CODEWITHSUNDEEP
c++11lambdaconditional-operator
Soumyajit Roy
Soumyajit Roy

Reputation: 463

Use of lambda expression in C++ ternary operator : error type mismatch

I am just exploring lambda expression and trying to use it to return from a function using C++ ternary operator. However the below code does not work with error!

this is a simplified version of the program to explain my actual problem so kindly post answer using a lambda only, not just return (x+y+offset).

In member function 'int Calculator::add(int, int)': 12:26: error: no match for ternary 'operator?:' (operand types are 'bool', 'int', and 'Calculator::add(int, int)::<lambda()>') 16:3: warning: control reaches end of non-void function [-Wreturn-type]

    #include <iostream>
    #include <string>
    
    class Calculator
    {
      public:
      bool initalized = false;
      int offset = 0;
      
      int add(int x, int y)
      {
          return !initalized ? -1 : [this, x, y]()
          {
              return ((x+y) + offset);
          };
      }
    };
    
    int main()
    {
        Calculator c;
        c.initalized = true;
        c.offset = 1;
    
        printf("%d", c.add(2,10));
    }

What am I doing wrong here?

Upvotes: 0

Views: 284

Answers (2)

R Sahu
R Sahu

Reputation: 206607

What am I doing wrong here?

The true/false parts of the ternary expression are not compatible. You cannot convert -1 to a lambda expression and you cannot convert the lambda expression to an int.

It's not clear to me what you are trying to do with the lambda expression but it can be:

return !initalized ? -1 : (x + y + this->offset);

If you have the need to use a lambda expression for some reason, you'll have to invoke the lambda expression.

return !initalized ? -1 : [this, x, y]() { return ((x+y) + this->offset); }();
//                                                                 missing ^^

Upvotes: 0

Amir Kirsh
Amir Kirsh

Reputation: 13760

Your ternary operator tries to return -1 or a lambda, which do not have a common type.

You need to actually call the lambda:

return !initalized ? -1 : [this, x, y]()
      {
          return ((x+y) + offset);
      }(); // <= added ()

Or without a lambda:

return !initalized ? -1 : x+y+offset;

Upvotes: 2

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