为什么double型函数返回不用return直接用cout

Question

错误❎

#include <iostream>
#include <iomanip>
using namespace std;
double add(double num1,double num2)
{
   double c=num1+num2;
    return c;

}
double sub(double num3,double num4)
{
    double c=num3-num4;
    return c;
}
double mul(double num5,double num6)
{
    double c=num5*num6;
    return c;
}
double divi(double num7,double num8)
{
    
    double c=num7/num8;
    return ;
}

int main() {
    cout << fixed << setprecision(2);
    double n1, n2; char sign;
    cin >> n1 >> sign >> n2;
    switch (sign) {
        case '+': add(n1, n2); break;
        case '-': sub(n1, n2); break;
        case '*': mul(n1, n2); break;
        case '/': divi(n1, n2); break;
        default: cout << "Unknown operation..." << endl;
    }
   return 0;
}
#include <iostream>
#include <iomanip>

using namespace std;

double add(double num1,double num2)
{
    double sum=num1+num2;
    return sum;
}
double sub(double num3,double num4)
{
    double c=num3-num4;
    return c;
}
double mul(double num5,double num6)
{
    double mu=num5*num6;
    return mu;
}
double divi(double num7,double num8)
{
    
    double div=num7/num8;
    return div;
}

int main() {
    cout << fixed << setprecision(2);
    double n1, n2; char sign;
    cin >> n1 >> sign >> n2;
    switch (sign) {
        case '+': add(n1, n2); break;
        case '-': sub(n1, n2); break;
        case '*': mul(n1, n2); break;
        case '/': divi(n1, n2); break;
        default: cout << "Unknown operation..." << endl;
    }
   return 0;
}

正确✔️1

double add(double x, double y)
{
    double z = x + y;
    return z;
}
double sub(double x, double y)
{
    double z = x - y;
    return z;
}
double mul(double x, double y)
{
    double z = x * y;
    return z;
}
double divi(double x, double y)
{
    double z = x / y;
    return z;}
int main() {
    cout << fixed << setprecision(2);
    double n1, n2; char sign;
    cin>>n1>>sign>>n2;
    
    switch (sign) {
        cout << setiosflags(ios::fixed) << setprecision(2);
    case '+':cout <<add(n1, n2); break;
    case '-': cout<<sub(n1, n2); break;
    case '*': cout<<mul(n1, n2); break;

正确的✔️2

double add(double a,double b)
{
    cout<<a+b;
}
double sub(double a,double b)
{
    cout<<a-b;

Answer 1

这个是为了看到结果吧 一般为了看到结果就是打印。在哪里打印都可以。