Posted on 2000-05-01
Last Modified: 2008-03-03
I am having problems converting a string that has uppercase letters to all lower case. this is my program. it is for a cgi application:





#include "libcgi++.h"

// this is a good cgi form parsing class, compliant with the program
and Unix systems

//This program is in charge of the search done by location.

//What it does is that if the user enters the name of the town

//it open that town file,if that file exists it outputs all

//the info in that file.[ Complexity is O(1)]

void main()



cout << "Content-type: text/html\n\n"; // because we will generate HTML

cout << "<HTML><HEAD></HEAD><BODY>"; // start of the HTML page

Cgi *input = new Cgi();

// it uses exceptions



  input = new Cgi();


catch (Cgi::CgiException& e)


   cerr << "Error while decoding" << endl << "(maybe called from
command-line?)." << endl;

   // do some other error processing


// here: instead of using cin, now you read from the form

// LOOK at the access to input[]: IT MEANS THAT THE VARIABLE NAME IN

string townName = (*input)["town"];

string x("mayaguez"), y("aguada"), z("aguadilla"), a("rincon"), b("cabo
rojo"), c("san juan");


strlwr( (char *)townName.c_str());




fstream in(townName.c_str(),ios::in);



       //if this happens the user should be sent to a page

       //that tells to hit back button and start search again

       cout<<"Hit back button"<<"<BR>";

           cout<<"Place not found"<<"<BR>";

       return 0;


    string read;




       //This are the results of this search which have to

       //be output to the screen.If possible a link should be

       //provided to a page with restaurant menu*/


        // Ok, here I have put the "<BR>" linebreak instead of "\n" because we
are writing HTML!!!


        // if you want to have a link, just output this commented line instead

        // note that I have to double the HTML quotes (to get them in the

        //cout << "<a href="""">" << read << 



cout << "</BODY></HTML>"; // ends the HTML page properly


I am using this to change it to lowercase but I am getting and error when I compile it in g++. I am using Unix:

strlwr( (char *)townName.c_str());

what I am doing wrong?
Question by:milalik
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LVL 14

Expert Comment

ID: 2766085
LVL 22

Expert Comment

ID: 2766107
The code

strlwr( (char *)townName.c_str());

is very dangerious.  The pointer returned by c_str() is a constant character pointer.  That is, it is a pointer to characters that must not be changed!  The reason you have to perform that cast (the "(char *)") is that the compiler won't allow you to change those characters because it is unsafe.

A quick fix is to just do

townName = strlwr(townName.c_str());
LVL 22

Expert Comment

ID: 2766110
Opps that fix is not good.  No you have to use tolower().  But I see Alex has already answered....
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Author Comment

ID: 2768803
std::transform(s.begin(), s.end(), s.begin(), ::tolower); ???

what does that means? and where should it be placed?
LVL 14

Expert Comment

ID: 2769265
instead of
>>strlwr( (char *)townName.c_str());
std::transform(townName.begin(), townName.end(), townName.begin(), ::tolower);

That meebs: make some function (tolower)
from begin to end and store result
from begin od townName

Next is text from BC Help:



Applies an operation to a range of values in a collection and stores the result.


#include <algorithm>
template <class InputIterator,

          class OutputIterator,
          class UnaryOperation>
OutputIterator transform (InputIterator first,
                          InputIterator last,
                          OutputIterator result,

                          UnaryOperation op);

template <class InputIterator1,

          class InputIterator2,
          class OutputIterator,
          class BinaryOperation>
OutputIterator transform (InputIterator1 first1,
                          InputIterator1 last1,
                          InputIterator2 first2,
                          OutputIterator result,

                          BinaryOperation binary_op);


The transform algorithm has two forms. The first form applies unary operation op to each element of the range [first,last), and sends the result to the output iterator result. For example, this version of transform, could be used to square each element in a vector. If the output iterator (result) is the same as the input iterator used to traverse the range, transform, performs its transformation inplace.
The second form of transform applies a binary operation, binary_op, to corresponding elements in the range [first1, last1) and the range that begins at first2, and sends the result to result. For example, transform can be used to add corresponding elements in two sequences, and store the set of sums in a third. The algorithm assumes, but does not check, that the second sequence has at least as many elements as the first sequence. Note that the output iterator result can be a third sequence, or either of the two input sequences.

Formally, transform assigns through every iterator i in the range [result, result + (last1 - first1)) a new corresponding value equal to:

op(*(first1 + (i - result))


binary_op(*(first1 + (i - result), *(first2 + (i - result)))

transform returns result + (last1 - first1). op and binary_op must not have any side effects. result may be equal to first in case of unary transform, or to first1 or first2 in case of binary transform.
Exactly last1 - first1 applications of op or binary_op are performed.




 using namespace std;

 int main()

   //Initialize a deque with an array of ints
   int arr1[5] = {99, 264, 126, 330, 132};
   int arr2[5] = {280, 105, 220, 84, 210};

   deque<int> d1(arr1, arr1+5), d2(arr2, arr2+5);

   //Print the original values

   cout << "The following pairs of numbers: "
        << endl << "     ";
   deque<int>::iterator i1;
   for(i1 = d1.begin(); i1 != d1.end(); i1++)
      cout << setw(6) << *i1 << " ";
   cout << endl << "     ";
   for(i1 = d2.begin(); i1 != d2.end(); i1++)

      cout << setw(6) << *i1 << " ";

   // Transform the numbers in the deque to their

   // factorials and store in the vector
   transform(d1.begin(), d1.end(), d2.begin(),

             d1.begin(), times<int>());

   //Display the results

   cout << endl << endl;
   cout << "Have the products: " << endl << "     ";
   for(i1 = d1.begin(); i1 != d1.end(); i1++)

     cout << setw(6) << *i1 << " ";

   return 0;


Rogue Wave Standard C++ Library User's Guide and Tutorial

Author Comment

ID: 2771424
LVL 14

Accepted Solution

AlexVirochovsky earned 50 total points
ID: 2771939
If the discussion helps you, i am very glad.

Author Comment

ID: 2772856
Yes it does help.

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