How to Solve Control Reaches End of Non-Void Function Error in C++

1. Use the return Statement at the End of the Function Body
2. Use the return Statements at the End of Each Code Path of the Function Body

This article will explain several methods of solving control reaches end of non-void function error C++.

Use the return Statement at the End of the Function Body

Non-void functions are required to have a return type. Consequently, the function needs to have a statement that returns the object of the corresponding type. If certain compiler flags are passed, this type of error or warning might get suppressed entirely, which will lead to run-time faults if the given function gets called in the program.

The following example code has the reverseString function defined, which takes a reference to a string and returns the string value. If we look into a function body, there is no return statement. Even though the reverseString does not pass any arguments to the caller function, the compiler only displays the warning message, and the executable program is built anyway. If the function is called, then the control flow would most likely lead to a segmentation fault.

#include <algorithm>
#include <iostream>
#include <iterator>

using std::cout;
using std::endl;
using std::reverse;
using std::string;

string reverseString(string &s) { string rev(s.rbegin(), s.rend()); }

int main() {
  string str = "This string is arbitrary";
  int cond = -1;

  cout << str << endl;
  cout << reverseString(str, cond) << endl;

  return EXIT_SUCCESS;
}

Use the return Statements at the End of Each Code Path of the Function Body

Another scenario where the control reaches the end of a non-void function is when the conditional blocks don’t have the return statements in every path. Thus, if the execution in the non-void function is branched and if statements don’t cover every possible path, then there needs to be an explicit return call at the end of the function body.

The next example demonstrates the string manipulation function with two conditional paths that pass the return value to the caller function. However, some cases are left not evaluated for the given condition, which implies that control flow might reach the end of the function block and cause run-time errors.

#include <algorithm>
#include <iostream>
#include <iterator>

using std::cout;
using std::endl;
using std::reverse;
using std::string;

string reverseString(string &s, int condition) {
  if (condition == -1) {
    string rev(s.rbegin(), s.rend());
    return s;
  } else if (condition == 0) {
    return s;
  }
}

int main() {
  string str = "This string is arbitrary";
  int cond = -1;

  cout << str << endl;
  cout << reverseString(str, cond) << endl;

  return EXIT_SUCCESS;
}

You will possibly see the warning below.

Main.cpp:15:1: warning: non-void function does not return a value in all control paths [-Wreturn-type]
}
^
1 warning generated.