如何在C ++中返回foreach循环的值？

Question

I need to take multiple lines of string and then prepends with spaces.

I am using boost::split to split the string and assign the value to a vector. And then inside for-each loop, I add spaces to the vector value.

static string test(){

    std::string text = "This is a sentence.\nAnd another sentence";
    string nl = "\n";
    string pad="     ";
    string txt;
    std::vector<std::string> results;

    boost::split(results, text, boost::is_any_of(nl));

    for(string line : results){
      txt = pad + line + nl;
      cout << txt;
    }
    return txt;
  }

I should get the return value as below. cout result is correct but the return value txt is unable to get.

      This is a sentence.
      And another sentence

Answer 1

having

  for(string line : results){ txt = pad + line + nl; cout << txt; } return txt; 

the return txt is the last value of pad + line + nl from the loop or an empty string if results is empty

Saying

cout result is correct but the return value txt is unable to get.

probably means you just need to collapse each string and return the result :

string r;

for(string line : results){
  txt = pad + line + nl;
  cout << txt;
  r += txt;
}
return r;

or something similar may be without pad

Answer 2

Another way:

static string test(){
    std::string text = "This is a sentence.\nAnd another sentence\n";
    std::string result = "   " + boost::algorithm::replace_all_copy(s, "\n", "\n    ");
    std::cout << result;
    return result;
}

Answer 3

Your loop:

for(string line : results){
  txt = pad + line + nl;
  cout << txt;
}
return txt;

resets txt on every loop iteration (and you print that iteration's text one at a time). You then only have the final iteration's value remaining after the loop.

Instead, append to txt to keep all values. As a consequence you will also need to hoist the cout out of the loop:

for(string line : results){
  txt += pad + line + nl;
}
cout << txt;
return txt;

Answer 4

I'll show the part of the function where the vector is already built. So the function I'll show is simplified but demonstrates how you can build and return the result string.

#include <iostream>
#include <string>
#include <vector>

static std::string test( const std::vector<std::string> &results )
{    
    const std::string pad( "     " );

    std::string::size_type n = 0;
    for ( const std::string &s : results ) n += s.length();

    n += results.size() * ( pad.size() + sizeof( '\n' ) );

    std::string txt;
    txt.reserve( n );

    for ( const std::string &s : results ) txt += pad + s + '\n';

    return txt;
}

int main()
{
    std::vector<std::string> results = { "This is a sentence.", "This is a sentence." };

    std::cout << test( results );
}

The program output is

 This is a sentence.
 This is a sentence.

To make the code more effective you should reserve enough space for the string txt . The loop

    for ( const std::string &s : results ) n += s.length();

calculates the total size of all strings stored in the vector.

Note: You could use the standard algorithm std::accumulate declared in the header <numeric> instead of the loop as for example

std::string::size_type n = std::accumulate( std::begin( results ), std::end( results ), 
                                            std::string::size_type( 0 ),
                                            []( const std::string::size_type acc, const std::string &s )
                                            {
                                                return acc + s.size();
                                            } );

Then you add the length of padding spaces for each string and the length of the character '\\n' .

Thus all is prepared to build the result string that is returned from the function.

std::string txt;
txt.reserve( n );

for ( const std::string &s : results ) txt += pad + s + '\n';

return txt;