Hex convertor but only 10/15 cases work and the non working are in the middle; 0010, 0011, 0100, 0101, 0110, 0111;;
// Test Code.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include <iostream>
#include <cmath>
#include <conio.h>
using namespace std;
int main(void)
{int A;
cout << "Enter the binary starting with the MSB
[code]....
The output should be something like:
5
45
543
5432
54321
This is a program to print upper half of the matrix.
#include <iostream>
using namespace std;
int upperhalf(int n, int apu[n][n]) {
cout<<"The upper half of the matrix is :
[Code] ....
the compiler is giving these errors:-
sh-4.2# g++ -o main *.cpp
main.cpp:4:31: error: use of parameter outside fun
ction body before ']' token
int upperhalf(int n, int apu[n][n])
[Code] ....
I dont know where i am wrong.
i am writing a program that accepts a decimal number from the user and convert it to binary numbers. After the conversion, i should count the number of 1's and 0's in the said binary number. I got upto converting and counting 1's using Brian Kernighan’s Algorithm. But, i can't seem to get it to count the number of 0's.
#include <iostream>
#include<bitset>
using namespace std;
int main() {
int num,count=0,Zero,count1 =0;
cout<<"Enter the number:";
cin>>num;
string binary;
[code].....
I am trying solve this a problem: [URL] .....
In short:
1. Input is an N x N grid with a non-negative integer in each cell.
2. We can move from one cell to any of its adjacent cell in all four directions.
3. The 'cost' needed to move from one cell to another is the positive difference of their values.
4. We have to find the minimum cost such that we can visit atleast half of the cells with that cost;
I am not being able to think of any solution that runs in time O(N2) or better (N2 because of the size of N, any algorithm worse than it will not run within the time limit) for this problem.
So I need a hint on how to solve this problem optimally.
I am very new to programming and have been working on a program that can receive decimals or binary numbers and convert them. The decimal --> binary works fine. For some reason I cannot figure out I cannot get the "BinaryToDecimal" function to perform. By putting a "printf" into the for-loop.
Code:
#include <stdio.h>#include <string.h>
#include <math.h>
char* ReverseString (char _result[]) {
int start, end, length = strlen(_result);
char swap;
for (start = 0, end = length-1; start < end; start++, end--)
[code]....
I have a code and am asked to modify it so that it will take as input as unsigned binary number up to 16 digits in length and convert it into its equivalent decimal number and output that decimal number.
All I know is that I use library function strlen() in <cstring> to calculate the length of the input string.
I also know I have to use something called pow(2,4);
//pow (); is found in cmath
I was told to use sum = sum >>16-l; (l is the length of />/>
#include <iostream>
using namespace std;
int main() {
[Code]....
how to take binary number as an input, generate partial products by bit-wise multiplication and in last step to add all the partial products to generate final products".
View 5 Replies View RelatedI had an exercise that required me to convert a number to binary (base 2) which as simple enough.
Code:
#include <iostream>#include <iomanip>
#include <cmath>
using namespace std;
void Conversion (int n);
int main () {
[Code] .....
I now have a follow on exercise that requires me to convert to binary from ant base up to 10, i thought this would just be replacing the 2 with a variable obtained form the user, but i am having problems as within the function i am getting an error that i haven't passed enough arguments and i cant see why i get this. I did the following:
Code:
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
float Conversion (int n, int b);
[Code] ....
I have to do a code for a encoder, decoder and converter for binary number and hexadecimal...
View 1 Replies View RelatedThe following piece of code is supposed to output the binary representation of a given integer and it does exactly that. However, if the given integer is 2, then output is 01. Is there a way to make the program output 0001. I am working on a C program that outputs 4-bit gray code.
#include <stdio.h>
#include <math.h>
int main(void) {
long int n=2;
while (n) {
if (n & 1)
printf("1");
[Code] ......
I recently wrote a program to convert numbers to binary in c++, Well here it is:
#include <iostream>
void recur(int convert) {
if(convert == 0) //if input is 0 , return nothing. {
return;
}
recur(convert/2); // divide convert by 2, get only a 1 or 0
[Code] ....
Write a program to print out the binary value of a 16 bit number.
Create integers i, count, and mask.
Set 'i' to a hex value of 0x1b53.
Set mask to a value of 0x8000. Why?
print a line to show the hex value of i and then the leader for the binary value like this: Hex value = 1b53 Binary=
Use a for loop to loop 16 times and print 16 digits, using count as the loop counter
To test for each digit value, bitwise and 'i' with 'mask'
when the result for the bitwise and is true, print the number '1'
when the result for the bitwise and is false, print the number '0'
then shift mask one place to the right
print a new line and then quit
Use prtscrn and make a hard copy of the code with the console output.
Extra: use the modulus of count and print a space after every 4th digit to make the binary easier to read
The output should look like this: Hex value = 1b53, Binary= 0001 1011 0101 0011
so far this is what i have
#include <stdio.h>
#include <stdlib.h>
int main() {
int i, count, mask;
// 1B53 0001 1011 0101 0011
// 8000 1000 0000 0000 0000
i = 0x1b53;
[Code] ....
it is telling me that there is an "else" without previous "if", also is the program that I wrote correct?
Write a program to determine the number of binary palindromes in a given range [a;b]. A binary palindrome is a number whose binary representation is reading the same in either forward or reverse direction (leading zeros not accounted for). Example: the decimal number 5 (binary 101) is palindromic.
View 2 Replies View RelatedBasically i need to make a number guessing game where user thinks of a numbver from 1 - 100 and the machine will try to guess it in the least number of times. Once it guesses the number it will also say how many tries it took to guess.
My code so far is
#include<iostream>
using namespace std;
const int MAX = 100;
int main() {
char ch;
cout << "Think of an integer number between 0 and " << MAX<<endl;
cout << "Write it down on a piece of paper then hit a key to continue"<<endl<<endl;
cin.get(ch);
[Code] ....
If I have a number 117, represented in binary as : 01110101 and I wanted to grab the top nibble. What would be the decimal value I would be extracting?
Would it be 0111 or 0101 decimal values 112 or 5 or is my understanding completely wrong?
Windows 7, 64 bits, Visual Studio 10.
I have a problem to read a large number of binary files, process them and store them under a new name. The program and routines go very well for 505 files. After reading 506 files, the program now refuses to read the next file. I have 16 Gb of memory and tried to close all other programs and restart the PC. it always stops after 506 files (512 files would be more understanding in a way...).
Here is my code. I have tried many things without success. This is only part of the loop that stops. The if test if (myfile.is_open() returns false by some reason. I can start the process again starting with the file that does not open and then it stops again after 506 files.
char * tfiBlock;
ifstream myfile (OrigFilename, ios::in|ios::binary|ios::ate);
if (myfile.is_open()) {
int lengde = myfile.tellg();
tfiBlock = new char [lengde];
//static char memblock [size];
[Code] .....
Clean up procedure:
delete[] tfiBlock;
Are there any limits to how many files that can be opened, or is it maybe someting to be set in the compiler?
So basically I have an array of hex numbers representing a binary number. Each binary number is 1/5th layer of the over all font.
For example... the letter A
........ B00000000 0x00
.****... B01111000 0x78
...*.*.. B00010100 0x14
...*..*. B00010010 0x12
...*.*.. B00010100 0x14
.****... B01111000 0x78
........ B00000000 0x00
As you can see each HEX number is a layer in the font which consists of in the above example 7 layers.
Now what I would like to do is create a C++ program, so I can visualize a HEX font array that I got off the internet.
#include <iostream>
using namespace std;
const char font[][5] = {
{0x00,0x00,0x00,0x00,0x00}, // 0x20 32
{0x00,0x00,0x6f,0x00,0x00}, // ! 0x21 33
{0x00,0x07,0x00,0x07,0x00}, // " 0x22 34
{0x14,0x7f,0x14,0x7f,0x14}, // # 0x23 35
[Code]...
Basically I am asking two things. How can I make this display the correct representing letter in the array when a user inputs his own text.
And secondly, how can I output the HEX numbers that is in the array as a binary number.
My size of binary file is 1920 KB and my struct size is 124 kb. now to find number of records in file I divided 1920/124 and it gives me 15.4.... do I add 1 to 15.4 and make it 16 or do i take it as 15?
View 9 Replies View Related