I was going through the following statement in a c book:
"interpreting -1 as unsigned int leads to a big +ve number"
-1 is already unsigned .... then what does the above statement mean ??
I have an embedded microcontroller system communicating with a similar system by radio. The api for the radio requires data to be transmitted as an unsigned char array. It will always transmit a positive integer in the range 0 to 255.When I receive the data I am having difficult in extracting this positive integer.
Code:
unsigned char rxData[4]={'1','2','3',''};
int inVal=0;
//want to assign inVal whatever number was transmitted
E.g. 123
I've been at this for a week and have tried at least 10 different approaches including the use of the atoi(), copying the absolute value of each element of rxData into another char array, reinterpret_cast, and others.
Consider a new data type, the mikesint, which can hold 9 bits.
(a) What is the largest integer that an unsigned mikesint can hold?
(b) What is the largest positive integer that a signed mikesint can hold?
(c) What is the largest negative integer that a signed mikesint can hold?
Not sure how to determine this. I'm stuck.
Create a program that keeps on reading positive numbers until user enters a zero value, and then outputs the last positive number entered.
I do not know how to output the 'last positive number' and I do not even know am I doing things correctly so far....
Code:
#include <stdio.h>
int main() {
double num;
for(;;) {
printf("Enter a positive number: ");
scanf("%lf",&num);
[Code] ....
Assume you want to make sure that the user enters a positive number that is divisible by 10 with no remainder. Write the condition you would use in the following do-while loop.
do {
cout << “Enter a positive number that is divisible by 10 with no remainder” << endl;
cin >> number;
}
while ( ____________________________________________________________);
My program uses a while loop to eventually get to an error of zero and a root of sqrt(3). I'm not understand why after the third iteration the program fails to compute a new x value. I'm using Visual Studio 2013. The code tag instructions were dubious.
Code:
#include <stdio.h>
#include <math.h>
main() {
/*This program uses the Newton-Raphson method to solve y = (x^3)-3 for it's roots.*/
printf("This program uses the Newton-Raphson method to solve y = (x^3)-3 for it's roots. Enter your estimate of the root.
");
float x,y,z;
int num;
num = 0;
[Code]...
#include <iostream>
using namespace std;
int main() {
int h;
double A[10][10];
[Code] .....
Write a C++ program that will input from the user a positive number n and find its factorial. Don’t forget to validate the input. The factorial of a positive integer n (denoted by n!) is defines as the product of the integers from 1 to n.
n! = 1* 2 * 3 * ... * (n - 1) * n
You should allow the user to continue working with your program for additional data sets.
Sample output:
Please enter a number: 5
5! = 120
Would you like to continue (Y/N)?Y
Please enter a number: 3
3! = 6
Would you like to continue (Y/N)?N
Good Bye!!
My code for what i think I'm doing is as follows:
#include <iostream>
using namespace std;
int main(){
int i=1;
int n;
[Code] ....
So I tried creating a test code for reading a negative number and positive number but whenever I enter a negative number it read it as being positive.
#include <stdio.h>
#include <iostream>
#include <iomanip>
[Code].....
PS: I am using char over int because the program that I am testing for requires me to use 8 bit variable.
I'm having an issue with the following assignment:
Program is supposed to take in an unsigned number and afterwards return the 'n-th' member of the following recursive function:
f(1) = 1, f(2) = 2; f(3) = 3; f(n+3) = f(n+2) + f(n+1) + f(n), n>0;
When I worked with Fibonacci it was pretty easy since I just had to decrement the next member for each step. I used the following:
#include <stdio.h>
#include <stdlib.h>
#define MAX 100
int fib(int n) {
static int memorize[MAX] = {1,1};
if(memorize[n])
return memorize[n];
[Code] ....
My main problem is that I have no visual of the current function, as well as the fact that it takes the f(n+3) = f(n+2) + f(n+1) + f(n), whilist I've only got f(n) to begin with.
Here is the code:
int unsigned long a,b;
scanf("%lu",&a);
scanf("%lu",&b);
printf("%lu",a*b);
suppose we input a very large number on a and b (no more than 1000 digit)
like a = 1234567890123456789012345678901234567890
and b = 9876543210987654321098765432109876543210
it outputs 1, why is it?
It is probably the answer is too big, is there any way to show a*b?
How can i write a function that will read an "unsigned integer" into a variable of type "unsigned short int"? i can not use cin >> inside the function.. so i am looking for atleast a hint!
View 16 Replies View Relatedhere's one more thing id like to do to make the input even better able to handle user error, but im not sure if its possible or at least easy. I need the function to return a large positive number. As of right now, it can handle users entering characters, but what if the user enters a negative number? is there a way to check to see if what is coming in is negative before the sign gets lost in conversion to unsigned"ness"?
Code:
unsigned long getNum(char prompt[80])
{
unsigned long darts;
printf("%s", prompt);
while((scanf("%lu", &darts)) != 1)
{
[code]....
unsigned char key[32];
139 unsigned char rfseed[32];
173 f = fopen("/dev/urandom","rb");
174 fread(key,1,32,f);
175 fread(rfseed,1,32,f);
I am having problems copying outputs of the above code into other unsigned char other[32]. I need to keep the output of dev/urandom for backup. But, when I try to assign the values by memcpy(other, key, 32), the values do not match. The same problem happens by assigning values index by index in a loop.
Is it really needed to specify 0 as an unsigned integer? I mean 0 is always 0 regardless it's signed or not, no? In the below example is the 0U really needed?
#include <stdio.h>
unsigned invert(unsigned x, int p, int n) {
return x ^ (~(~0U << n) << p);
}
int main(void) {
[Code]...
What are positive and negative infinity for different data types in c++, are they represent maximum and minimum limit of a type? or positive infinity is not a finite value.can some explain this positive and negative infinity paradigm
View 3 Replies View RelatedConsider this piece of code from the following website: [URL] .....
Code:
unsigned intx = 50;
x += (x << 2) + 1;
The website above says the following about the code:
Although this is a valid manipulation, the result of the shift depends on the underlying representation of the integer type and is consequently implementation-defined.
How exactly would a legal left shift operation on an unsigned integer result in implementation-defined behaviour?
I need to convert a string IP to an unsigned int (uint32), but however solutions I've found elsewhere have not worked (such as `atoi`).
When using `(uint32)"54.171.82.217 ";` : [URL] ....
When using `atoi("54.171.82.217");`: [URL] .....
How can I correctly convert the string version of the IP to the uint32 type?
I'm confused about the actual value of 8 bytes for unsigned integers.
The below code suggests the value is 13217906525252912201:
#include <stdio.h>
#include <inttypes.h>
typedef uint64_t byte_int_t;
int main(void){
byte_int_t t;
printf("%" PRIu64 "
", t);
}
./runprogram
13217906525252912201
However, when I use a calculator, I get a different value: 2^64= 1.8446744e+19
So I was wondering is this really 8 bytes? So I try below test and it produces 8, as expected:
#include <stdio.h>
#include <inttypes.h>
typedef uint64_t byte_int_t;
int main(void) {
byte_int_t t;
printf("%u
", sizeof(t));
return 0;
}
So why does C and my calculator provide two different results?
How do I print an unsigned char in c++? E.g.
unsigned char a[100] = "acdef";
wprintf(L"a is %u
", a);
wcout << "a is " << a << endl;
gives
a is 2880488
a is 002BF3E8
and not
a is acdef
a is acdef
??
what is the difference between unsigned char and char?
I have a double variable and depending on certain conditions I need to set certain bits of an unsigned short Variable. For example, if double var is odd I need to set the 15th bit of the unsigned short variable.
View 4 Replies View Relatedwhen you convert 1.7 to unsigned int, it becomes 1 or 2?
View 9 Replies View RelatedI've sometimes encountered unexpected runtime issues caused by unsigned values being decremented below zero.
Example 1: "unsigned_value += negative_integer_value;"
Example 2: "for( size_t i = size - 1; i >= 0; --i )"
My compiler doesn't provide any compile-time or run-time warnings.
As far as I know, it's not possible to overload operators of primitive data types to check if the unsigned value is decremented below zero.
Any clever strategy to trace such cases at debug runtime, without having to add asserts all over the code? It's important that it does not affect performance in release mode.
I have written a function that takes in a positive decimal and returns its Binary equivalent; however, the output always adds an additional zero to the binary. What could I do to get rid of it?
If the number is 7, it outputs 0111 instead of 111.
Code:
#include <stdio.h>
void Dec(int n) {
if(n > 0)
Dec(n/2);
printf("%i", n%2);
[Code] ....
I have task to make program.. To enter 10 positive numbers in an array. Find three numbers that stand near each other and that have the maximum sum.
I understand how to make array, enter 10 number i find only one biggest number from 10.. How to find three max number sum that stand near each other..
int Fib1 = 1;
int Fib2 = 2;
int Fib3 = 0;
int randomynumber;
int Loop;
[code].....
this returns negative numbers sometimes.what did i do wrong side note this is not the complete program it is only the part with the problem because the complete code is sort of longish and very confusing