Or operator not working

问题:

When I enter start then the program outputs the else function even though I fulfilled the criteria, I have tried with && as well and it still didn't work. Any answers would be appreciated.

#include <iostream>
#include <string>
#include <windows.h>
using namespace std;

int main ()
{
    float timer;
    bool end;
    std::string input;

    end = false;

    cout << "Enter start then a number to count down from" << ".\n";

    while (end == false){
        cin >> input;

        if (input.find("end" || "End") != std::string::npos)
        end = true;

        else if (input.find("start" || "restart" || "Start" || "Restart") != std::string::npos)
        {
            cin >> timer;

            while (timer>0){
                timer -= 0.1;

                Sleep(100);

                cout << timer << ".\n";
            }

            cout << "Finished! Enter restart then another number to perform another countdown, or enter end to close the program" << ".\n";
        }

        else
        cout << "Enter start" << ".\n";
    }

    return 0;
}

回答1:

Replace

if (input.find("end" || "End") != std::string::npos)

with:

if (input.find("end") != std::string::npos || input.find("End") != std::string::npos)

Similarly for your other if.

It seems obvious what your expression means, but when you break it down it really doesn't make sense. find expects a string, and "end" || "End" is not a string.

回答2:

the Logical or operator, || only works in boolean expressions.

For instance, if you had

bool A = true
bool B = false
bool C = A||B;  

than you will have set bool C to be True. IT just takes 2 booleans, and returns true if either of those booleans is true. That's all logical or does.

You might want to try something like

if (input.find("end") != std::string::npos || input.find("End") != std::string::npos)

回答3:

The || works only in logical boolean expression.

From the standard (emphasis is mine):

5.15 Logical OR operator [expr.log.or]

The || operator groups left-to-right. The operands are both contextually converted to bool (Clause 4). It returns true if either of its operands is true, and false otherwise.

So in input.find("end" || "End"), it tries to convert "end" and "End" to bool. And the operator || will return a bool also.

---

Here to solve your problem you need to replace:

if (input.find("end" || "End") != std::string::npos)

by

if ( input.find("End") != std::string::npos ||
     input.find("End") != std::string::npos )

And do the same in the second find.

回答4:

C++ simply doesn't work that way. When you write

input.find("end" || "End") != std::string::npos

the compiler sees the logical or on two non-null const char pointers, which results in the boolean value true. This is then interpreted as a char with the value 1 ('\1') which is then searched in the string - certainly not what you intended. If you want to know if you string is in a set of strings, you could use:

static std::set<std::string> s = { "end", "End" };
s.find( input ) != s.end();

While maybe not the most efficient code in the world, but with a C++11 compiler you can also condense it into a single line like this:

if( std::set<std::string>{ "end", "End" }.count( input ) ) {
    // found...
}

回答5:

 if (input.find("end" || "End") != std::string::npos)
 //             ^^^^^^^^^^^^^^

The || operator is not being used correctly here. The righthand expression will return true because it is non-zero, then it will be returned. So the statement resolves to input.find("end"). You need to use two separate conditional statements there:

 if (input.find("end") != std::string::npos ||
     input.find("End") != std::string::npos)

回答6:

I'd recommend using regex instead for things like that: regex

回答7:

The argument of the function call

input.find("end" || "End") 

has type bool and means that addess of string literal "end" or/and address of string literal "End" is not equal to zero. It is obvious that the both string literals have addresses that are not equal to zero. So the call is equivalent to

input.find(true) 

The compiler finds an overloaded function find that is the most suitable for this argument. This function is

find( charT, c, size_tipe pos = 0 );

Value true is implicitly converted to value charT( 1 ) and the function tries to find char with value 1 in your string.