Assume that the given input are 5,+,5 in C.
ex)
5
+
5
And I would like to get the answer 10.
I thought code like the below.
#include
using namespace std;
int main(){
char a,b,symbol;
cin >> a >> symbol >> b;
printf("%d", (a-'0') symbol (b-'0'));
return 0;
}
Expected value is 10. However, I got an error in symbol
How would I get the solution?
There are multiple things that I noticed:
First, you are printing integer, but a and b are both chars.
Knowing that there are only a few symbols that you can use, maybe a switch case would be more appropriate?
int main(){
int a, b;
char symbol;
switch(symbol){
case('-'): print("%d", a - b);
case('+'): print("%d", a + b);
case('*'): print("%d", a * b);
case('/'): print("%d", a / b);
case('%'): print("%d", a % b);
default : print("Not a known symbol");
}
Second: Why are you substracting a from the character 0?
Related
I have two variable bit-shifting code fragments that I want to SSE-vectorize by some means:
1) a = 1 << b (where b = 0..7 exactly), i.e. 0/1/2/3/4/5/6/7 -> 1/2/4/8/16/32/64/128/256
2) a = 1 << (8 * b) (where b = 0..7 exactly), i.e. 0/1/2/3/4/5/6/7 -> 1/0x100/0x10000/etc
OK, I know that AMD's XOP VPSHLQ would do this, as would AVX2's VPSHLQ. But my challenge here is whether this can be achieved on 'normal' (i.e. up to SSE4.2) SSE.
So, is there some funky SSE-family opcode sequence that will achieve the effect of either of these code fragments? These only need yield the listed output values for the specific input values (0-7).
Update: here's my attempt at 1), based on Peter Cordes' suggestion of using the floating point exponent to do simple variable bitshifting:
#include <stdint.h>
typedef union
{
int32_t i;
float f;
} uSpec;
void do_pow2(uint64_t *in_array, uint64_t *out_array, int num_loops)
{
uSpec u;
for (int i=0; i<num_loops; i++)
{
int32_t x = *(int32_t *)&in_array[i];
u.i = (127 + x) << 23;
int32_t r = (int32_t) u.f;
out_array[i] = r;
}
}
I have a problem, and I don't know how to deal with it. I made a simple calculator with few functions. Every option is working fine, but I have problem when it comes to quadratic equation. When I put any numbers the response that I get is "The roots of the equations are : x=1.79599e-307. I don't know to deal with it - can you help me ? I will post parts of my code that includes the problem. Thanks You for any feedback and help!
main:
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include "kalkulator.cpp"
...
using namespace std;
case 10:
cout<<"\nQuadratic Equation \n";
cout<<"\nSelect number \n";
cin>>a;
cout<<"\nSelect number \n";
cin>>b;
cout<<"\nSelect bumber \n";
cin>>c;
int r;
r=equation(a, b, c, x1, x2);
if (r==0) cout << "\n No roots\n" << endl;
else if (r==1) cout << "\n 1 root x = " << x1 << endl;
else if (r==2) cout << "\n 2 roots x1 = " << x1 << " and x2 = " << x2 << endl;
break;
kalkulator.cpp
#include "kalkulator.h"
using namespace std;
...
double equation(double a, double b, double c, double x1, double
x2)
{
double delta=b*b-4*a*c;
if (delta<0.0)
{
return 0;}
if (delta==0.0)
{
x1=-b/(2*a);
return 1;}
else
{
delta=sqrt(delta);
x1==(-b-delta)/(2*a);
x2==(-b+delta)/(2*a);
return 2;}}
kalkulator.h
#ifndef kalkulator_H
#define kalkulator_H
class kalkulator
{
private:
double a;
double b;
double c;
double x1;
double x2;
public:
double equation(double, double, double, double, double);
};
#endif
If you want to set x1 and x2 through the supplied parameters, you need to pass them by reference:
double equation(double, double, double, double&, double&);
and
double equation(double a, double b, double c, double & x1, double & x2)
Also, in the last lines of your code, "x1==" and "x2==" should be "x1=" and "x2=".
Got stuck on an problem getting more familiar with C(++) on Hackerrank. Using scanf() I want to read a line with a bunch of different values.
The issue is that the value after char is ignored. Read solutions to this was to add a space before the char, but I've tried running a scanf() on a seperate line per value, and only keeping the space infront of the char but still the problem persisted.
The lesson to be learned seems to be using scanf() and not any other type of input method. So no reading with cin or fscanf(). And reason for this being one can read all kinds of values on the same line. So I don't want to make more than this one line for reading the input either.
#include <iostream>
#include <cstdio>
using namespace std;
int main() {
int i;
long l;
long long ll;
char c;
float f;
double d;
/** Read "3 444 12345678912345 a 334.23 14049.30493" from stdin */
scanf("%i %ld %lld %c %f %lf", &i, &l, &ll, &c, &f, &d);
printf("%d\n", i); // 3, as expected
printf("%ld\n", l); // 444 also as expected
printf("%lld\n", ll); // 12345678912345 ...
printf("%c\n", c); // 'a'
printf("%f\n", d); // 14049.30493 ?? It should be: 334.23
printf("%lf\n", d); // 14049.30493 but this time as it should be.
return 0;
}
scanf("%i %ld %lld %c %f %lf", &i, &l, &ll, &c, &f, &d);
^
!F! D
printf("%d\n", i);
printf("%ld\n", l);
printf("%lld\n", ll);
printf("%c\n", c);
printf("%f\n", d); // < D!
printf("%lf\n", d); // < D!
You are printing the d variable twice and never f.
I want to write the following matlab code in Eigen (where K is pxp and W is pxb):
H = (K*W)>0;
However the only thing that I came up so far is:
H = ((K*W.array() > 0).select(1,0));
This code doesn't work as explained here, but replacing 0 with VectorXd::Constant(p,0) (as suggested in the link question) generates a runtime error:
Eigen::internal::variable_if_dynamic<T, Value>::variable_if_dynamic(T) [with T = long int; int Value = 1]: Assertion `v == T(Value)' failed.
How can I solve this?
You don't need .select(). You just need to cast an array of bool to an array of H's component type.
H = ((K * W).array() > 0.0).cast<double>();
Your original attempt failed because the size of your constant 1/0 array is not match with the size of H. Using VectorXd::Constant is not a good choice when H is MatrixXd. You also have a problem with parentheses. I think you want * rather than .* in matlab notation.
#include <iostream>
#include <Eigen/Eigen>
using namespace Eigen;
int main() {
const int p = 5;
const int b = 10;
MatrixXd H(p, b), K(p, p), W(p, b);
K.setRandom();
W.setRandom();
H = ((K * W).array() > 0.0).cast<double>();
std::cout << H << std::endl << std::endl;
H = ((K * W).array() > 0).select(MatrixXd::Constant(p, b, 1),
MatrixXd::Constant(p, b, 0));
std::cout << H << std::endl;
return 0;
}
When calling a template member function in a template, you need to use the template keyword.
#include <iostream>
#include <Eigen/Eigen>
using namespace Eigen;
template<typename Mat, typename Vec>
void createHashTable(const Mat &K, Eigen::MatrixXi &H, Mat &W, int b) {
Mat CK = K;
H = ((CK * W).array() > 0.0).template cast<int>();
}
int main() {
const int p = 5;
const int b = 10;
Eigen::MatrixXi H(p, b);
Eigen::MatrixXf W(p, b), K(p, p);
K.setRandom();
W.setRandom();
createHashTable<Eigen::MatrixXf, Eigen::VectorXf>(K, H, W, b);
std::cout << H << std::endl;
return 0;
}
See this for some explanation.
Issue casting C++ Eigen::Matrix types via templates
I want to convert integers to double in the following piece of code:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int a , b;
double c;
cout<<"Enter two integers: ";
cin>>a>>b;
try
{
if (b == 0)
throw 0;
}
catch (int a)
{
cout<<"You made a division by zero?"<<endl<<a;
}
c = static_cast<double>(a/b);
cout<<"The division is: "<<fixed<<setprecision(2)<<c;
}
How to change the code to output the double value?
As mentioned you should cast not a result of division but variables itself, to get double value. All three options works.
int a, b;
double c;
std::cout << "Enter two integers: ";
std::cin >> a >> b;
try
{
if (b == 0)
throw 0;
}
catch (int a)
{
std::cout << "You made a division by zero?" << std::endl << a;
}
c = static_cast<double>(a) / b;
c = a / static_cast<double>(b);
c = static_cast<double>(a) / static_cast<double>(b);
std::cout << "The division is: " << std::fixed << std::setprecision(2) << c;
std::cin >> a >> b;
You are casting the result of the division, you should cast the operands instead.
Don't use an exception to catch a condition that you already catch using a simple if. Just use the if, and use an else to skip the division if the second operand is zero.
int a , b;
double c;
cout<<"Enter two integers: ";
cin>>a>>b;
if (b == 0) {
cout<<"You made a division by zero?"<<endl;
} else {
c = static_cast<double>(a) / static_cast<double>(b);
cout<<"The division is: "<<fixed<<setprecision(2)<<c;
}